Elementary Linear Algebra

Kuttler

January 10, 2012

Elementary Linear Algebra was written by Dr. Kenneth Kuttler of Brigham Young University for teaching Linear Algebra I. After The Saylor Foundation accepted his submission to Wave I of the Open Textbook Challenge, this textbook was relicensed as CC-BY 3.0.

Information on The Saylor Foundation’s Open Textbook Challenge can be found at www.saylor.org/otc/.

Elementary Linear Algebra © January 10, 2012 by Kenneth Kuttler, is licensed under a Creative Commons Attribution (CC BY) license made possible by funding from The Saylor Foundation's Open Textbook Challenge in order to be incorporated into Saylor.org's collection of open courses available at: http:// www.saylor.org "Full license terms may be viewed at: http://creativecommons.org/licenses/by/3.0/legalcode

Contents

1 Some Prerequisite Topics

1.1 Sets And Set Notation

1.2 Functions

1.3 Graphs Of Functions

1.4 The Complex Numbers

1.5 Polar Form Of Complex Numbers

1.6 Roots Of Complex Numbers

1.7 The Quadratic Formula

1.8 Exercises

2 Fn

2.1 Algebra in Fn

2.2 Geometric Meaning Of Vectors

2.3 Geometric Meaning Of Vector Addition

2.4 Distance Between Points In Rn Length Of A Vector

2.5 Geometric Meaning Of Scalar Multiplication

2.6 Exercises

2.7 Vectors And Physics

2.8 Exercises

3 Vector Products

3.1 The Dot Product

3.2 The Geometric Signiﬁcance Of The Dot Product

3.2.1 The Angle Between Two Vectors

3.2.2 Work And Projections

3.2.3 The Inner Product And Distance In Cn

3.3 Exercises

3.4 The Cross Product

3.4.1 The Distributive Law For The Cross Product

3.4.2 The Box Product

3.4.3 A Proof Of The Distributive Law

3.5 The Vector Identity Machine

3.6 Exercises

4 Systems Of Equations

4.1 Systems Of Equations, Geometry

4.2 Systems Of Equations, Algebraic Procedures

4.2.1 Elementary Operations

4.2.2 Gauss Elimination

4.3 Exercises

5 Matrices

5.1 Matrix Arithmetic

5.1.1 Addition And Scalar Multiplication Of Matrices

5.1.2 Multiplication Of Matrices

5.1.3 The ijth Entry Of A Product

5.1.4 Properties Of Matrix Multiplication

5.1.5 The Transpose

5.1.6 The Identity And Inverses

5.1.7 Finding The Inverse Of A Matrix

5.2 Exercises

6 Determinants

6.1 Basic Techniques And Properties

6.1.1 Cofactors And 2 × 2 Determinants

6.1.2 The Determinant Of A Triangular Matrix

6.1.3 Properties Of Determinants

6.1.4 Finding Determinants Using Row Operations

6.2 Applications

6.2.1 A Formula For The Inverse

6.2.2 Cramer’s Rule

6.3 Exercises

7 The Mathematical Theory Of Determinants*

7.1 The Function sgnn

7.2 The Determinant

7.2.1 The Deﬁnition

7.2.2 Permuting Rows Or Columns.

7.2.3 A Symmetric Deﬁnition

7.2.4 The Alternating Property Of The Determinant

7.2.5 Linear Combinations And Determinants

7.2.6 The Determinant Of A Product

7.2.7 Cofactor Expansions

7.2.8 Formula For The Inverse

7.2.9 Cramer’s Rule

7.2.10 Upper Triangular Matrices

7.3 The Cayley Hamilton Theorem∗

8 Rank Of A Matrix

8.1 Elementary Matrices

8.2 The Row Reduced Echelon Form Of A Matrix

8.3 The Rank Of A Matrix

8.3.1 The Deﬁnition Of Rank

8.3.2 Finding The Row And Column Space Of A Matrix

8.4 Linear Independence And Bases

8.4.1 Linear Independence And Dependence

8.4.2 Subspaces

8.4.3 Basis Of A Subspace

8.4.4 Extending An Independent Set To Form A Basis

8.4.5 Finding The Null Space Or Kernel Of A Matrix

8.4.6 Rank And Existence Of Solutions To Linear Systems

8.5 Fredholm Alternative

8.5.1 Row, Column, And Determinant Rank

8.6 Exercises

9 Linear Transformations

9.1 Linear Transformations

9.2 Constructing The Matrix Of A Linear Transformation

9.2.1 Rotations in R²

9.2.2 Rotations About A Particular Vector

9.2.3 Projections

9.2.4 Matrices Which Are One To One Or Onto

9.2.5 The General Solution Of A Linear System

9.3 Exercises

10 The LU Factorization

10.1 Deﬁnition Of An LU factorization

10.2 Finding An LU Factorization By Inspection

10.3 Using Multipliers To Find An LU Factorization

10.4 Solving Systems Using The LU Factorization

10.5 Justiﬁcation For The Multiplier Method

10.6 The PLU Factorization

10.7 The QR Factorization

10.8 Exercises

11 Linear Programming

11.1 Simple Geometric Considerations

11.2 The Simplex Tableau

11.3 The Simplex Algorithm

11.3.1 Maximums

11.3.2 Minimums

11.4 Finding A Basic Feasible Solution

11.5 Duality

11.6 Exercises

12 Spectral Theory

12.1 Eigenvalues And Eigenvectors Of A Matrix

12.1.1 Deﬁnition Of Eigenvectors And Eigenvalues

12.1.2 Finding Eigenvectors And Eigenvalues

12.1.3 A Warning

12.1.4 Triangular Matrices

12.1.5 Defective And Nondefective Matrices

12.1.6 Diagonalization

12.1.7 The Matrix Exponential

12.1.8 Complex Eigenvalues

12.2 Some Applications Of Eigenvalues And Eigenvectors

12.2.1 Principle Directions

12.2.2 Migration Matrices

12.3 The Estimation Of Eigenvalues

12.4 Exercises

13 Matrices And The Inner Product

13.1 Symmetric And Orthogonal Matrices

13.1.1 Orthogonal Matrices

13.1.2 Symmetric And Skew Symmetric Matrices

13.1.3 Diagonalizing A Symmetric Matrix

13.2 Fundamental Theory And Generalizations

13.2.1 Block Multiplication Of Matrices

13.2.2 Orthonormal Bases, Gram Schmidt Process

13.2.3 Schur’s Theorem

13.3 Least Square Approximation

13.3.1 The Least Squares Regression Line

13.3.2 The Fredholm Alternative

13.4 The Right Polar Factorization∗

13.5 The Singular Value Decomposition

13.6 Approximation In The Frobenius Norm∗

13.7 Moore Penrose Inverse∗

13.8 Exercises

14 Numerical Methods For Solving Linear Systems

14.1 Iterative Methods For Linear Systems

14.1.1 The Jacobi Method

14.1.2 The Gauss Seidel Method

14.2 The Operator Norm*

14.3 The Condition Number*

14.4 Exercises

15 Numerical Methods For Solving The Eigenvalue Problem

15.1 The Power Method For Eigenvalues

15.2 The Shifted Inverse Power Method

15.2.1 Complex Eigenvalues

15.3 The Rayleigh Quotient

15.4 The QR Algorithm

15.4.1 Basic Considerations

15.4.2 The Upper Hessenberg Form

15.5 Exercises

16 Vector Spaces

16.1 Algebraic Considerations

16.1.1 The Deﬁnition

16.2 Exercises

16.2.1 Linear Independence And Bases

16.3 Vector Spaces And Fields∗

16.3.1 Irreducible Polynomials

16.3.2 Polynomials And Fields

16.3.3 The Algebraic Numbers

16.3.4 The Lindemann Weierstrass Theorem And Vector Spaces

16.4 Exercises

16.5 Inner Product Spaces

16.5.1 Basic Deﬁnitions And Examples

16.5.2 The Cauchy Schwarz Inequality And Norms

16.5.3 The Gram Schmidt Process

16.5.4 Approximation And Least Squares

16.5.5 Fourier Series

16.5.6 The Discreet Fourier Transform

16.6 Exercises

17 Linear Transformation

17.1 Matrix Multiplication As A Linear Transformation

17.2 L(V, W ) As A Vector Space

17.3 Eigenvalues And Eigenvectors Of Linear Transformations

17.4 Block Diagonal Matrices

17.5 The Matrix Of A Linear Transformation

17.5.1 Some Geometrically Deﬁned Linear Transformations

17.5.2 Rotations About A Given Vector

17.6 Exercises

A The Jordan Canonical Form*

B The Fundamental Theorem Of Algebra

C Answers To Selected Exercises

C.1 Exercises 21

C.2 Exercises 36

C.3 Exercises 47

C.4 Exercises 56

C.5 Exercises 72

C.6 Exercises 92

C.7 Exercises 111

C.8 Exercises 158

C.9 Exercises 175

C.10 Exercises 190

C.11 Exercises 214

C.12 Exercises 239

C.13 Exercises 276

C.14 Exercises 296

C.15 Exercises 322

C.16 Exercises 326

C.17 Exercises 345

C.18 Exercises 363

C.19 Exercises 388

Copyright © 2011,

Preface

This is an introduction to linear algebra. The main part of the book features row operations and everything is done in terms of the row reduced echelon form and speciﬁc algorithms. At the end, the more abstract notions of vector spaces and linear transformations on vector spaces are presented. However, this is intended to be a first course in linear algebra for students who are sophomores or juniors who have had a course in one variable calculus and a reasonable background in college algebra. I have given complete proofs of all the fundamental ideas, but some topics such as Markov matrices are not complete in this book but receive a plausible introduction. The book contains a complete treatment of determinants and a simple proof of the Cayley Hamilton theorem although these are optional topics. The Jordan form is presented as an appendix. I see this theorem as the beginning of more advanced topics in linear algebra and not really part of a beginning linear algebra course. There are extensions of many of the topics of this book in my on line book [11]. I have also not emphasized that linear algebra can be carried out with any field although there is an optional section on this topic, most of the book being devoted to either the real numbers or the complex numbers. It seems to me this is a reasonable specialization for a first course in linear algebra.

Linear algebra is a wonderful interesting subject. It is a shame when it degenerates into nothing more than a challenge to do the arithmetic correctly. It seems to me that the use of a computer algebra system can be a great help in avoiding this sort of tedium. I don’t want to over emphasize the use of technology, which is easy to do if you are not careful, but there are certain standard things which are best done by the computer. Some of these include the row reduced echelon form, PLU factorization, and QR factorization. It is much more fun to let the machine do the tedious calculations than to suffer with them yourself. However, it is not good when the use of the computer algebra system degenerates into simply asking it for the answer without understanding what the oracular software is doing. With this in mind, there are a few interactive links which explain how to use a computer algebra system to accomplish some of these more tedious standard tasks. These are obtained by clicking on the symbol . I have included how to do it using maple and scientiﬁc notebook because these are the two systems I am familiar with and have on my computer. Other systems could be featured as well. It is expected that people will use such computer algebra systems to do the exercises in this book whenever it would be helpful to do so, rather than wasting huge amounts of time doing computations by hand. However, this is not a book on numerical analysis so no effort is made to consider many important numerical analysis issues.

Some Prerequisite Topics

The reader should be familiar with most of the topics in this chapter. However, it is often the case that set notation is not familiar and so a short discussion of this is included first. Complex numbers are then considered in somewhat more detail. Many of the applications of linear algebra require the use of complex numbers, so this is the reason for this introduction.

1.1   Sets And Set Notation

A set is just a collection of things called elements. Often these are also referred to as points in calculus. For example {1, 2, 3, 8} would be a set consisting of the elements 1,2,3, and 8. To indicate that 3 is an element of {1, 2, 3, 8} , it is customary to write 3 ∈ {1, 2, 3, 8} . 9 ∉ {1, 2, 3, 8} means 9 is not an element of {1, 2, 3, 8}. Sometimes a rule specifies a set. For example you could specify a set as all integers larger than 2. This would be written as S = {x ∈ Z : x > 2}. This notation says: the set of all integers, x, such that x > 2.

If A and B are sets with the property that every element of A is an element of B, then A is a subset of B. For example, {1, 2, 3, 8} is a subset of {1, 2, 3, 4, 5, 8}, in symbols, {1, 2, 3, 8} ⊆ {1, 2, 3, 4, 5, 8}. It is sometimes said that “A is contained in B” or even “B contains A”. The same statement about the two sets may also be written as {1, 2, 3, 4, 5, 8} ⊇ {1, 2, 3, 8}.

The union of two sets is the set consisting of everything which is an element of at least one of the sets, A or B. As an example of the union of two sets {1, 2, 3, 8}∪{3, 4, 7, 8} = {1, 2, 3, 4, 7, 8} because these numbers are those which are in at least one of the two sets. In general

A ∪ B ≡ {x : x ∈A or x ∈B}.

Be sure you understand that something which is in both A and B is in the union. It is not an exclusive or.

The intersection of two sets, A and B consists of everything which is in both of the sets. Thus {1, 2, 3, 8} ∩ {3, 4, 7, 8} = {3, 8} because 3 and 8 are those elements the two sets have in common. In general,

A ∩ B ≡ {x : x ∈ A and x ∈B}.

The symbol [a, b] where a and b are real numbers, denotes the set of real numbers x, such that a ≤ x ≤ b and [a, b) denotes the set of real numbers such that a ≤ x < b. (a, b) consists of the set of real numbers x such that a < x < b and (a, b] indicates the set of numbers x such that a < x ≤ b. [a,∞) means the set of all numbers x such that x ≥ a and (−∞, a] means the set of all real numbers which are less than or equal to a. These sorts of sets of real numbers are called intervals. The two points a and b are called endpoints of the interval. Other intervals such as (−∞, b) are defined by analogy to what was just explained. In general, the curved parenthesis indicates the end point it sits next to is not included while the square parenthesis indicates this end point is included. The reason that there will always be a curved parenthesis next to ∞ or −∞ is that these are not real numbers. Therefore, they cannot be included in any set of real numbers.

A special set which needs to be given a name is the empty set also called the null set, denoted by . Thus is defined as the set which has no elements in it. Mathematicians like to say the empty set is a subset of every set. The reason they say this is that if it were not so, there would have to exist a set A, such that has something in it which is not in A. However, has nothing in it and so the least intellectual discomfort is achieved by saying ⊆ A.

If A and B are two sets, A \ B denotes the set of things which are in A but not in B. Thus

A \ B ≡ {x ∈A : x ∉ B}.

Set notation is used whenever convenient.

To illustrate the use of this notation relative to intervals consider three examples of inequalities. Their solutions will be written in the notation just described.

Example 1.1.1 Solve the inequality 2x + 4 ≤ x − 8 

x ≤ −12 is the answer. This is written in terms of an interval as (−∞, −12].

Example 1.1.2 Solve the inequality (x + 1) (2x − 3) ≥ 0.

The solution is x ≤ −1 or x ≥ . In terms of set notation this is denoted by (−∞,−1]∪[,∞).

Example 1.1.3 Solve the inequality x (x +2) ≥ −4. 

This is true for any value of x. It is written as R or (−∞, ∞) .

1.2 Functions

The concept of a function is that of something which gives a unique output for a given input.

Definition 1.2.1 Consider two sets, D and R along with a rule which assigns a unique element of R to every element of D. This rule is called a function and it is denoted by a letter such as f. Given x ∈ D, f (x) is the name of the thing in R which results from doing f to x. Then D is called the domain of f. In order to specify that D pertains to f, the notation D (f) may be used. The set R is sometimes called the range of f. The set of all elements of R which are of the form f (x) for some x ∈ D is therefore, a subset of R. This is sometimes referred to as the image of f. When this set equals R, the function f is said to be onto, also surjective. If whenever x ≠ y it follows f (x) ≠ f (y), the function is called one to one., also injective It is common notation to write f : D  R to denote the situation just described in this definition where f is a function defined on a domain D which has values in a codomain R. Sometimes you may also see something like D  R to denote the same thing.

Example 1.2.2 Let D consist of the set of people who have lived on the earth except for Adam and for d ∈ D, let f (d) ≡ the biological father of d. Then f is a function.

This function is not the sort of thing studied in calculus but it is a function just the same.

Example 1.2.3 Consider the list of numbers, {1, 2, 3, 4, 5, 6, 7} ≡ D. Define a function which assigns an element of D to R ≡ {2, 3, 4, 5, 6, 7, 8} by f (x) ≡ x + 1 for each x ∈ D.

This function is onto because every element of R is the result of doing f to something in D. The function is also one to one. This is because if x + 1 = y + 1, then it follows x = y. Thus different elements of D must go to different elements of R.

In this example there was a clearly defined procedure which determined the function. However, sometimes there is no discernible procedure which yields a particular function.

Example 1.2.4 Consider the ordered pairs, (1, 2) , (2,−2) , (8, 3) , (7, 6) and let 

D ≡ {1, 2, 8, 7},

the set of first entries in the given set of ordered pairs and let

R ≡ {2,−2, 3, 6}, 

the set of second entries, and let f (1) = 2, f (2) = −2, f (8) = 3, and f (7) = 6. 

This specifies a function even though it does not come from a convenient formula.

1.3 Graphs Of Functions

Recall the notion of the Cartesian coordinate system you probably saw earlier. It involved an x axis, a y axis, two lines which intersect each other at right angles and one identifies a point by specifying a pair of numbers. For example, the number (2, 3) involves going 2 units to the right on the x axis and then 3 units directly up on a line perpendicular to the x axis. For example, consider the following picture.

Because of the simple correspondence between points in the plane and the coordinates of a point in the plane, it is often the case that people are a little sloppy in referring to these things. Thus, it is common to see (x, y) referred to as a point in the plane. In terms of relations, if you graph the points as just described, you will have a way of visualizing the relation.

The reader has likely encountered the notion of graphing relations of the form y = 2x + 3 or y = x² + 5. The meaning of such an expression in terms of defining a relation is as follows. The relation determined by the equation y = 2x + 3 means the set of all ordered pairs (x, y) which are related by this formula. Thus the relation can be written as

{(x, y) : y = 2x + 3}.

The relation determined by y = x² + 5 is

{(x, y) : y = x² + 5}.

Note that these relations are also functions. For the first, you could let f (x) = 2x+3 and this would tell you a rule which tells what the function does to x. However, some relations are not functions. For example, you could consider x² + y² = 1. Written more formally, the relation it defines is

{(x, y) : x² + y² = 1}

Now if you give a value for x, there might be two values for y which are associated with the given value for x. In fact

y = ± √1-x²  

Thus this relation would not be a function. 

Recall how to graph a relation. You first found lots of ordered pairs which satisfied the relation. For example (0, 3),(1, 5), and (−1, 1) all satisfy y = 2x + 3 which describes a straight line. Then you connected them with a curve. Here are some simple examples which you should see that you understand. First here is the graph of y = x² + 1.

Now here is the graph of the relation y = 2x + 1 which is a straight line.

Sometimes a relation is defined using different formulas depending on the location of one of the variables. For example, consider

Then the graph of this relation is sketched below.

A very important type of relation is one of the form y − y₀ = m (x −  x₀) , where m, x₀, and y₀ are numbers. The reason this is important is that if there are two points, (x₁, y₁) , and (x₂, y₂) which satisfy this relation, then

Remember from high school, the slope of the line segment through two points is always the difference in the y values divided by the difference in the x values, taken in the same order. Sometimes this is referred to as the rise divided by the run. This shows that there is a constant slope m, the slope of the line, between any pair of points satisfying this relation. Such a relation is called a straight line. Also, the point (x₀, y₀) satisfies the relation. This is more often called the equation of the straight line.

Geometrically, this means the graph of the relation is a straight line because the slope between any two points is always the same.

Example 1.3.1 Find the relation for a straight line which contains the point (1, 2) and has constant slope equal to 3.

From the above discussion, (y − 2) = 3 (x − 1) .

Definition 1.3.2 Let f : D (f) R (f) be a function. The graph of f consists of the set

{(x, y) : y = f (x) for x ∈ D ( f )}.

Note that knowledge of the graph of a function is equivalent to knowledge of the function. To find f (x) , simply observe the ordered pair which has x as its first element and the value of y equals f (x) .

Here is the graph of the function, f (x) = x² − 2

1.4 The Complex Numbers

Recall that a real number is a point on the real number line. Just as a real number should be considered as a point on the line, a complex number is considered a point in the plane which can be identified in the usual way using the Cartesian coordinates of the point. Thus (a, b) identifies a point whose x coordinate is a and whose y coordinate is b. In dealing with complex numbers, such a point is written as a + ib. For example, in the following picture, I have graphed the point 3 + 2i. You see it corresponds to the point in the plane whose coordinates are (3, 2) .

Multiplication and addition are defined in the most obvious way subject to the convention that i² = −1. Thus,

(a + ib) + (c + id) = (a + c) + i (b + d)

and

(a + ib) (c + id) = ac + iad + ibc + i²bd 

= (ac − bd) + i (bc + ad) . 

Every non zero complex number a + ib, with a² + b² ≠ 0, has a unique multiplicative inverse.

You should prove the following theorem. 

Theorem 1.4.1 The complex numbers with multiplication and addition defined as above form a field satisfying all the field axioms. These are the following list of properties.

x + y = y + x, (commutative law for addition)

x + 0= x,(additive identity).

For each x ∈ R, there exists −x ∈ R such that x + (−x= 0, (existence of additive inverse).

(x + y) + z = x + (y + z) , (associative law for addition).

xy = yx, (commutative law for multiplication). You could write this as x × y = y × x.

(xy) z = x (yz) , (associative law for multiplication).

1x = x, (multiplicative identity).

For each x ≠ 0, there exists x⁻¹ such that xx⁻¹ = 1.(existence of multiplicative inverse).

x (y + z) = xy + xz.(distributive law).

Something which satisfies these axioms is called a field. Linear algebra is all about fields, although in this book, the field of most interest will be the field of complex numbers or the field of real numbers. You have seen in earlier courses that the real numbers also satisfies the above axioms. The field of complex numbers is denoted as C and the field of real numbers is denoted as R. An important construction regarding complex numbers is the complex conjugate denoted by a horizontal line above the number. It is defined as follows.

What it does is reflect a given complex number across the x axis. Algebraically, the following formula is easy to obtain.

Definition 1.4.2 Define the absolute value of a complex number as follows.

Thus, denoting by z the complex number z = a + ib,

.

Also from the definition, if z = x + iy and w = u + iv are two complex numbers, then you can verify that:

|zw| = |z||w|

The triangle inequality holds for the absolute value for complex numbers just as it does for the
ordinary absolute value.

Proposition 1.4.3 Let z, w be complex numbers. Then the triangle inequality holds.

|z + w| ≤ |z| + |w|, ||z| − |w|| ≤ |z − w|. 

Proof: Let z = x + iy and w = u + iv. First note that 

and so

so this shows the first version of the triangle inequality. To get the second,

z = z − w + w, w = w − z + z

and so by the first form of the inequality

|z| ≤ |z − w| +|w| , |w| ≤ |z − w| + |z| 

and so both |z| − |w| and |w| − |z| are no larger than |z −w| and this proves the second version because ||z| − |w|| is one of |z| − |w| or |w| − |z|.

With this definition, it is important to note the following. Be sure to verify this. It is not too hard but you need to do it.

Remark 1.4.4 : Let z = a + ib and w = c + id. Then Thus the distance between the point in the plane determined by the ordered pair (a, b) and the ordered pair (c, d) equals |z −w| where z and w are as just described.

For example, consider the distance between (2, 5) and (1, 8) . From the distance formula this distance equals On the other hand, letting z = 2 + i5 and w = 1 + i8, z − w = 1 −i3 and so (z − w  = (1− i3) (1 + i3) = 10 so |z − w| = the same thing obtained with the distance formula.

1.5 Polar Form Of Complex Numbers

Complex numbers, are often written in the so called polar form which is described next. Suppose z = x + iy is a complex number. Then

Now note that

and so

 is a point on the unit circle. Therefore, there exists a unique angle ∈ [0, 2π) such that

The polar form of the complex number is then

 r (cos + i sin )

where is this angle just described and

1.6 Roots Of Complex Numbers

A fundamental identity is the formula of De Moivre which follows. 

Theorem 1.6.1 Let r > 0 be given. Then if n is a positive integer, 

[r (cos t + i sin t)]n = rn (cos nt + i sin nt) . 

Proof: It is clear the formula holds if n = 1. Suppose it is true for n. 

[r (cos t + i sin t)]n+1 = [r (cos t + i sin t)]n [r (cos t + i sin t)] 

which by induction equals

   = rn+1 (cos nt + i sin nt) (cos t + i sin t)

                                               = rn+1 ((cos nt cos t − sin nt sin t) + i (sin nt cos t + cos nt sin t))

= rn+1 (cos (n + 1) t + i sin (n + 1) t) 

by the formulas for the cosine and sine of the sum of two angles. 

Corollary 1.6.2 Let z be a non zero complex number. Then there are always exactly k kth roots of z in C.

Proof: Let z = x + iy and let z = |z|(cos t + i sin t) be the polar form of the complex number. By De Moivre’s theorem, a complex number

r (cos α + i sin α) ,

is a kth root of z if and only if

rk (cos kα + i sin kα) = |z|(cos t + i sin t) .

This requires rk = |z| and so r = |z|1/k and so also both cos (kα) = cos t and sin (kα) = sin t. This can only happen if

kα = t + 2lπ

for l an integer. Thus

and so the kth roots of z are of the form

Since the cosine and sine are periodic of period 2π, there are exactly k distinct numbers which result from this formula.

Example 1.6.3 Find the three cube roots of i.

First note that . Using the formula in the proof of the above corollary, the cube roots of i are

where l = 0, 1, 2. Therefore, the roots are

and

Thus the cube roots of i are and −i.

The ability to find kth roots can also be used to factor some polynomials.

Example 1.6.4 Factor the polynomial x³ − 27.

First find the cube roots of 27. By the above procedure using De Moivre’s theorem, these cube roots are 3, 3

Therefore, x³ + 27 =

Note also = x² + 3x + 9 and so

x³ -27= (x - 3)(x² + 3x + 9)

where the quadratic polynomial x² + 3x + 9 cannot be factored without using complex numbers.

Note that even though the polynomial x³−27 has all real coefficients, it has some complex zeros,

 These zeros are complex conjugates of each other. It is always this way.

You should show this is the case. To see how to do this, see Problems 13 and 14 below.

Another fact for your information is the Fundamental Theorem of Algebra. This theorem says that any polynomial of degree at least 1 having any complex coefficients always has a root in C. This is sometimes referred to by saying C is algebraically complete. Gauss is usually credited with giving a proof of this theorem in 1797 but many others worked on it and the first completely correct proof was due to Argand in 1806. For more on this theorem, you can google fundamental theorem of algebra and look at the interesting Wikipedia article on it. Proofs of this theorem usually involve the use of techniques from calculus even though it is really a result in algebra.

1.7 The Quadratic Formula

The quadratic formula

gives the solutions x to

ax² + bx + c = 0

where a, b, c are real numbers. It holds even if b² − 4ac < 0. This is easy to show from the above. There are exactly two square roots to this number b²−4ac from the above methods using De Moivre’s theorem. These roots are of the form

and

Thus the solutions, according to the quadratic formula are still given correctly by the above formula. 

Do these solutions predicted by the quadratic formula continue to solve the quadratic equation? Yes, they do. You only need to observe that when you square a square root of a complex number z, you recover z. Thus

Similar reasoning shows directly that  also solves the quadratic equation.

What if the coefficients of the quadratic equation are actually complex numbers? Does the formula hold even in this case? The answer is yes. This is a hint on how to do Problem 23 below, a special case of the fundamental theorem of algebra, and an ingredient in the proof of some versions of this theorem.

Example 1.7.1 Find the solutions to x² − 2ix− 5 = 0.

Formally, from the quadratic formula, these solutions are

Now you can check that these really do solve the equation. In general, this will be the case. See Problem 23 below.

1.8 Exercises

Let z = 5 + i9. Find z⁻¹. 

Let z = 2 + i7 and let w = 3 − i8. Find zw, z + w, z², and w/z.

Give the complete solution to x⁴ + 16 = 0.

Graph the complex cube roots of 8 in the complex plane. Do the same for the four fourth roots of 16.

If z is a complex number, show there exists ω a complex number with |ω| = 1 and ωz = |z|.

De Moivre’s theorem says [r (cos t + i sin t)]n = rn (cos nt + i sin nt) for n a positive integer. Does this formula continue to hold for all integers n, even negative integers? Explain.

You already know formulas for cos (x + y) and sin (x + y) and these were used to prove De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sin (5x) and one for cos (5x).

If z and w are two complex numbers and the polar form of z involves the angle while the polar form of w involves the angle , show that in the polar form for zw the angle involved is + . Also, show that in the polar form of a complex number z, r = |z|. 

Factor x³ + 8 as a product of linear factors.

Write x³+ 27 in the form (x + 3)(x² + ax + b) where x²+ ax + b cannot be factored any more using only real numbers.

Completely factor x⁴ + 16 as a product of linear factors.

Factor x⁴ + 16 as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers.

If z, w are complex numbers prove  and then show by induction that = Also verify that  In words this says the conjugate of a product equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates. 

Suppose p (x) = anxn + an−1xn−1 +· · ·+ a₁x + a₀ where all the ak are real numbers. Suppose also that p (z) = 0 for some z ∈ C. Show it follows that  = 0 also.

Show that 1 + i, 2 + i are the only two zeros to

p (x) = x² − (3 + 2i) x + (1 + 3i)

so the zeros do not necessarily come in conjugate pairs if the coefficients are not real. 

I claim that 1 = −1. Here is why.

This is clearly a remarkable result but is there something wrong with it? If so, what is wrong? 

De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents, not just integers. 

Therefore, squaring both sides it follows 1 = −1 as in the previous problem. What does this tell you about De Moivre’s theorem? Is there a profound difference between raising numbers to integer powers and raising numbers to non integer powers?

Review Problem 6 at this point. Now here is another question: If n is an integer, is it always true that (cos − i sin )n = cos () − i sin (n)? Explain.

Suppose you have any polynomial in cos and sin . By this I mean an expression of the form  where aαβ ∈ C. Can this always be written in the form? Explain.

Suppose p (x) = anxn + an−1xn−1 +· · · + a₁x + a₀ is a polynomial and it has n zeros,

z₁, z₂,· · ·, zn

listed according to multiplicity. (z is a root of multiplicity m if the polynomial f (x) = (x − z)m divides p (x) but (x − z) f (x) does not.) Show that

p (x) = an (x − z₁) (x − z₂) · · · (x − zn) .

Give the solutions to the following quadratic equations having real coefficients.

(a) x² − 2x + 2 = 0

(b) 3x² + x + 3 = 0

(c) x² − 6x + 13 = 0

(d) x² + 4x + 9 = 0

(e) 4x² + 4x + 5 = 0

Give the solutions to the following quadratic equations having complex coefficients. Note how the solutions do not come in conjugate pairs as they do when the equation has real coefficients.

(a) x² + 2x + 1 + i = 0
(b) 4x² + 4ix− 5 = 0
(c) 4x² + (4 + 4i) x + 1 + 2i = 0 

(d) x² − 4ix− 5 = 0
(e) 3x² + (1−i) x + 3i = 0

Prove the fundamental theorem of algebra for quadratic polynomials having coefficients in C. That is, show that an equation of the form ax²+ bx + c = 0 where a, b, c are complex numbers, a ≠ 0 has a complex solution. Hint: Consider the fact, noted earlier that the expressions given from the quadratic formula do in fact serve as solutions.

Fn

The notation, Cn refers to the collection of ordered lists of n complex numbers. Since every real number is also a complex number, this simply generalizes the usual notion of Rn, the collection of all ordered lists of n real numbers. In order to avoid worrying about whether it is real or complex numbers which are being referred to, the symbol F will be used. If it is not clear, always pick C.

Definition 2.0.1 Define Fn ≡ {(x₁, · · · , xn) : xj ∈ F for j = 1, · · · , n}. 

(x₁, · · · , xn) = (y₁, · · · , yn)

if and only if for all j = 1, · · · , n, xj = yj. When (x₁, · · · , xn) ∈ Fn, it is conventional to denote (x₁, · · · , xn) by the single bold face letter, x. The numbers, xj are called the coordinates. Elements in Fn are called vectors. The set

{(0, · · · , 0, t, 0, · · · , 0) : t ∈ R}

for t in the ith slot is called the ith coordinate axis in the case of Rn. The point 0 ≡ (0, · · · , 0) is called the origin.

Thus (1, 2, 4i) ∈ F³ and (2, 1, 4i) ∈ F³ but (1, 2, 4i) ≠ (2, 1, 4i) because, even though the same numbers are involved, they don’t match up. In particular, the first entries are not equal.

The geometric significance of Rn for n ≤ 3 has been encountered already in calculus or in precalculus. Here is a short review. First consider the case when n = 1. Then from the definition, R¹ = R. Recall that R is identified with the points of a line. Look at the number line again. Observe that this amounts to identifying a point on this line with a real number. In other words a real number determines where you are on this line. Now suppose n = 2 and consider two lines which intersect each other at right angles as shown in the following picture.

Notice how you can identify a point shown in the plane with the ordered pair, (2,6) . You go to the right a distance of 2 and then up a distance of 6. Similarly, you can identify another point in the plane with the ordered pair (–8, 3) . Go to the left a distance of 8 and then up a distance of 3. The reason you go to the left is that there is a – sign on the eight. From this reasoning, every ordered pair determines a unique point in the plane. Conversely, taking a point in the plane, you could draw two lines through the point, one vertical and the other horizontal and determine unique points, x₁ on the horizontal line in the above picture and x₂ on the vertical line in the above picture, such that the point of interest is identified with the ordered pair, (x₁, x₂) . In short, points in the plane can be identified with ordered pairs similar to the way that points on the real line are identified with real numbers. Now suppose n = 3. As just explained, the first two coordinates determine a point in a plane. Letting the third component determine how far up or down you go, depending on whether this number is positive or negative, this determines a point in space. Thus, (1, 4,–5) would mean to determine the point in the plane that goes with (1, 4) and then to go below this plane a distance of 5 to obtain a unique point in space. You see that the ordered triples correspond to points in space just as the ordered pairs correspond to points in a plane and single real numbers correspond to points on a line.

You can’t stop here and say that you are only interested in n ≤ 3. What if you were interested in the motion of two objects? You would need three coordinates to describe where the first object is and you would need another three coordinates to describe where the other object is located. Therefore, you would need to be considering R⁶. If the two objects moved around, you would need a time coordinate as well. As another example, consider a hot object which is cooling and suppose you want the temperature of this object. How many coordinates would be needed? You would need one for the temperature, three for the position of the point in the object and one more for the time. Thus you would need to be considering R⁵. Many other examples can be given. Sometimes n is very large. This is often the case in applications to business when they are trying to maximize profit subject to constraints. It also occurs in numerical analysis when people try to solve hard problems on a computer.

There are other ways to identify points in space with three numbers but the one presented is the most basic. In this case, the coordinates are known as Cartesian coordinates after Descartes¹ who invented this idea in the first half of the seventeenth century. I will often not bother to draw a distinction between the point in space and its Cartesian coordinates.

The geometric significance of Cn for n > 1 is not available because each copy of C corresponds to the plane or R².

2.1 Algebra in Fn

There are two algebraic operations done with elements of Fn. One is addition and the other is multiplication by numbers, called scalars. In the case of Cn the scalars are complex numbers while in the case of Rn the only allowed scalars are real numbers. Thus, the scalars always come from F in either case.

Definition 2.1.1 If x ∈ Fn and a ∈ F, also called a scalar, then ax ∈ Fn is defined by

| ax = a (x₁, · · · , xn) ≡ (ax₁, · · · , axn) . | (2.1)

This is known as scalar multiplication. If x, y ∈ Fn then x + y ∈ Fn and is defined by

| x + y = (x₁, · · · , xn) + (y₁, · · · , yn) ≡ (x₁ + y₁, · · · , xn + yn) | (2.2)

Fn is often called n-dimensional space. With this definition, vector addition and scalar multiplication satisfy the conclusions of the following theorem. More generally, these properties are called the vector space axioms.

Theorem 2.1.2 For v, w ∈ Fn and α, β scalars, (real numbers), the following hold. 

| v + w = w + v, | (2.3)

the commutative law of addition,

| (v + w) + z = v+ (w + z) , | (2.4)

the associative law for addition, 

| v + 0 = v, | (2.5)

the existence of an additive identity, 

| v + (–v) = 0, | (2.6)

the existence of an additive inverse, Also

α (v + w) = αv+αw,

(α + β) v =αv+βv,

α (βv) = αβ (v) ,

1v = v.

(2.7)

(2.8)

(2.9)

(2.10)

In the above 0 = (0, · · · , 0).

You should verify these properties all hold. For example, consider (2.7)

α (v + w) = α (v₁ + w₁, · · · , vn + wn)

= (α (v₁ + w₁) , · · · , α (vn + wn)) 

= (αv₁ + αw₁, · · · , αvn + αwn)

= (αv₁, · · · , αvn) + (αw₁, · · · , αwn) = αv + αw. 

As usual subtraction is defined as x–y ≡ x+ (–y) .

2.2 Geometric Meaning Of Vectors

The geometric meaning is especially significant in the case of Rn for n = 2, 3. Here is a short discussion of this topic.

Definition 2.2.1 Let x = (x₁, · · · , xn) be the coordinates of a point in Rn. Imagine an arrow with its tail at 0 = (0, · · · , 0) and its point at x as shown in the following picture in the case of R³.

Then this arrow is called the position vector of the point x. Given two points, P, Q whose coordinates are (p₁, · · · , pn) and (q₁, · · · , qn) respectively, one can also determine the position vector from P to Q defined as follows.

Thus every point determines a vector and conversely, every such vector (arrow) which has its tail at 0 determines a point of Rn, namely the point of Rn which coincides with the point of the vector. Also two different points determine a position vector going from one to the other as just explained.

Imagine taking the above position vector and moving it around, always keeping it pointing in the same direction as shown in the following picture.

After moving it around, it is regarded as the same vector because it points in the same direction and has the same length.² Thus each of the arrows in the above picture is regarded as the same vector. The components of this vector are the numbers, x₁, · · · , xn. You should think of these numbers as directions for obtaining an arrow. Starting at some point (α₁, α₂, · · · , αn) in Rn, you move to the point (α₁ + x₁, · · · , αn) and from there to the point (α₁ + x₁, α₂ + x₂, α₃· · ·, αn) and then to (α₁ + x₁, α₂ + x₂, α₃ + x₃, · · · , αn) and continue this way until you obtain the point

(α₁ + x₁, α₂ + x₂, · · · , αn + xn) .

The arrow having its tail at (α₁, α₂, · · · , αn) and its point at 

(α₁ + x₁, α₂ + x₂, · · · , αn + xn)

looks just like the arrow which has its tail at 0 and its point at (x₁, · · · , xn) so it is regarded as the same vector.

2.3 Geometric Meaning Of Vector Addition

It was explained earlier that an element of Rn is an n tuple of numbers and it was also shown that this can be used to determine a point in three dimensional space in the case where n = 3 and in two dimensional space, in the case where n = 2. This point was specified relative to some coordinate axes.

Consider the case where n = 3 for now. If you draw an arrow from the point in three dimensional space determined by (0, 0, 0) to the point (α, b, c) with its tail sitting at the point (0, 0, 0) and its point at the point (α, b, c) , this arrow is called the position vector of the point determined by u ≡ (α, b, c) . One way to get to this point is to start at (0, 0, 0) and move in the direction of the x₁ axis to (α, 0, 0) and then in the direction of the x₂ axis to (α, b, 0) and finally in the direction of the x₃ axis to (α, b, c) . It is evident that the same arrow (vector) would result if you began at the point v ≡ (d, e, f) , moved in the direction of the x₁ axis to (d + α, e, f) , then in the direction of the x₂ axis to (d + α, e + b, f) , and finally in the x₃ direction to (d + α, e + b, f + c) only this time, the arrow would have its tail sitting at the point determined by v ≡ (d, e, f) and its point at (d + α, e + b, f + c) . It is said to be the same arrow (vector) because it will point in the same direction and have the same length. It is like you took an actual arrow, the sort of thing you shoot with a bow, and moved it from one location to another keeping it pointing the same direction. This is illustrated in the following picture in which v + u is illustrated. Note the parallelogram determined in the picture by the vectors u and v.

Thus the geometric significance of (d, e, f) + (a, b, c) = (d + a, e + b, f + c) is this. You start with the position vector of the point (d, e, f) and at its point, you place the vector determined by (a, b, c) with its tail at (d, e, f) . Then the point of this last vector will be (d + a, e + b, f + c) . This is the geometric significance of vector addition. Also, as shown in the picture, u + v is the directed diagonal of the parallelogram determined by the two vectors u and v. A similar interpretation holds in Rn, n > 3 but I can’t draw a picture in this case.

Since the convention is that identical arrows pointing in the same direction represent the same vector, the geometric significance of vector addition is as follows in any number of dimensions.

Procedure 2.3.1 Let u and v be two vectors. Slide v so that the tail of v is on the point of u. Then draw the arrow which goes from the tail of u to the point of the slid vector v. This arrow represents the vector u + v.

2.4 Distance Between Points In Rn Length Of A Vector

How is distance between two points in Rn defined?

Definition 2.4.1 Let x = (x₁, · · · , xn) and y = (y₁, · · · , yn) be two points in Rn. Then |x – y| to indicates the distance between these points and is defined as

distance between x and y ≡ |x–y| ≡

This is called the distance formula. Thus |x| ≡ |x – 0|. The symbol, B (a, r) is defined by

B (a, r) ≡ {x ∈ Rn : |x – a|< r}.

This is called an open ball of radius r centered at a. It means all points in Rn which are closer to a than r. The length of a vector x is the distance between x and 0. 

First of all, note this is a generalization of the notion of distance in R. There the distance between two points, x and y was given by the absolute value of their difference. Thus |x – y| is equal to the distance between these two points on R. Now  where the square root is always the positive square root. Thus it is the same formula as the above definition except there is only one term in the sum. Geometrically, this is the right way to define distance which is seen from the Pythagorean theorem. Often people use two lines to denote this distance, ||x – y||. However, I want to emphasize this is really just like the absolute value. Also, the notation I am using is fairly standard.

Consider the following picture in the case that n = 2.

There are two points in the plane whose Cartesian coordinates are (x₁, x₂) and (y₁, y₂) respectively. Then the solid line joining these two points is the hypotenuse of a right triangle which is half of the rectangle shown in dotted lines. What is its length? Note the lengths of the sides of this triangle are |y₁ –x₁| and |y₂ –x₂|. Therefore, the Pythagorean theorem implies the length of the hypotenuse equals

which is just the formula for the distance given above. In other words, this distance defined above is the same as the distance of plane geometry in which the Pythagorean theorem holds.

Now suppose n = 3 and let (x₁, x₂, x₃) and (y₁, y₂, y₃) be two points in R³. Consider the following picture in which one of the solid lines joins the two points and a dotted line joins the points (x₁, x₂, x₃) and (y₁, y₂, y₃) .

By the Pythagorean theorem, the length of the dotted line joining (x₁, x₂, x₃) and (y₁, y₂, x₃) equals

while the length of the line joining (y₁, y₂, x₃) to (y₁, y₂, y₃) is just |y₃ –x₃|. Therefore, by the Pythagorean theorem again, the length of the line joining the points (x₁, x₂, x₃) and (y₁, y₂, y₃) equals

which is again just the distance formula above.

This completes the argument that the above definition is reasonable. Of course you cannot continue drawing pictures in ever higher dimensions but there is no problem with the formula for distance in any number of dimensions. Here is an example.

Example 2.4.2 Find the distance between the points in R⁴, 

a = (1, 2,–4, 6)

and

b = (2, 3,–1, 0)

Use the distance formula and write

|a–b|² = (1– 2)² + (2– 3)² + (–4– (–1))² + (6– 0)² = 47

Therefore,  |a – b| =

All this amounts to defining the distance between two points as the length of a straight line joining these two points. However, there is nothing sacred about using straight lines. One could define the distance to be the length of some other sort of line joining these points. It won’t be done in this book but sometimes this sort of thing is done.

Another convention which is usually followed, especially in R² and R³ is to denote the first component of a point in R² by x and the second component by y. In R³ it is customary to denote the first and second components as just described while the third component is called z.

Example 2.4.3 Describe the points which are at the same distance between (1, 2, 3) and (0, 1, 2) .

Let (x, y, z) be such a point. Then

Squaring both sides

(x– 1)² + (y – 2)² + (z – 3)² = x² + (y – 1)² + (z – 2)²

and so 

x² – 2x + 14 + y² – 4y + z² – 6z = x² + y² – 2y + 5 + z² – 4z

 which implies

–2x + 14– 4y – 6z = –2y + 5– 4z 

and so

| 2x + 2y + 2z = –9. | (2.11)

Since these steps are reversible, the set of points which is at the same distance from the two given points consists of the points, (x, y, z) such that (2.11) holds.

There are certain properties of the distance which are obvious. Two of them which follow directly from the definition are

| x – y| = |y – x|,

|x – y| ≥ 0 and equals 0 only if y = x.

The third fundamental property of distance is known as the triangle inequality. Recall that in any triangle the sum of the lengths of two sides is always at least as large as the third side. I will show you a proof of this later. This is usually stated as

|x + y| ≤ |x| +|y|.

Here is a picture which illustrates the statement of this inequality in terms of geometry.

2.5 Geometric Meaning Of Scalar Multiplication

As discussed earlier, x = (x₁, x₂, x₃) determines a vector. You draw the line from 0 to x placing the point of the vector on x. What is the length of this vector? The length of this vector is defined to equal |x| as in Definition 2.4.1. Thus the length of x equals When you multiply x by a scalar α, you get (αx₁, αx₂, αx₃) and the length of this vector is defined as

Thus the following holds. 

|αx| = |α| |x|.

In other words, multiplication by a scalar magnifies the length of the vector. What about the direction? You should convince yourself by drawing a picture that if α is negative, it causes the resulting vector to point in the opposite direction while if α > 0 it preserves the direction the vector points.

You can think of vectors as quantities which have direction and magnitude, little arrows. Thus any two little arrows which have the same length and point in the same direction are considered to be the same vector even if their tails are at different points.

You can always slide such an arrow and place its tail at the origin. If the resulting point of the vector is (a, b, c), it is clear the length of the little arrow is Geometrically, the way you add two geometric vectors is to place the tail of one on the point of the other and then to form the vector which results by starting with the tail of the first and ending with this point as illustrated in the following picture. Also when (a, b, c) is referred to as a vector, you mean any of the arrows which have the same direction and magnitude as the position vector of this point. Geometrically, for u = (u₁, u₂, u₃) , αu is any of the little arrows which have the same direction and magnitude as (αu₁, αu₂, αu₃) .

The following example is art which illustrates these definitions and conventions.

Exercise 2.5.1 Here is a picture of two vectors, u and v.

Sketch a picture of u + v, u–v, and u+2v.

First here is a picture of u + v. You first draw u and then at the point of u you place the tail of v as shown. Then u + v is the vector which results which is drawn in the following pretty picture.

Next consider u –v. This means u+ (–v) . From the above geometric description of vector addition, –v is the vector which has the same length but which points in the opposite direction to v. Here is a picture.

Finally consider the vector u+2v. Here is a picture of this one also.

2.6 Exercises

Verify all the properties (2.3)-(2.10).

Compute 5 (1, 2 + 3i, 3,–2) + 6 (2 – i, 1,–2, 7) .

Draw a picture of the points in R² which are determined by the following ordered pairs.

(a) (1 , 2)

(b) (–2,–2)

(c) (–2, 3)

(d) (2,–5)

Does it make sense to write (1, 2) + (2, 3, 1)? Explain.

Draw a picture of the points in R³ which are determined by the following ordered triples.

(a) (1, 2, 0)

(b) (–2,–2, 1) 

(c) (–2, 3,–2)

2.7 Vectors And Physics

Suppose you push on something. What is important? There are really two things which are important, how hard you push and the direction you push. This illustrates the concept of force.

Definition 2.7.1 Force is a vector. The magnitude of this vector is a measure of how hard it is pushing. It is measured in units such as Newtons or pounds or tons. Its direction is the direction in which the push is taking place.

Vectors are used to model force and other physical vectors like velocity. What was just described would be called a force vector. It has two essential ingredients, its magnitude and its direction. Geometrically think of vectors as directed line segments or arrows as shown in the following picture in which all the directed line segments are considered to be the same vector because they have the same direction, the direction in which the arrows point, and the same magnitude (length).

Because of this fact that only direction and magnitude are important, it is always possible to put a vector in a certain particularly simple form. Let  be a directed line segment or vector. Then it follows that  consists of the points of the form

p + t (q–p)

where t ∈ [0, 1] . Subtract p from all these points to obtain the directed line segment consisting of the points

0 + t (q–p) , t ∈ [0, 1] .

The point in Rn, q – p, will represent the vector. 

Geometrically, the arrow, , was slid so it points in the same direction and the base is at the origin 0. For example, see the following picture.

In this way vectors can be identified with points of Rn.

Definition 2.7.2 Let x = (x₁, · · · , xn) ∈ Rn. The position vector of this point is the vector whose point is at x and whose tail is at the origin, (0, · · · , 0). If x = (x₁, · · · , xn) is called a vector, the vector which is meant is this position vector just described. Another term associated with this is standard position. A vector is in standard position if the tail is placed at the origin.

It is customary to identify the point in Rn with its position vector.

The magnitude of a vector determined by a directed line segment  is just the distance between the point p and the point q. By the distance formula this equals

and for v any vector in Rn the magnitude of v equals .

Example 2.7.3 Consider the vector v ≡ (1, 2, 3) in Rn. Find |v|.

First, the vector is the directed line segment (arrow) which has its base at 0 ≡ (0, 0, 0) and its point at (1, 2, 3) . Therefore,

What is the geometric significance of scalar multiplication? If a represents the vector v in the sense that when it is slid to place its tail at the origin, the element of Rn at its point is a, what is rv?

Thus the magnitude of rv equals |r| times the magnitude of v. If r is positive, then the vector represented by rv has the same direction as the vector v because multiplying by the scalar r, only has the effect of scaling all the distances. Thus the unit distance along any coordinate axis now has length r and in this rescaled system the vector is represented by a. If r < 0 similar considerations apply except in this case all the ai also change sign. From now on, a will be referred to as a vector instead of an element of Rn representing a vector as just described. The following picture illustrates the effect of scalar multiplication.

Note there are n special vectors which point along the coordinate axes. These are

ei ≡ (0, · · · , 0, 1, 0, · · · , 0)

where the 1 is in the ith slot and there are zeros in all the other spaces. See the picture in the case of R³.

The direction of ei is referred to as the ith direction. Given a vector v = (a₁, · · · , an) , aiei is the ith component of the vector. Thus

aiei = (0, · · · , 0, ai, 0, · · · , 0)

and so this vector gives something possibly nonzero only in the ith direction. Also, knowledge of the ith component of the vector is equivalent to knowledge of the vector because it gives the entry in the ith slot and for v = (a₁, · · · , an) ,

What does addition of vectors mean physically? Suppose two forces are applied to some object. Each of these would be represented by a force vector and the two forces acting together would yield an overall force acting on the object which would also be a force vector known as the resultant. Suppose the two vectors are   and  Then the vector a involves a component in direction, aiei while the component in the ith direction of b is biei. Then it seems physically reasonable that the resultant vector should have a component in the ith direction equal to (ai + bi) ei. This is exactly what is obtained when the vectors, a and b are added.

a + b = (a₁ + b₁, · · · , an + bn) .

Thus the addition of vectors according to the rules of addition in Rn which were presented earlier, yields the appropriate vector which duplicates the cumulative effect of all the vectors in the sum.

What is the geometric significance of vector addition? Suppose u, v are vectors,

u = (u₁, · · · , un) , v = (v₁, · · · , vn)

Then u + v = (u₁ + v₁, · · · , un + vn) . How can one obtain this geometrically? Consider the directed line segment,   and then, starting at the end of this directed line segment, follow the directed line segment  to its end, u + v. In other words, place the vector u in standard position with its base at the origin and then slide the vector v till its base coincides with the point of u. The point of this slid vector determines u + v. To illustrate, see the following picture

Note the vector u + v is the diagonal of a parallelogram determined from the two vectors u and v and that identifying u + v with the directed diagonal of the parallelogram determined by the vectors u and v amounts to the same thing as the above procedure.

An item of notation should be mentioned here. In the case of Rn where n ≤ 3, it is standard notation to use i for e₁, j for e₂, and k for e₃. Now here are some applications of vector addition to some problems.

Example 2.7.4 There are three ropes attached to a car and three people pull on these ropes. The first exerts a force of 2i+3j–2k Newtons, the second exerts a force of 3i+5j + k Newtons and the third exerts a force of 5i – j+2k. Newtons. Find the total force in the direction of i.

To find the total force add the vectors as described above. This gives 10i+7j + k Newtons. Therefore, the force in the i direction is 10 Newtons.

As mentioned earlier, the Newton is a unit of force like pounds.

Example 2.7.5 An airplane flies North East at 100 miles per hour. Write this as a vector.

A picture of this situation follows.

The vector has length 100. Now using that vector as the hypotenuse of a right triangle having equal sides, the sides should be each of length Therefore, the vector would be  + 

This example also motivates the concept of velocity.

Definition 2.7.6 The speed of an object is a measure of how fast it is going. It is measured in units of length per unit time. For example, miles per hour, kilometers per minute, feet per second. The velocity is a vector having the speed as the magnitude but also specifying the direction.

Thus the velocity vector in the above example is  +

Example 2.7.7 The velocity of an airplane is 100i + j + k measured in kilometers per hour and at a certain instant of time its position is (1, 2, 1) . Here imagine a Cartesian coordinate system in which the third component is altitude and the first and second components are measured on a line from West to East and a line from South to North. Find the position of this airplane one minute later.

Consider the vector (1, 2, 1) , is the initial position vector of the airplane. As it moves, the position vector changes. After one minute the airplane has moved in the i direction a distance of kilometer. In the j direction it has moved kilometer during this same time, while it moves  kilometer in the k direction. Therefore, the new displacement vector for the airplane is

Example 2.7.8 A certain river is one half mile wide with a current flowing at 4 miles per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of 3 miles per hour. How far down the river does he find himself when he has swam across? How far does he end up swimming?

Consider the following picture.

You should write these vectors in terms of components. The velocity of the swimmer in still water would be 3j while the velocity of the river would be –4i. Therefore, the velocity of the swimmer is –4i + 3j. Since the component of velocity in the direction across the river is 3, it follows the trip takes 1/6 hour or 10 minutes. The speed at which he travels is miles per hour and so he travels  miles. Now to find the distance downstream he finds himself, note that if x is this distance, x and 1/2 are two legs of a right triangle whose hypotenuse equals 5/6 miles. Therefore, by the Pythagorean theorem the distance downstream is

2.8 Exercises

The wind blows from West to East at a speed of 50 miles per hour and an airplane which travels at 300 miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

In the situation of Problem 1 how many degrees to the West of North should the airplane head in order to fly exactly North. What will be the speed of the airplane relative to the ground?

In the situation of 2 suppose the airplane uses 34 gallons of fuel every hour at that air speed and that it needs to fly North a distance of 600 miles. Will the airplane have enough fuel to arrive at its destination given that it has 63 gallons of fuel?

An airplane is flying due north at 150 miles per hour. A wind is pushing the airplane due east at 40 miles per hour. After 1 hour, the plane starts flying 30° East of North. Assuming the plane starts at (0, 0) , where is it after 2 hours? Let North be the direction of the positive y axis and let East be the direction of the positive x axis.

City A is located at the origin while city B is located at (300, 500) where distances are in miles. An airplane flies at 250 miles per hour in still air. This airplane wants to fly from city A to city B but the wind is blowing in the direction of the positive y axis at a speed of 50 miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city B having flown the shortest possible distance. How long will it take to get there?

A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West. A man swims directly toward the opposite shore from the South bank of the river at a speed of 3 miles per hour. How far down the river does he find himself when he has swam across? How far does he end up swimming?

A certain river is one half mile wide with a current flowing at 2 miles per hour from East to West. A man can swim at 3 miles per hour in still water. In what direction should he swim in order to travel directly across the river? What would the answer to this problem be if the river flowed at 3 miles per hour and the man could swim only at the rate of 2 miles per hour?

Three forces are applied to a point which does not move. Two of the forces are 2i + j + 3k Newtons and i – 3j + 2k Newtons. Find the third force.

The total force acting on an object is to be 2i + j + k Newtons. A force of –i + j + k Newtons is being applied. What other force should be applied to achieve the desired total force?

A bird flies from its nest 5 km. in the direction 60° north of east where it stops to rest on a tree. It then flies 10 km. in the direction due southeast and lands atop a telephone pole. Place an xy coordinate system so that the origin is the bird’s nest, and the positive x axis points east and the positive y axis points north. Find the displacement vector from the nest to the telephone pole.

A car is stuck in the mud. There is a cable stretched tightly from this car to a tree which is 20 feet long. A person grasps the cable in the middle and pulls with a force of 100 pounds perpendicular to the stretched cable. The center of the cable moves two feet and remains still. What is the tension in the cable? The tension in the cable is the force exerted on this point by the part of the cable nearer the car as well as the force exerted on this point by the part of the cable nearer the tree.

¹ Ren´e Descartes 1596-1650 is often credited with inventing analytic geometry although it seems the ideas were actually known much earlier. He was interested in many different subjects, physiology, chemistry, and physics being some of them. He also wrote a large book in which he tried to explain the book of Genesis scientifically. Descartes ended up dying in Sweden.

²I will discuss how to define length later. For now, it is only necessary to observe that the length should be defined in such a way that it does not change when such motion takes place.

Vector Products

3.1 The Dot Product

There are two ways of multiplying vectors which are of great importance in applications. The first of these is called the dot product, also called the scalar product and sometimes the inner product.

Definition 3.1.1 Let a, b be two vectors in Rn define a · b as

The dot product a · b is sometimes denoted as (a, b) where a comma replaces ·.

With this definition, there are several important properties satisfied by the dot product. In the statement of these properties, α and β will denote scalars and a, b, c will denote vectors. 

Proposition 3.1.2 The dot product satisfies the following properties.

a · b = b · a

a · a ≥ 0 and equals zero if and only if a = 0

(αa + βb) · c =α (a · c) + β (b · c)

c · (αa + βb) = α (c · a) + β (c · b)

|a|² = a · a

(3.1)

(3.2)

(3.3)

(3.4)

(3.5)

You should verify these properties. Also be sure you understand that (3.4) follows from the first three and is therefore redundant. It is listed here for the sake of convenience.

Example 3.1.3 Find (1, 2, 0,−1) · (0, 1, 2, 3) .

This equals 0 + 2 + 0 + −3 = −1.

Example 3.1.4 Find the magnitude of a = (2, 1, 4, 2) . That is, find |a|.

This is 

The dot product satisfies a fundamental inequality known as the Cauchy-Schwarz inequality.

Theorem 3.1.5 The dot product satisfies the inequality

| |a · b| ≤ |a| |b|. | (3.6)

Furthermore equality is obtained if and only if one of a or b is a scalar multiple of the other.

Proof: First note that if b = 0 both sides of (3.6) equal zero and so the inequality holds in this case. Therefore, it will be assumed in what follows that b ≠ 0.

Define a function of t ∈ R

f (t) = (a + tb) · (a + tb) .

Then by (3.2), f (t) ≥ 0 for all t ∈ R. Also from (3.3),(3.4),(3.1), and (3.5)

f (t) = a · (a + tb) + tb · (a + tb)

 = a · a + t (a · b) + tb · a + t²b · b 

 = |a|² + 2t (a · b) +|b|2 t².

Now this means the graph, y = f (t) is a polynomial which opens up and either its vertex touches the t axis or else the entire graph is above the x axis. In the first case, there exists some t where f (t) = 0 and this requires a + tb = 0 so one vector is a multiple of the other. Then clearly equality holds in (3.6). In the case where b is not a multiple of a, it follows f (t) > 0 for all t which says f (t) has no real zeros and so from the quadratic formula,

(2 (a · b))² − 4|a|²|b|²< 0

which is equivalent to |(a · b)|< |a| |b|. 

You should note that the entire argument was based only on the properties of the dot product listed in (3.1) - (3.5). This means that whenever something satisfies these properties, the Cauchy-Schwarz inequality holds. There are many other instances of these properties besides vectors in Rn.

The Cauchy-Schwarz inequality allows a proof of the triangle inequality for distances in Rn in much the same way as the triangle inequality for the absolute value.

Theorem 3.1.6 (Triangle inequality) For a, b ∈ Rn

| |a + b| ≤ |a| + |b| | (3.7)

and equality holds if and only if one of the vectors is a nonnegative scalar multiple of the other. Also

| ||a| − |b|| ≤ |a − b| | (3.8)

Proof: By properties of the dot product and the Cauchy-Schwarz inequality,

|a + b|² = (a + b) · (a + b)
= (a · a) + (a · b) + (b · a) + (b · b)
= |a|² + 2 (a · b) + |b|²
≤ |a|² + 2|a · b| + |b|² 
≤ |a|² + 2|a| |b| + |b|²
= (|a| +|b|)2 .

Taking square roots of both sides you obtain (3.7).

It remains to consider when equality occurs. If either vector equals zero, then that vector equals zero times 6the other vector and the claim about when equality occurs is verified. Therefore, it can be assumed both vectors are nonzero. To get equality in the second inequality above, Theorem 3.1.5 implies one of the vectors must be a multiple of the other. Say b = αa. If α < 0 then equality cannot occur in the first inequality because in this case

(a · b) = α|a| 2 < 0 < |α| |a|² = |a · b|

Therefore, α ≥ 0. 

To get the other form of the triangle inequality, 

a = a − b + b

so 

|a| = |a − b + b|
≤ |a − b| + |b| .

Therefore,

| |a| − |b| ≤ |a − b| | (3.9)

Similarly,

| |b| − |a| ≤ |b − a| = |a − b|. | (3.10)

It follows from (3.9) and (3.10) that (3.8) holds. This is because ||a| − |b|| equals the left side of either (3.9) or (3.10) and either way, ||a| − |b|| ≤ |a−b|.

3.2 The Geometric Significance Of The Dot Product

3.2.1 The Angle Between Two Vectors

Given two vectors, a and b, the included angle is the angle between these two vectors which is less than or equal to 180 degrees. The dot product can be used to determine the included angle between two vectors. To see how to do this, consider the following picture.

By the law of cosines,

|a - b|2 = |a|2 + |b|-2 - 2|a| |b|

Also from the properties of the dot product,

|a − b|2 = |(a − b) · (a − b)

= |a|² + |b|² − 2a · b 

and so comparing the above two formulas,

| a · b = |a| |b|  | (3.11)

In words, the dot product of two vectors equals the product of the magnitude of the two vectors multiplied by the cosine of the included angle. Note this gives a geometric description of the dot product which does not depend explicitly on the coordinates of the vectors.

Example 3.2.1 Find the angle between the vectors 2i + j − k and 3i + 4j + k. 

The dot product of these two vectors equals 6 + 4− 1 = 9 and the norms are  and Therefore, from (3.11) the cosine of the included angle equals

Now the cosine is known, the angle can be determines by solving the equation, = . 720 58. This will involve using a calculator or a table of trigonometric functions. The answer is = . 766 16 radians or in terms of degrees, = . 766 16×  = 43. 898°. Recall how this last computation is done. Set up a proportion,  because 360° corresponds to 2π radians. However, in calculus, you should get used to thinking in terms of radians and not degrees. This is because all the important calculus formulas are defined in terms of radians.

Example 3.2.2 Let u, v be two vectors whose magnitudes are equal to 3 and 4 respectively and such that if they are placed in standard position with their tails at the origin, the angle between u and the positive x axis equals 30° and the angle between v and the positive x axis is -30°. Find u·v.

From the geometric description of the dot product in (3.11)

u · v = 3 × 4 × cos (60°) = 3 × 4 × 1/2 = 6.

Observation 3.2.3 Two vectors are said to be perpendicular if the included angle is π/2 radians (90°). You can tell if two nonzero vectors are perpendicular by simply taking their dot product. If the answer is zero, this means they are perpendicular because cos = 0.

Example 3.2.4 Determine whether the two vectors, 2i + j − k and 1i + 3j + 5k are perpendicular.

When you take this dot product you get 2 + 3−5 = 0 and so these two are indeed perpendicular.

Definition 3.2.5 When two lines intersect, the angle between the two lines is the smaller of the two angles determined.

Example 3.2.6 Find the angle between the two lines, (1, 2, 0)+t (1, 2, 3) and (0, 4,−3)+t (−1, 2,−3) .

These two lines intersect, when t = 0 in the first and t = −1 in the second. It is only a matter of finding the angle between the direction vectors. One angle determined is given by

| [picture] | (3.12)

We don’t want this angle because it is obtuse. The angle desired is the acute angle given by

 .

It is obtained by using replacing one of the direction vectors with −1 times it.

3.2.2 Work And Projections

Our first application will be to the concept of work. The physical concept of work does not in any way correspond to the notion of work employed in ordinary conversation. For example, if you were to slide a 150 pound weight off a table which is three feet high and shuffle along the floor for 50 yards, sweating profusely and exerting all your strength to keep the weight from falling on your feet, keeping the height always three feet and then deposit this weight on another three foot high table, the physical concept of work would indicate that the force exerted by your arms did no work during this project even though the muscles in your hands and arms would likely be very tired. The reason for such an unusual definition is that even though your arms exerted considerable force on the weight, enough to keep it from falling, the direction of motion was at right angles to the force they exerted. The only part of a force which does work in the sense of physics is the component of the force in the direction of motion (This is made more precise below.). The work is defined to be the magnitude of the component of this force times the distance over which it acts in the case where this component of force points in the direction of motion and (−1) times the magnitude of this component times the distance in case the force tends to impede the motion. Thus the work done by a force on an object as the object moves from one point to another is a measure of the extent to which the force contributes to the motion. This is illustrated in the following picture in the case where the given force contributes to the motion.

In this picture the force, F is applied to an object which moves on the straight line from p₁ to p₂. There are two vectors shown, F|| and F⊥ and the picture is intended to indicate that when you add these two vectors you get F while F|| acts in the direction of motion and F⊥ acts perpendicular to the direction of motion. Only F|| contributes to the work done by F on the object as it moves from p₁ to p₂. F|| is called the component of the force in the direction of motion. From trigonometry, you see the magnitude of F|| should equal |F| |cos |. Thus, since F|| points in the direction of the vector from p₁ to p₂, the total work done should equal

If the included angle had been obtuse, then the work done by the force, F on the object would have been negative because in this case, the force tends to impede the motion from p₁ to p₂ but in this case, cos would also be negative and so it is still the case that the work done would be given by the above formula. Thus from the geometric description of the dot product given above, the work equals

|F| |p₂ − p₁|cos = F·(p₂−p₁) .

This explains the following definition.

Definition 3.2.7 Let F be a force acting on an object which moves from the point p₁ to the point p₂. Then the work done on the object by the given force equals F·(p₂ −p₁) .

The concept of writing a given vector F in terms of two vectors, one which is parallel to a given vector D and the other which is perpendicular can also be explained with no reliance on trigonometry, completely in terms of the algebraic properties of the dot product. As before, this is mathematically more significant than any approach involving geometry or trigonometry because it extends to more interesting situations. This is done next.

Theorem 3.2.8 Let F and D be nonzero vectors. Then there exist unique vectors F|| and F⊥ such that

| F = F|| + F⊥ | (3.13)

where F|| is a scalar multiple of D, also referred to as

projD (F) ,

and F⊥ · D = 0. The vector projD (F) is called the projection of F onto D.

Proof: Suppose (3.13) and F|| = αD. Taking the dot product of both sides with D and using F⊥ · D = 0, this yields

F · D = α|D|²

which requires α = F · D/| D|2 . Thus there can be no more than one vector F||. It follows F⊥ must equal F − F||. This verifies there can be no more than one choice for both F|| and F⊥. 

Now let

and let

Then F|| = α D where αIt only remains to verify F⊥ · D = 0. But

Example 3.2.9 Let F = 2i+7j − 3k Newtons. Find the work done by this force in moving from the point (1, 2, 3) to the point (−9,−3, 4) along the straight line segment joining these points where distances are measured in meters.

According to the definition, this work is

(2i+7j − 3k) · (−10i − 5j + k) = −20 + (−35) + (−3)
= −58 Newton meters.

Note that if the force had been given in pounds and the distance had been given in feet, the units on the work would have been foot pounds. In general, work has units equal to units of a force times units of a length. Instead of writing Newton meter, people write joule because a joule is by definition a Newton meter. That word is pronounced “jewel” and it is the unit of work in the metric system of units. Also be sure you observe that the work done by the force can be negative as in the above example. In fact, work can be either positive, negative, or zero. You just have to do the computations to find out.

Example 3.2.10 Find proju (v) if u = 2i + 3j− 4k and v = i− 2j + k.

From the above discussion in Theorem 3.2.8, this is just

Example 3.2.11 Suppose a, and b are vectors and b⊥ = b−proja (b) . What is the magnitude of b⊥ in terms ofthe included angle?

where is the included angle between a and b which is less than π radians. Therefore, taking square roots,

|b⊥| = |b| sin .

3.2.3 The Inner Product And Distance In Cn

It is necessary to give a generalization of the dot product for vectors in Cn. This is often called the inner product. It reduces to the definition of the dot product in the case the components of the vector are real.

Definition 3.2.12 Let x, y ∈ Cn. Thus x = (x₁,· · ·, xn) where each xk ∈ C and a similar formula holding for y. Then the inner product of these two vectors is defined to be

The inner product is often denoted as (x, y) or (x, y).

Notice how you put the conjugate on the entries of the vector y. It makes no difference if the vectors happen to be real vectors but with complex vectors you must do it this way. The reason for this is that when you take the inner product of a vector with itself, you want to get the square of the length of the vector, a positive number. Placing the conjugate on the components of y in the above definition assures this will take place. Thus

If you didn’t place a conjugate as in the above definition, things wouldn’t work out correctly. For example,

(1 + i)² + 2² = 4 + 2i

and this is not a positive number.

The following properties of the inner product follow immediately from the definition and you should verify each of them.

Properties of the inner product:

u·v =

If a, b are numbers and u, v, z are vectors then (au + bv) · z = a (u · z) + b (v · z) .

u·u ≥ 0 and it equals 0 if and only if u = 0.

Note this implies (x·αyb) = (x · y) because 

The norm is defined in the usual way.

Definition 3.2.13 For x ∈ Cn

Here is a fundamental inequality called the Cauchy-Schwarz inequality which is stated here in Cn. First here is a simple lemma.

Lemma 3.2.14 If z ∈ C there exists ∈ C such that z = |z| and || = 1.

Proof: Let = 1 if z = 0 and otherwise, let = . Recall that for
z = x + iy,  = x − iy and z = ||².

I will give a proof of this important inequality which depends only on the above list of properties of the inner product. It will be slightly different than the earlier proof.

Theorem 3.2.15 (Cauchy-Schwarz)The following inequality holds for x and y ∈ Cn.

| |(x · y)| ≤ (x · x)1/2 (y · y)1/2 | (3.14)

Equality holds in this inequality if and only if one vector is a multiple of the other. 

Proof: Let ∈ C such that || = 1 and

(x · y) = |(x · y)|

Consider  where t ∈ R. Then from the above list of properties of the dot  product,

and this must hold for all t ∈ R. Therefore, if (y · y) = 0 it must be the case that |(x · y)| = 0 also since otherwise the above inequality would be violated. Therefore, in this case,

|(x · y)| ≤ (x · x)1/2 (y · y)1/2.

On the other hand, if (y · y) ≠ 0, then p (t) ≥ 0 for all t means the graph of y = p (t) is a parabola which opens up and it either has exactly one real zero in the case its vertex touches the t axis or it has no real zeros.

From the quadratic formula this happens exactly when

4|(x · y)|² − 4 (x · x) (y · y) ≤ 0

which is equivalent to (3.14).

It is clear from a computation that if one vector is a scalar multiple of the other that equality holds in (3.14). Conversely, suppose equality does hold. Then this is equivalent to saying 4|(x · y)|²− 4 (x · x) (y · y) = 0 and so from the quadratic formula, there exists one real zero to p (t) = 0. Call it t₀. Then

and so .

Note that I only used part of the above properties of the inner product. It was not necessary to use the one which says that if (x · x) = 0 then x = 0.

By analogy to the case of Rn, length or magnitude of vectors in Cn can be defined.

Definition 3.2.16 Let z ∈ Cn. Then |z| ≡ (z · z)1/2.

The conclusions of the following theorem are also called the axioms for a norm.

Theorem 3.2.17 For length defined in Definition 3.2.16, the following hold.

| |z| ≥ 0 and |z| = 0 if and only if z = 0 | (3.16)

| If α is a scalar, |αz| = |α| |z| |z + w| ≤ |z| + |w|. | (3.17) (3.18)

Proof: The first two claims are left as exercises. To establish the third, you use the same argument which was used in Rn.

|z + w|² = (z + w, z + w)
= z · z + w · w + w · z + z · w
= |z|² + |w|² + 2 Re w · z
≤ |z|² + |w|² + 2|w · z|
≤ |z|² + |w|² + 2|w| |z| = (|z| + |w|)² . 

Occasionally, I may refer to the inner product in Cn as the dot product. They are the same thing for Rn. However, it is convenient to draw a distinction when discussing matrix multiplication a little later.

3.3 Exercises

Use formula (3.11) to verify the Cauchy-Schwarz inequality and to show that equality occurs if and only if one of the vectors is a scalar multiple of the other.

For u, v vectors in R³, define the product, u ∗ v ≡u₁v₁ + 2u₂v₂ + 3u₃v₃. Show the axioms for a dot product all hold for this funny product. Prove

|u ∗ v| ≤ (u ∗ u)1/2 (v ∗ v)1/2.

Hint: Do not try to do this with methods from trigonometry.

Find the angle between the vectors 3i − j − k and i + 4j + 2k.

Find the angle between the vectors i − 2j + k and i + 2j − 7k.

Find proju (v) where v = (1, 0,−2) and u = (1, 2, 3) .

Find proju (v) where v = (1, 2,−2) and u = (1, 0, 3) .

Find proju (v) where v = (1, 2,−2, 1) and u = (1, 2, 3, 0) .

Does it make sense to speak of proj₀ (v)?

If F is a force and D is a vector, show projD (F) = (|F|cos ) u where u is the unit vector in the direction of D, u = D/|D| and is the included angle between the two vectors, F and D. |F| cos is sometimes called the component of the force, F in the direction, D.

Prove the Cauchy-Schwarz inequality in Rn as follows. For u, v vectors, consider

(u − projv u)·(u − projv u) ≥ 0

Now simplify using the axioms of the dot product and then put in the formula for the projection. Of course this expression equals 0 and you get equality in the Cauchy-Schwarz inequality if and only if u = projv u. What is the geometric meaning of u = projv u?

A boy drags a sled for 100 feet along the ground by pulling on a rope which is 20 degrees from the horizontal with a force of 40 pounds. How much work does this force do?

A girl drags a sled for 200 feet along the ground by pulling on a rope which is 30 degrees from the horizontal with a force of 20 pounds. How much work does this force do?

A large dog drags a sled for 300 feet along the ground by pulling on a rope which is 45 degrees from the horizontal with a force of 20 pounds. How much work does this force do?

How much work in Newton meters does it take to slide a crate 20 meters along a loading dock by pulling on it with a 200 Newton force at an angle of 30° from the horizontal?

An object moves 10 meters in the direction of j. There are two forces acting on this object, F₁ = i + j + 2k, and F₂ = −5i + 2j−6k. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

An object moves 10 meters in the direction of j + i. There are two forces acting on this object, F₁ = i + 2j + 2k, and F₂ = 5i + 2j−6k. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?

An object moves 20 meters in the direction of k + j. There are two forces acting on this object, F₁ = i + j + 2k, and F₂ = i + 2j−6k. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force.

If a, b, c are vectors. Show that (b + c)⊥ = b⊥ + c⊥ where b⊥ = b− proja (b) .

Find (1, 2, 3, 4)· (2, 0, 1, 3) .

Show that   

Prove from the axioms of the dot product the parallelogram identity, |a + b|² + |a − b|² = 2|a|² + 2|b|².

3.4 The Cross Product

The cross product is the other way of multiplying two vectors in R³. It is very different from the dot product in many ways. First the geometric meaning is discussed and then a description in terms of coordinates is given. Both descriptions of the cross product are important. The geometric description is essential in order to understand the applications to physics and geometry while the coordinate description is the only way to practically compute the cross product.

Definition 3.4.1 Three vectors, a, b, c form a right handed system if when you extend the fingers of your right hand along the vector a and close them in the direction of b, the thumb points roughly in the direction of c.

For an example of a right handed system of vectors, see the following picture.

In this picture the vector c points upwards from the plane determined by the other two vectors. You should consider how a right hand system would differ from a left hand system. Try using your left hand and you will see that the vector c would need to point in the opposite direction as it would for a right hand system.

From now on, the vectors, i, j, k will always form a right handed system. To repeat, if you extend the fingers of your right hand along i and close them in the direction j, the thumb points in the direction of k.

The following is the geometric description of the cross product. It gives both the direction and the magnitude and therefore specifies the vector.

Definition 3.4.2 Let a and b be two vectors in R³. Then a × b is defined by the following two rules.

1. |a × b| = |a| |b| sin where is the included angle.

2. a × b·a = 0, a × b · b = 0, and a, b, a × b forms a right hand system.

Note that |a×b| is the area of the parallelogram determined by a and b.

The cross product satisfies the following properties.

| a × b = −(b × a) , a × a = 0, | (3.19)

For α a scalar,

| (αa)×b = α (a × b) = a× (αb) , | (3.20)

For a, b, and c vectors, one obtains the distributive laws,

| a× (b + c) = a × b + a × c, | (3.21)

| (b + c) × a = b × a + c × a. | (3.22)

Formula (3.19) follows immediately from the definition. The vectors a×b and b × a have the same magnitude, |a| |b| sin , and an application of the right hand rule shows they have opposite direction. Formula (3.20) is also fairly clear. If α is a nonnegative scalar, the direction of (αa) ×b is the same as the direction of a×b,α (a×b) and a×(αb) while the magnitude is just α times the magnitude of a×b which is the same as the magnitude of α (a × b) and a× (αb) . Using this yields equality in (3.20). In the case where α < 0, everything works the same way except the vectors are all pointing in the opposite direction and you must multiply by |α| when comparing their magnitudes. The distributive laws are much harder to establish but the second follows from the first quite easily. Thus, assuming the first, and using (3.19),

(b + c)× a = −a× (b + c) = − (a × b + a × c)

= b × a + c × a.

A proof of the distributive law is given in a later section for those who are interested.

Now from the definition of the cross product,

i × j = k  j × i = −k
k × i = j  i × k = −j
j × k = i  k × j = −i

With this information, the following gives the coordinate description of the cross product.

Proposition 3.4.3 Let a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k be two vectors. Then

| a × b = (a₂b₃ − a₃b₂) i+ (a₃b₁ − a₁b₃) j + (a₁b₂ − a₂b₁) k. | (3.23)

Proof: From the above table and the properties of the cross product listed,

( a₁i + a₂j + a₃k) × (b₁i + b₂j + b₃k) = 

a₁b₂i × j + a₁b₃i × k + a₂b₁j × i + a₂b₃j × k+ a₃b₁k × i + a₃b₂k × j

= a₁b₂k−a₁b₃j−a₂b₁k + a₂b₃i + a₃b₁j − a₃b₂i

|      = (a₂b₃ − a₃b₂) i+ (a₃b₁ − a₁b₃) j+ (a₁b₂ −a₂b₁) k | (3.24)

It is probably impossible for most people to remember (3.23). Fortunately, there is a somewhat easier way to remember it. Define the determinant of a 2 × 2 matrix as follows

Then

| [picture] | (3.25)

where you expand the determinant along the top row. This yields

Note that to get the scalar which multiplies i you take the determinant of what is left after deleting the first row and the first column and multiply by (−1)¹⁺¹ because i is in the first row and the first column. Then you do the same thing for the j and k. In the case of the j there is a minus sign because j is in the first row and the second column and so(−1)¹⁺² = −1 while the k is multiplied by (−1)³⁺¹ = 1. The above equals

| (a₂b₃ − a₃b₂) i− (a₁b₃ − a₃b₁) j+ (a₁b₂ − a₂b¹) k | (3.26)

which is the same as (3.24). There will be much more presented on determinants later. For now, consider this an introduction if you have not seen this topic.

Example 3.4.4 Find (i − j + 2k) × (3i − 2j + k) .

Use (3.25) to compute this.

Example 3.4.5 Find the area of the parallelogram determined by the vectors,

(i − j + 2k) , (3i − 2j + k) .

These are the same two vectors in Example 3.4.4.

From Example 3.4.4 and the geometric description of the cross product, the area is just the norm of the vector obtained in Example 3.4.4. Thus the area is .

Example 3.4.6 Find the area of the triangle determined by (1, 2, 3) , (0, 2, 5) , (5, 1, 2) .

This triangle is obtained by connecting the three points with lines. Picking (1, 2, 3) as a starting point, there are two displacement vectors, (−1, 0, 2) and (4, −1, −1) such that the given vector added to these displacement vectors gives the other two vectors. The area of the triangle is half the area of the parallelogram determined by (−1, 0, 2) and (4, −1, −1) . Thus (−1, 0, 2)× (4,−1,−1) = (2, 7, 1) and so the area of the triangle is .

Observation 3.4.7 In general, if you have three points (vectors) in R³, P,Q,R the area of the triangle is given by

3.4.1 The Distributive Law For The Cross Product

This section gives a proof for (3.21), a fairly difficult topic. It is included here for the interested student. If you are satisfied with taking the distributive law on faith, it is not necessary to read this section. The proof given here is quite clever and follows the one given in [3]. Another approach, based on volumes of parallelepipeds is found in [14] and is discussed a little later.

Lemma 3.4.8 Let b and c be two vectors. Then b × c = b × c⊥ where c||+ c⊥ = c and c⊥·b = 0. 

Proof: Consider the following picture.

Now  and so c⊥ is in the plane determined by c and b. Therefore, from the geometric definition of the cross product, b × c and b × c⊥ have the same direction. Now, referring to the picture,

|b × c⊥| = |b| |c⊥| = |b| |c|sin = |b × c|.

Therefore, b × c and b × c⊥ also have the same magnitude and so they are the same vector. 

With this, the proof of the distributive law is in the following theorem.

Theorem 3.4.9 Let a, b, and c be vectors in R³. Then

| a× (b + c) = a × b + a × c | (3.27)

Proof: Suppose first that a · b = a · c = 0. Now imagine a is a vector coming out of the page and let b, c and b + c be as shown in the following picture.

Then a × b, a × (b + c) , and a × c are each vectors in the same plane, perpendicular to a as shown. Thus a × c · c = 0, a × (b + c)·(b + c) = 0, and a × b · b = 0. This implies that to get a × b you move counterclockwise through an angle of π/2 radians from the vector b. Similar relationships exist between the vectors a×(b + c) and b + c and the vectors a×c and c. Thus the angle between a × b and a × (b + c) is the same as the angle between b + c and b and the angle between a × c and a × (b + c) is the same as the angle between c and b + c. In addition to this, since a is perpendicular to these vectors,

|a × b| = |a| |b|,|a× (b + c)| = |a| |b + c|, and

|a×c| = |a| |c|. 

Therefore,

and so

showing the triangles making up the parallelogram on the right and the four sided figure on the left in the above picture are similar. It follows the four sided figure on the left is in fact a parallelogram and this implies the diagonal is the vector sum of the vectors on the sides, yielding (3.27).

Now suppose it is not necessarily the case that a · b = a · c = 0. Then write b = b|| + b⊥ where b⊥ · a = 0. Similarly c = c|| + c⊥. By the above lemma and what was just shown,

a × (b + c) = a × (b + c)⊥ = a × (b⊥ + c⊥)

= a × b⊥ + a × c⊥ = a × b + a × c. 

The result of Problem 18 of the exercises 3.3 is used to go from the first to the second line.

3.4.2 The Box Product

Definition 3.4.10 A parallelepiped determined by the three vectors, a, b, and c consists of

{ra+sb + tc : r, s, t ∈ [0, 1]}.

That is, if you pick three numbers, r, s, and t each in [0, 1] and form ra+sb + tc, then the collection of all such points is what is meant by the parallelepiped determined by these three vectors.

The following is a picture of such a thing.

You notice the area of the base of the parallelepiped, the parallelogram determined by the vectors, a and b has area equal to |a × b| while the altitude of the parallelepiped is |c|cos where is the angle shown in the picture between c and a × b. Therefore, the volume of this parallelepiped is the area of the base times the altitude which is just

|a × b| |c| cos = a × b · c.

This expression is known as the box product and is sometimes written as [a, b, c] . You should consider what happens if you interchange the b with the c or the a with the c. You can see geometrically from drawing pictures that this merely introduces a minus sign. In any case the box product of three vectors always equals either the volume of the parallelepiped determined by the three vectors or else minus this volume.

Example 3.4.11 Find the volume of the parallelepiped determined by the vectors, i + 2j− 5k, i + 3j− 6k,3i + 2j + 3k.

According to the above discussion, pick any two of these, take the cross product and then take the dot product of this with the third of these vectors. The result will be either the desired volume or minus the desired volume.

Now take the dot product of this vector with the third which yields

(3i + j + k)· (3i + 2j + 3k) = 9 + 2 + 3 = 14.

This shows the volume of this parallelepiped is 14 cubic units.

There is a fundamental observation which comes directly from the geometric definitions of the cross product and the dot product.

Lemma 3.4.12 Let a, b, and c be vectors. Then (a × b)·c = a·(b × c) .

Proof: This follows from observing that either (a × b)·c and a·(b × c) both give the volume of the parallelepiped or they both give −1 times the volume. 

Notation 3.4.13 The box product a × b·c = a·b × c is denoted more compactly as [a, b, c].

3.4.3 A Proof Of The Distributive Law

Here is another proof of the distributive law for the cross product. Let x be a vector. From the above observation,

x · a× (b + c) = (x × a) · (b + c)
= (x × a) · b+ (x × a) · c
= x · a × b + x · a × c
= x · (a × b + a × c) .

Therefore,

x· [a× (b + c) − (a × b + a × c)] = 0

for all x. In particular, this holds for x = a×(b + c) − (a × b + a × c) showing that a× (b + c) = a × b + a × c and this proves the distributive law for the cross product another way. 

Observation 3.4.14 Suppose you have three vectors, u = (a, b, c) , v = (d, e, f) , and w = (g, h, i) . Then u · v × w is given by the following.

The message is that to take the box product, you can simply take the determinant of the matrix which results by letting the rows be the rectangular components of the given vectors in the order in which they occur in the box product. More will be presented on determinants later.

3.5 The Vector Identity Machine

In practice, you often have to deal with combinations of several cross products mixed in with dot products. It is extremely useful to have a technique which will allow you to discover vector identities and simplify expressions involving cross and dot products in three dimensions. This involves two special symbols, δij and εijk which are very useful in dealing with vector identities. To begin with, here is the definition of these symbols.

Definition 3.5.1 The symbol δij, called the Kronecker delta symbol is defined as follows.

With the Kronecker symbol i and j can equal any integer in {1, 2,· · ·, n} for any n ∈ N. 

Definition 3.5.2 For i, j, and k integers in the set, {1, 2, 3}, εijk is defined as follows.

The subscripts ijk and ij in the above are called indices. A single one is called an index. This symbol εijk is also called the permutation symbol.

The way to think of εijk is that ε₁₂₃ = 1 and if you switch any two of the numbers in the list i, j, k, it changes the sign. Thus εijk = −εjik and εijk = −εkji etc. You should check that this rule reduces to the above definition. For example, it immediately implies that if there is a repeated index, the answer is zero. This follows because εiij = −εiij and so εiij = 0.

It is useful to use the Einstein summation convention when dealing with these symbols. Simply stated, the convention is that you sum over the repeated index. Thus aibi means ∑i aibi. Also, δijxj means ∑iiδijxj = xi. Thus δijxj = xi, δii = 3. When you use this convention, there is one very important thing to never forget. It is this: Never have an index be repeated more than once. Thus aibi is all right but aiibi is not. The reason for this is that you end up getting confused about what is meant. If you want to write ∑i aibici it is best to simply use the summation notation.i There is a very important reduction identity connecting these two symbols.

Lemma 3.5.3 The following holds.

εijkεirs = (δjrδks − δkrδjs) .

Proof: If {j, k} ≠ {r, s} then every term in the sum on the left must have either εijk or εirs contains a repeated index. Therefore, the left side equals zero. The right side also equals zero in this case. To see this, note that if the two sets are not equal, then there is one of the indices in one of the sets which is not in the other set. For example, it could be that j is not equal to either r or s. Then the right side equals zero.

Therefore, it can be assumed {j, k} = {r, s}. If i = r and j = s for s ≠ r, then there is exactly one term in the sum on the left and it equals 1. The right also reduces to 1 in this case. If i = s and j = r, there is exactly one term in the sum on the left which is nonzero and it must equal −1. The right side also reduces to −1 in this case. If there is a repeated index in {j, k}, then every term in the sum on the left equals zero. The right also reduces to zero in this case because then j = k = r = s and so the right side becomes (1) (1)− (−1) (−1) = 0.

Proposition 3.5.4 Let u, v be vectors in Rn where the Cartesian coordinates of u are (u₁,· · ·, un) and the Cartesian coordinates of v are (v₁,· · ·, vn). Then u · v = uivi. If u, v are vectors in R³, then

(u × v)i = εijkujvk.

Also, δikak = ai.

Proof: The first claim is obvious from the definition of the dot product. The second is verified by simply checking that it works. For example,

and so

(u × v)₁ = (u₂v₃ − u₃v₂) .

From the above formula in the proposition,

ε1jkujvk ≡u₂v₃ − u₃v₂,

the same thing. The cases for (u × v)₂ and (u × v)₃ are verified similarly. The last claim follows directly from the definition.

With this notation, you can easily discover vector identities and simplify expressions which involve the cross product.

Example 3.5.5 Discover a formula which simplifies (u × v) · (z × w) , u, v ∈ R³.

From the above description of the cross product and dot product, along with the reduction identity,

(u × v)· (z × w) =

εijkujvkεirszrws = (δjrδks −δjsδkr) ujvkzrws
= ujvkzjwk −ujvkzkwj 
= (u · z) (v · w)− (u · w) (v · z)

Example 3.5.6 Simplify u× (u × v) .

The ith component is

εijkuj (u × v)k = εijkujεkrsurvs = εkijεkrsujurvs
= (δirδjs −δjrδis) ujurvs
= ujuivj −ujujvi
= (u·v) ui − |u|²vi

Hence

u× (u × v) = (u·v) u− |u|²v

because the ith components of the two sides are equal for any i.

3.6 Exercises

Show that if a × u = 0 for all unit vectors, u, then a = 0.

Find the area of the triangle determined by the three points, (1, 2, 3) , (4, 2, 0) and (−3, 2, 1) .

Find the area of the triangle determined by the three points, (1, 0, 3) , (4, 1, 0) and (−3, 1, 1) .

Find the area of the triangle determined by the three points, (1, 2, 3) , (2, 3, 4) and (3, 4, 5) . Did something interesting happen here? What does it mean geometrically?

Find the area of the parallelogram determined by the vectors, (1 , 2, 3), (3,−2, 1) .

Find the area of the parallelogram determined by the vectors, (1, 0, 3), (4,−2, 1) .

Find the volume of the parallelepiped determined by the vectors, i−7j−5k, i−2j−6k,3i+2j+3k.

Suppose a, b, and c are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer?

What does it mean geometrically if the box product of three vectors gives zero?

Using Problem 9, find an equation of a plane containing the two position vectors, a and b and the point 0. Hint: If (x, y, z) is a point on this plane the volume of the parallelepiped determined by (x, y, z) and the vectors a, b equals 0.

Using the notion of the box product yielding either plus or minus the volume of the parallelepiped determined by the given three vectors, show that

(a × b)·c = a·(b × c)

In other words, the dot and the cross can be switched as long as the order of the vectors remains the same. Hint: There are two ways to do this, by the coordinate description of the dot and cross product and by geometric reasoning. It is better if you use geometric reasoning.

Is a× (b × c) = (a × b) ×c? What is the meaning of a × b × c? Explain. Hint: Try (i × j) ×j.

Discover a vector identity for (u × v) ×w and one for u× (v × w).

Discover a vector identity for (u × v)× (z × w).

Simplify (u × v)· (v × w)× (w × z) .

Simplify |u × v|²(u × v)² − |u| ²|v|².

For u, v, w functions of t, show the product rules

(u×v)′ = u′ ×v + u×v′
(u·v)′ = u′ ·v + u·v′

If u is a function of t, and the magnitude |u (t)| is a constant, show from the above problem that the velocity u′ is perpendicular to u.

When you have a rotating rigid body with angular velocity vector Ω, then the velocity vector v ≡ u′ is given by

v = Ω × u

where u is a position vector. The acceleration is the derivative of the velocity. Show that if Ω is a constant vector, then the acceleration vector a = v′ is given by the formula

a = Ω×(Ω × u) .

Now simplify the expression. It turns out this is centripetal acceleration.

Verify directly that the coordinate description of the cross product, a × b has the property that it is perpendicular to both a and b. Then show by direct computation that this coordinate description satisfies

|a × b|² = |a|²|b|² − (a · b)²
= |a|²|b|² (1− cos ² ())

where is the angle included between the two vectors. Explain why |a × b| has the correct magnitude. All that is missing is the material about the right hand rule. Verify directly from the coordinate description of the cross product that the right thing happens with regards to the vectors i, j, k. Next verify that the distributive law holds for the coordinate description of the cross product. This gives another way to approach the cross product. First define it in terms of coordinates and then get the geometric properties from this. However, this approach does not yield the right hand rule property very easily.

Systems Of Equations

4.1 Systems Of Equations, Geometry

As you know, equations like 2x + 3y = 6 can be graphed as straight lines in R². To find the solution to two such equations, you could graph the two straight lines and the ordered pairs identifying the point (or points) of intersection would give the x and y values of the solution to the two equations because such an ordered pair satisfies both equations. The following picture illustrates what can occur with two equations involving two variables.

In the first example of the above picture, there is a unique point of intersection. In the second, there are no points of intersection. The other thing which can occur is that the two lines are really the same line. For example, x + y = 1 and 2x + 2y = 2 are relations which when graphed yield the same line. In this case there are infinitely many points in the simultaneous solution of these two equations, every ordered pair which is on the graph of the line. It is always this way when considering linear systems of equations. There is either no solution, exactly one or infinitely many although the reasons for this are not completely comprehended by considering a simple picture in two dimensions, R².

Example 4.1.1 Find the solution to the system x + y = 3, y − x = 5.

You can verify the solution is (x, y) = (−1, 4) . You can see this geometrically by graphing the equations of the two lines. If you do so correctly, you should obtain a graph which looks something like the following in which the point of intersection represents the solution of the two equations.

Example 4.1.2 You can also imagine other situations such as the case of three intersecting lines having no common point of intersection or three intersecting lines which do intersect at a single point as illustrated in the following picture.

In the case of the first picture above, there would be no solution to the three equations whose graphs are the given lines. In the case of the second picture there is a solution to the three equations whose graphs are the given lines.

The points, (x, y, z) satisfying an equation in three variables like 2x + 4y − 5z = 8 form a plane ¹ and geometrically, when you solve systems of equations involving three variables, you are taking intersections of planes. Consider the following picture involving two planes.

Notice how these two planes intersect in a line. It could also happen the two planes could fail to intersect.

Now imagine a third plane. One thing that could happen is this third plane could have an intersection with one of the first planes which results in a line which fails to intersect the first line as illustrated in the following picture.

Thus there is no point which lies in all three planes. The picture illustrates the situation in which the line of intersection of the new plane with one of the original planes forms a line parallel to the line of intersection of the first two planes. However, in three dimensions, it is possible for two lines to fail to intersect even though they are not parallel. Such lines are called skew lines. You might consider whether there exist two skew lines, each of which is the intersection of a pair of planes selected from a set of exactly three planes such that there is no point of intersection between the three planes. You can also see that if you tilt one of the planes you could obtain every pair of planes having a nonempty intersection in a line and yet there may be no point in the intersection of all three.

It could happen also that the three planes could intersect in a single point as shown in the following picture.

In this case, the three planes have a single point of intersection. The three planes could also intersect in a line.

Thus in the case of three equations having three variables, the planes determined by these equations could intersect in a single point, a line, or even fail to intersect at all. You see that in three dimensions there are many possibilities. If you want to waste some time, you can try to imagine all the things which could happen but this will not help for more variables than 3 which is where many of the important applications lie.

Relations like x + y −2z + 4w = 8 are often called hyper-planes.² However, it is impossible to draw pictures of such things. The only rational and useful way to deal with this subject is through the use of algebra not art. Mathematics exists partly to free us from having to always draw pictures in order to draw conclusions.

4.2 Systems Of Equations, Algebraic Procedures

4.2.1 Elementary Operations

Consider the following example.

Example 4.2.1 Find x and y such that

| x + y = 7 and 2x − y = 8. | (4.1)

The set of ordered pairs, (x, y) which solve both equations is called the solution set.

You can verify that (x, y) = (5, 2) is a solution to the above system. The interesting question is this: If you were not given this information to verify, how could you determine the solution? You can do this by using the following basic operations on the equations, none of which change the set of solutions of the system of equations.

Definition 4.2.2 Elementary operations are those operations consisting of the following.

Interchange the order in which the equations are listed.

 Multiply any equation by a nonzero number.

 Replace any equation with itself added to a multiple of another equation.

Example 4.2.3 To illustrate the third of these operations on this particular system, consider the following.

x + y = 7
2 x − y = 8

The system has the same solution set as the system

x + y = 7
−3y = −6

To obtain the second system, take the second equation of the first system and add -2 times the first equation to obtain

−3y = −6.

Now, this clearly shows that y = 2 and so it follows from the other equation that x + 2 = 7 and so x = 5.

Of course a linear system may involve many equations and many variables. The solution set is still the collection of solutions to the equations. In every case, the above operations of Definition 4.2.2 do not change the set of solutions to the system of linear equations.

Theorem 4.2.4 Suppose you have two equations, involving the variables,

| (x₁,· · ·, xn) E₁ = f₁, E₂ = f₂ | (4.2)

where E₁ and E₂ are expressions involving the variables and f₁ and f₂ are constants. (In the above example there are only two variables, x and y and E₁ = x + y while E₂ = 2x − y.) Then the system E₁ = f₁, E₂ = f₂ has the same solution set as

| E₁ = f₁, E₂ + aE₁ = f₂ + af₁. | (4.3)

Also the system E₁ = f₁, E₂ = f₂ has the same solutions as the system, E₂ = f₂, E₁ = f₁. The system E₁ = f₁, E₂ = f₂ has the same solution as the system E₁ = f₁, aE₂ = af₂ provided a ≠ 0.

Proof: If (x₁,· · ·, xn) solves E₁ = f₁, E₂ = f₂ then it solves the first equation in E₁ = f₁, E₂ + aE₁ = f₂ + af₁. Also, it satisfies aE₁ = af₁ and so, since it also solves E₂ = f₂ it must solve E₂+aE₁ = f₂+af₁. Therefore, if (x₁,· · ·, xn) solves E₁ = f₁, E₂ = f₂ it must also solve E₂+aE₁ = f₂+af₁. On the other hand, if it solves the system E₁ = f₁ and E₂+aE₁ = f₂+af₁, then aE₁ = af₁ and so you can subtract these equal quantities from both sides of E₂ + aE₁ = f₂ + af₁ to obtain E₂ = f₂ showing that it satisfies E₁ = f₁, E₂ = f₂.

The second assertion of the theorem which says that the system E₁ = f₁, E₂ = f₂ has the same solution as the system, E₂ = f₂, E₁ = f₁ is seen to be true because it involves nothing more than listing the two equations in a different order. They are the same equations.

The third assertion of the theorem which says E₁ = f₁, E₂ = f₂ has the same solution as the system E₁ = f₁, aE₂ = af₂ provided a ≠ 0 is verified as follows: If (x₁,· · ·, xn) is a solution of E₁ = f₁, E₂ = f₂, then it is a solution to E₁ = f₁, aE₂ = af₂ because the second system only involves multiplying the equation, E₂ = f₂ by a. If (x₁,· · ·, xn) is a solution of E₁ = f₁, aE₂ = af₂, then upon multiplying aE₂ = af₂ by the number 1/a, you find that E₂ = f2 . 

Stated simply, the above theorem shows that the elementary operations do not change the solution set of a system of equations.

Here is an example in which there are three equations and three variables. You want to find values for x, y, z such that each of the given equations are satisfied when these values are plugged in to the equations.

Example 4.2.5 Find the solutions to the system,

| x + 3y + 6z = 25 2x + 7y + 14z = 58 2y + 5z = 19 | (4.4)

To solve this system replace the second equation by (−2) times the first equation added to the second. This yields the system

| x + 3y + 6z = 25 y + 2z = 8 2y + 5z = 19 | (4.5)

Now take (−2) times the second and add to the third. More precisely, replace the third equation with (−2) times the second added to the third. This yields the system

| x + 3y + 6z = 25 y + 2z = 8 z = 3 | (4.6)

At this point, you can tell what the solution is. This system has the same solution as the original system and in the above, z = 3. Then using this in the second equation, it follows y + 6 = 8 and so y = 2. Now using this in the top equation yields x + 6 + 18 = 25 and so x = 1. This process is called back substitution.

Alternatively, in (4.6) you could have continued as follows. Add (−2) times the bottom equation to the middle and then add (−6) times the bottom to the top. This yields

x + 3y = 7
y = 2
z = 3

Now add (−3) times the second to the top. This yields

x = 1
  y = 2 ,
z = 3

a system which has the same solution set as the original system. This avoided back substitution and led to the same solution set.

4.2.2 Gaussian Elimination

A less cumbersome way to represent a linear system is to write it as an augmented matrix. For example the linear system, (4.4) can be written as

It has exactly the same information as the original system but here it is understood there is an

x column, , a y column,  and a z column, .

The rows correspond to the equations in the system. Thus the top row in the augmented matrix corresponds to the equation,

x + 3y + 6z = 25. 

Now when you replace an equation with a multiple of another equation added to itself, you are just taking a row of this augmented matrix and replacing it with a multiple of another row added to it. Thus the first step in solving (4.4) would be to take (−2) times the first row of the augmented matrix above and add it to the second row,

Note how this corresponds to (4.5). Next take (−2) times the second row and add to the third,

This augmented matrix corresponds to the system

x + 3y + 6z = 25
y + 2z = 8
z = 3

which is the same as (4.6). By back substitution you obtain the solution x = 1, y = 6, and z = 3.

In general a linear system is of the form

a₁₁x₁ +· · · + a1nxn = b1,
.
·
·
am1x₁ +· · · + amnxn = bm

(4.7)

where the xi are variables and the aij and bi are constants. This system can be represented by the augmented matrix 

Changes to the system of equations in (4.7) as a result of an elementary operations translate into changes of the augmented matrix resulting from a row operation. Note that Theorem 4.2.4 implies that the row operations deliver an augmented matrix for a system of equations which has the same solution set as the original system.

Definition 4.2.6 The row operations consist of the following

Switch two rows.

Multiply a row by a nonzero number.

Replace a row by a multiple of another row added to it.

Gaussian elimination is a systematic procedure to simplify an augmented matrix to a reduced form. In the following definition, the term “leading entry” refers to the first nonzero entry of a row when scanning the row from left to right.

Definition 4.2.7 An augmented matrix is in echelon form if

All nonzero rows are above any rows of zeros.

Each leading entry of a row is in a column to the right of the leading entries of any rows above it.

Definition 4.2.8 An augmented matrix is in row reduced echelon form if

All nonzero rows are above any rows of zeros.

Each leading entry of a row is in a column to the right of the leading entries of any rows above it.

All entries in a column above and below a leading entry are zero.

Each leading entry is a 1, the only nonzero entry in its column.

Example 4.2.9 Here are some augmented matrices which are in row reduced echelon form.

Example 4.2.10 Here are augmented matrices in echelon form which are not in row reduced echelon form but which are in echelon form.

Example 4.2.11 Here are some augmented matrices which are not in echelon form.

Definition 4.2.12 A pivot position in a matrix is the location of a leading entry in an echelon form resulting from the application of row operations to the matrix. A pivot column is a column that contains a pivot position.

For example consider the following. 

Example 4.2.13 Suppose

Where are the pivot positions and pivot columns?

Replace the second row by −3 times the first added to the second. This yield

This is not in reduced echelon form so replace the bottom row by −4 times the top row added to the bottom. This yields

This is still not in reduced echelon form. Replace the bottom row by −1 times the middle row added to the bottom. This yields  

which is in echelon form, although not in reduced echelon form. Therefore, the pivot positions in the original matrix are the locations corresponding to the first row and first column and the second row and second columns as shown in the following:

Thus the pivot columns in the matrix are the first two columns.

The following is the algorithm for obtaining a matrix which is in row reduced echelon form.

Algorithm 4.2.14

This algorithm tells how to start with a matrix and do row operations on it in such a way as to end up with a matrix in row reduced echelon form.

Find the first nonzero column from the left. This is the first pivot column. The position at the top of the first pivot column is the first pivot position. Switch rows if necessary to place a nonzero number in the first pivot position.

Use row operations to zero out the entries below the first pivot position.

Ignore the row containing the most recent pivot position identified and the rows above it. Repeat steps 1 and 2 to the remaining sub-matrix, the rectangular array of numbers obtained from the original matrix by deleting the rows you just ignored. Repeat the process until there are no more rows to modify. The matrix will then be in echelon form.

Moving from right to left, use the nonzero elements in the pivot positions to zero out the elements in the pivot columns which are above the pivots.

Divide each nonzero row by the value of the leading entry. The result will be a matrix in row reduced echelon form.

This row reduction procedure applies to both augmented matrices and non augmented matrices. There is nothing special about the augmented column with respect to the row reduction procedure.

Example 4.2.15 Here is a matrix.

Do row reductions till you obtain a matrix in echelon form. Then complete the process by producing one in row reduced echelon form.

The pivot column is the second. Hence the pivot position is the one in the first row and second column. Switch the first two rows to obtain a nonzero entry in this pivot position.

.

Step two is not necessary because all the entries below the first pivot position in the resulting matrix are zero. Now ignore the top row and the columns to the left of this first pivot position. Thus you apply the same operations to the smaller matrix

The next pivot column is the third corresponding to the first in this smaller matrix and the second pivot position is therefore, the one which is in the second row and third column. In this case it is not necessary to switch any rows to place a nonzero entry in this position because there is already a nonzero entry there. Multiply the third row of the original matrix by −2 and then add the second row to it. This yields

The next matrix the steps in the algorithm are applied to is

The first pivot column is the first column in this case and no switching of rows is necessary because there is a nonzero entry in the first pivot position. Therefore, the algorithm yields for the next step

Now the algorithm will be applied to the matrix

There is only one column and it is nonzero so this single column is the pivot column. Therefore, the algorithm yields the following matrix for the echelon form.

To complete placing the matrix in reduced echelon form, multiply the third row by 3 and add −2 times the fourth row to it. This yields

Next multiply the second row by 3 and take 2 times the fourth row and add to it. Then add the fourth row to the first

Next work on the fourth column in the same way.

Take −1/2 times the second row and add to the first.

Finally, divide by the value of the leading entries in the nonzero rows. 

The above algorithm is the way a computer would obtain a reduced echelon form for a given matrix. It is not necessary for you to pretend you are a computer but if you like to do so, the algorithm described above will work. The main idea is to do row operations in such a way as to end up with a matrix in echelon form or row reduced echelon form because when this has been done, the resulting augmented matrix will allow you to describe the solutions to the linear system of equations in a meaningful way. When you do row operations until you obtain row reduced echelon form, the process is called the Gauss Jordan method. Otherwise, it is called Gaussian elimination.

Example 4.2.16 Give the complete solution to the system of equations, 5x + 10y − 7z = −2, 2x + 4y − 3z = −1, and 3x + 6y + 5z = 9.

The augmented matrix for this system is

Multiply the second row by 2, the first row by 5, and then take (−1) times the first row and add to the second. Then multiply the first row by 1/5. This yields

Now, combining some row operations, take (−3) times the first row and add this to 2 times the last row and replace the last row with this. This yields.

One more row operation, taking (−1) times the second row and adding to the bottom yields.

This is impossible because the last row indicates the need for a solution to the equation

0x + 0y + 0z = 20

and there is no such thing because 0 ≠ 20. This shows there is no solution to the three given equations. When this happens, the system is called inconsistent. In this case it is very easy to describe the solution set. The system has no solution.

Here is another example based on the use of row operations.

Example 4.2.17 Give the complete solution to the system of equations, 3x−y−5z = 9, y−10z = 0, and −2x + y = −6.

The augmented matrix of this system is

Replace the last row with 2 times the top row added to 3 times the bottom row. This gives 

The entry, 3 in this sequence of row operations is called the pivot. It is used to create zeros in the other places of the column. Next take −1 times the middle row and add to the bottom. Here the 1 in the second row is the pivot.

Take the middle row and add to the top and then divide the top row which results by 3.

This is in reduced echelon form. The equations corresponding to this reduced echelon form are y = 10z and x = 3 + 5z. Apparently z can equal any number. Lets call this number t. ³Therefore, the solution set of this system is x = 3 + 5t, y = 10t, and z = t where t is completely arbitrary. The system has an infinite set of solutions which are given in the above simple way. This is what it is all about, finding the solutions to the system.

There is some terminology connected to this which is useful. Recall how each column corresponds to a variable in the original system of equations. The variables corresponding to a pivot column are called basic variables. The other variables are called free variables. In Example 4.2.17 there was one free variable, z, and two basic variables, x and y. In describing the solution to the system of equations, the free variables are assigned a parameter. In Example 4.2.17 this parameter was t. Sometimes there are many free variables and in these cases, you need to use many parameters. Here is another example.

Example 4.2.18 Find the solution to the system

x + 2y − z + w = 3
x + y − z + w = 1
x + 3y − z + w = 5

The augmented matrix is

Take −1 times the first row and add to the second. Then take −1 times the first row and add to the third. This yields

Now add the second row to the bottom row

This matrix is in echelon form and you see the basic variables are x and y while the free variables are z and w. Assign s to z and t to w. Then the second row yields the equation, y = 2 while the top equation yields the equation, x + 2y −s + t = 3 and so since y = 2, this gives x + 4−s + t = 3 showing that x = −1 + s−t, y = 2, z = s, and w = t. It is customary to write this in the form

This is another example of a system which has an infinite solution set but this time the solution set depends on two parameters, not one. Most people find it less confusing in the case of an infinite solution set to first place the augmented matrix in row reduced echelon form rather than just echelon form before seeking to write down the description of the solution. In the above, this means we don’t stop with the echelon form (4.9). Instead we first place it in reduced echelon form as follows.

The solution is y = 2 from the second row and x = −1 + z −w from the first. Thus letting z = s and w = t, the solution is given in (4.10).

The number of free variables is always equal to the number of different parameters used to describe the solution. If there are no free variables, then either there is no solution as in the case where row operations yield an echelon form like

or there is a unique solution as in the case where row operations yield an echelon form like

Also, sometimes there are free variables and no solution as in the following:

There are a lot of cases to consider but it is not necessary to make a major production of this. Do row operations till you obtain a matrix in echelon form or reduced echelon form and determine whether there is a solution. If there is, see if there are free variables. In this case, there will be infinitely many solutions. Find them by assigning different parameters to the free variables and obtain the solution. If there are no free variables, then there will be a unique solution which is easily determined once the augmented matrix is in echelon or row reduced echelon form. In every case, the process yields a straightforward way to describe the solutions to the linear system. As indicated above, you are probably less likely to become confused if you place the augmented matrix in row reduced echelon form rather than just echelon form.

In summary,

Definition 4.2.19 A system of linear equations is a list of equations,

a₁₁x₁ + a₁₂x₂ +· · · + a1nxn = b₁
a₂₁x₁ + a₂₂x₂ +· · · + a2nxn = b₂

.
.
.

am1x₁ + am2x₂ +· · · + amnxn = bm

where aij are numbers, and bj is a number. The above is a system of m equations in the n variables, x₁, x₂· · ·, xn. Nothing is said about the relative size of m and n. Written more simply in terms of summation notation, the above can be written in the form

It is desired to find (x₁,· · ·, xn) solving each of the equations listed.

As illustrated above, such a system of linear equations may have a unique solution, no solution, or infinitely many solutions and these are the only three cases which can occur for any linear system. Furthermore, you do exactly the same things to solve any linear system. You write the augmented matrix and do row operations until you get a simpler system in which it is possible to see the solution, usually obtaining a matrix in echelon or reduced echelon form. All is based on the observation that the row operations do not change the solution set. You can have more equations than variables, fewer equations than variables, etc. It doesn’t matter. You always set up the augmented matrix and go to work on it.

Definition 4.2.20 A system of linear equations is called consistent if there exists a solution. It is called inconsistent if there is no solution.

These are reasonable words to describe the situations of having or not having a solution. If you think of each equation as a condition which must be satisfied by the variables, consistent would mean there is some choice of variables which can satisfy all the conditions. Inconsistent would mean there is no choice of the variables which can satisfy each of the conditions.

4.3 Exercises

Find the point (x₁, y₁) which lies on both lines, x + 3y = 1 and 4x − y = 3.

Solve Problem 1 graphically. That is, graph each line and see where they intersect.

Find the point of intersection of the two lines 3x + y = 3 and x + 2y = 1.

Solve Problem 3 graphically. That is, graph each line and see where they intersect.

Do the three lines, x+2y = 1, 2x−y = 1, and 4x+3y = 3 have a common point of intersection? If so, find the point and if not, tell why they don’t have such a common point of intersection.

Do the three planes, x + y − 3z = 2, 2x + y + z = 1, and 3x + 2y − 2z = 0 have a common point of intersection? If so, find one and if not, tell why there is no such point.

You have a system of k equations in two variables, k ≥ 2. Explain the geometric significance of

(a) No solution.

(b) A unique solution.

(c) An infinite number of solutions.

Here is an augmented matrix in which ∗ denotes an arbitrary number and  denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?

Here is an augmented matrix in which ∗ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?

Here is an augmented matrix in which ∗ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?

Here is an augmented matrix in which ∗ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?

Suppose a system of equations has fewer equations than variables. Must such a system be consistent? If so, explain why and if not, give an example which is not consistent.

If a system of equations has more equations than variables, can it have a solution? If so, give an example and if not, tell why not.

Find h such that

is the augmented matrix of an inconsistent matrix.

Find h such that 

 is the augmented matrix of a consistent matrix. 

Find h such that

 is the augmented matrix of a consistent matrix.

Choose h and k such that the augmented matrix shown has one solution. Then choose h and k such that the system has no solutions. Finally, choose h and k such that the system has infinitely many solutions.

Choose h and k such that the augmented matrix shown has one solution. Then choose h and k such that the system has no solutions. Finally, choose h and k such that the system has infinitely many solutions.

Determine if the system is consistent. If so, is the solution unique?

x + 2y + z − w = 2
x − y + z + w = 1
2x + y − z = 1
4x + 2y + z = 5

Determine if the system is consistent. If so, is the solution unique?

x + 2y + z − w = 2
x − y + z + w = 0
2x + y − z = 1
4x + 2y + z = 3

Find the general solution of the system whose augmented matrix is

Find the general solution of the system whose augmented matrix is

Find the general solution of the system whose augmented matrix is

Find the general solution of the system whose augmented matrix is

Find the general solution of the system whose augmented matrix is

Give the complete solution to the system of equations, 7x + 14y + 15z = 22, 2x + 4y + 3z = 5, and 3x + 6y + 10z = 13.

Give the complete solution to the system of equations, 3x − y + 4z = 6, y + 8z = 0, and −2i + y = −4.

Give the complete solution to the system of equations, 9x−2y+4z = −17, 13x−3y+6z = −25, and −2x − z = 3.

Give the complete solution to the system of equations, 65x+84y+16z = 546, 81x+105y+20z = 682, and 84x + 110y + 21z = 713.

Give the complete solution to the system of equations, 8x + 2y + 3z = −3, 8x + 3y + 3z = −1, and 4x + y + 3z = −9.

Give the complete solution to the system of equations, −8x+2y+5z = 18,−8x+3y +5z = 13, and −4x + y + 5z = 19.

Give the complete solution to the system of equations, 3x − y − 2z = 3, y − 4z = 0, and −2x + y = −2.

Give the complete solution to the system of equations, −9x + 15y = 66,−11x + 18y = 79 ,−x + y = 4, and z = 3.

Give the complete solution to the system of equations, −19x+8y = −108, −71x+30y = −404, −2x + y = −12, 4x + z = 14.

Consider the system −5x + 2y−z = 0 and −5x−2y−z = 0. Both equations equal zero and so −5x + 2y−z = −5x − 2y − z which is equivalent to y = 0. Thus x and z can equal anything. But when x = 1, z = −4, and y = 0 are plugged in to the equations, it doesn’t work. Why?

Four times the weight of Gaston is 150 pounds more than the weight of Ichabod. Four times the weight of Ichabod is 660 pounds less than seventeen times the weight of Gaston. Four times the weight of Gaston plus the weight of Siegfried equals 290 pounds. Brunhilde would balance all three of the others. Find the weights of the four sisters.

The steady state temperature, u in a plate solves Laplace’s equation, Δu = 0. One way to approximate the solution which is often used is to divide the plate into a square mesh and require the temperature at each node to equal the average of the temperature at the four adjacent nodes. This procedure is justified by the mean value property of harmonic functions. In the following picture, the numbers represent the observed temperature at the indicated nodes. Your task is to find the temperature at the interior nodes, indicated by x, y, z, and w. One of the equations is z = 

³ In this context t is called a parameter.

Matrices

5.1 Matrix Arithmetic

5.1.1 Addition And Scalar Multiplication Of Matrices

You have now solved systems of equations by writing them in terms of an augmented matrix and then doing row operations on this augmented matrix. It turns out such rectangular arrays of numbers are important from many other different points of view. Numbers are also called scalars. In these notes numbers will always be either real or complex numbers. I will refer to the set of numbers as F sometimes when it is not important to worry about whether the number is real or complex. Thus F can be either the real numbers, R or the complex numbers, C.

A matrix is a rectangular array of numbers. Several of them are referred to as matrices. For example, here is a matrix.

The size or dimension of a matrix is defined as m × n where m is the number of rows and n is the number of columns. The above matrix is a 3 × 4 matrix because there are three rows and four columns. The first row is (1 2 3 4) , the second row is (5 2 8 7) and so forth. The first column is

When specifying the size of a matrix, you always list the

number of rows before the number of columns. Also, you can remember the columns are like columns in a Greek temple. They stand upright while the rows just lay there like rows made by a tractor in a plowed field. Elements of the matrix are identified according to position in the matrix. For example, 8 is in position 2, 3 because it is in the second row and the third column. You might remember that you always list the rows before the columns by using the phrase Rowman Catholic. The symbol, (aij) refers to a matrix. The entry in the ith row and the jth column of this matrix is denoted by aij. Using this notation on the above matrix, a₂₃ = 8, a₃₂ = −9, a₁₂ = 2, etc.

There are various operations which are done on matrices. Matrices can be added multiplied by a scalar, and multiplied by other matrices. To illustrate scalar multiplication, consider the following example in which a matrix is being multiplied by the scalar 3.

The new matrix is obtained by multiplying every entry of the original matrix by the given scalar. If A is an m × n matrix, −A is defined to equal (−1) A.

Two matrices must be the same size to be added. The sum of two matrices is a matrix which is obtained by adding the corresponding entries. Thus

Two matrices are equal exactly when they are the same size and the corresponding entries are identical. Thus

because they are different sizes. As noted above, you write (cij) for the matrix C whose ijth entry is cij. In doing arithmetic with matrices you must define what happens in terms of the cij sometimes called the entries of the matrix or the components of the matrix.

The above discussion stated for general matrices is given in the following definition.

Definition 5.1.1 (Scalar Multiplication) If A = (aij) and k is a scalar, then kA = (kaij) .

Example 5.1.2 

Definition 5.1.3 (Addition) If A = (aij) and B = (bij) are two m × n matrices. Then A + B = C where

C = (cij)

for cij = aij + bij.

Example 5.1.4

To save on notation, we will often use Aij to refer to the ijth entry of the matrix A.

Definition 5.1.5 (The zero matrix) The m × n zero matrix is the m × n matrix having every entry equal to zero. It is denoted by 0.

Example 5.1.6 The 2× 3 zero matrix is

Note there are 2 × 3 zero matrices, 3 × 4 zero matrices, etc. In fact there is a zero matrix for every size.

Definition 5.1.7 (Equality of matrices) Let A and B be two matrices. Then A = B means that the two matrices are of the same size and for A = (aij) and B = (bij) , aij = bij for all 1 ≤ i ≤ m and 1 ≤ j ≤ n.

The following properties of matrices can be easily verified. You should do so.

Commutative Law Of Addition.

| A + B = B + A, | (5.1)

Associative Law for Addition.

| (A + B) + C = A + (B + C) | (5.2)

Existence of an Additive Identity

| A + 0 = A, | (5.3)

Existence of an Additive Inverse

| A + (−A) = 0, | (5.4)

Also for α, β scalars, the following additional properties hold.

Distributive law over Matrix Addition.

| α (A + B) = αA + αB, | (5.5)

Distributive law over Scalar Addition

| (α + β) A = αA + βA, | (5.6)

Associative law for Scalar Multiplication

| α (βA) = αβ (A) , | (5.7)

Rule for Multiplication by 1.

| 1A = A. | (5.8)

As an example, consider the Commutative Law of Addition. Let A + B = C and B + A = D. Why is D = C?

Cij = Aij + Bij = Bij + Aij = Dij.

Therefore, C = D because the ijth entries are the same. Note that the conclusion follows from the commutative law of addition of numbers.

5.1.2 Multiplication Of Matrices

Definition 5.1.8 Matrices which are n × 1 or 1 × n are called vectors and are often denoted by a bold letter. Thus the n × 1 matrix

is also called a column vector. The 1 × n matrix

(x₁· · ·xn)

is called a row vector.

Although the following description of matrix multiplication may seem strange, it is in fact the most important and useful of the matrix operations. To begin with consider the case where a matrix is multiplied by a column vector. First consider a special case.

One way to remember this is as follows. Slide the vector, placing it on top the two rows as shown and then do the indicated operation.

multiply the numbers on the top by the numbers on the bottom and add them up to get a single number for each row of the matrix as shown above.

In more general terms,

Another way to think of this is

Thus you take x₁ times the first column, add to x₂ times the second column, and finally x₃ times the third column. In general, here is the definition of how to multiply an (m × n) matrix times a (n × 1) matrix.

Definition 5.1.9 Let A = Aij be an m × n matrix and let v be an n × 1 matrix,

Then Av is an m× 1 matrix and the ith component of this matrix is

Thus

| [picture] | (5.9)

In other words, if

A = (a₁,· · ·, an) 

where the ak are the columns,

This follows from (5.9) and the observation that the jth column of A is  

so (5.9) reduces to

Note also that multiplication by an m × n matrix takes an n × 1 matrix, and produces an m × 1 matrix.

Here is another example. 

Example 5.1.10 Compute

First of all this is of the form (3 × 4) (4 × 1) and so the result should be a (3 × 1) . Note how the inside numbers cancel. To get the element in the second row and first and only column, compute

You should do the rest of the problem and verify 

The next task is to multiply an m × n matrix times an n × p matrix. Before doing so, the following may be helpful.

For A and B matrices, in order to form the product, AB the number of columns of A must equal the number of rows of B.

Note the two outside numbers give the size of the product. Remember:

If the two middle numbers don't match, you can't multiply the matrices!

Definition 5.1.11 When the number of columns of A equals the number of rows of B the two matrices are said to be conformable and the product, AB is obtained as follows. Let A be an m × n matrix and let B be an n × p matrix. Then B is of the form

B = (b₁,· · ·, bp)

where bk is an n × 1 matrix or column vector. Then the m × p matrix AB is defined as follows:

| AB ≡ (Ab₁,· · ·, Abp) | (5.10)

where Abk is an m × 1 matrix or column vector which gives the kth column of AB.

Example 5.1.12 Multiply the following.

The first thing you need to check before doing anything else is whether it is possible to do the multiplication. The first matrix is a 2 × 3 and the second matrix is a 3 × 3. Therefore, is it possible to multiply these matrices. According to the above discussion it should be a 2 × 3 matrix of the form

You know how to multiply a matrix times a vector and so you do so to obtain each of the three columns. Thus

Example 5.1.13 Multiply the matrices.

First check if it is possible. This is of the form (3 × 3) (2 × 3).The inside numbers do not match therefore multiplying the matricies is impossible. Aren’t they the same two matrices considered in the previous example? Yes they are. It is just that here they are in a different order. This shows something you must always remember about matrix multiplication.

This is very different than multiplication of numbers!

5.1.3 The ijth Entry Of A Product

It is important to describe matrix multiplication in terms of entries of the matrices. What is the ijth entry of AB? It would be the ith entry of the jth column of AB. Thus it would be the ith entry of Abj. Now

and from the above definition, the ith entry is

| [picture] | (5.11)

In terms of pictures of the matrix, you are doing

Then as explained above, the jth column is of the form

which is a m × 1 matrix or column vector which equals

The second entry of this m × 1 matrix is

Similarly, the ith entry of this m × 1 matrix is

This shows the following definition for matrix multiplication in terms of the ijth entries of the product coincides with Definition 5.1.11.

Definition 5.1.14 Let A = (Aij) be an m × n matrix and let B = (Bij) be an n × p matrix. Then AB is an m × p matrix and

| [picture] | (5.12)

Another way to write this is

Note that to get (AB)ij you involve the ith row of A and the jth column of B. Specifically, the ijth entry of AB is the dot product of the ith row of A with the jth column of B. This is what the formula in (5.12) says. (Note that here the dot product does not involve taking conjugates.)

Example 5.1.15 Multiply if possible

First check to see if this is possible. It is of the form (3 × 2) (2 × 3) and since the inside numbers match, the two matrices are conformable and it is possible to do the multiplication. The result should be a 3 × 3 matrix. The answer is of the form

where the commas separate the columns in the resulting product. Thus the above product equals

a 3 × 3 matrix as desired. In terms of the ijth entries and the above definition, the entry in the third row and second column of the product should equal

You should try a few more such examples to verify the above definition in terms of the ijth entries works for other entries.

Example 5.1.16 Multiply if possible

This is not possible because it is of the form (3 × 2) (3 × 3) and the middle numbers don’t match. In other words the two matrices are not conformable in the indicated order.

Example 5.1.17 Multiply if possible

This is possible because in this case it is of the form (3 × 3) (3 × 2) and the middle numbers do match so the matrices are conformable. When the multiplication is done it equals 

Check this and be sure you come up with the same answer.

Example 5.1.18 Multiply if possible 

In this case you are trying to do (3 × 1) (1 × 4) . The inside numbers match so you can do it. Verify

5.1.4 Properties Of Matrix Multiplication

As pointed out above, sometimes it is possible to multiply matrices in one order but not in the other order. What if it makes sense to multiply them in either order? Will the two products be equal then?

Example 5.1.19 Compare

The first product is

The second product is

You see these are not equal. Again you cannot conclude that AB = BA for matrix multiplication even when multiplication is defined in both orders. However, there are some properties which do hold.

Proposition 5.1.20 If all multiplications and additions make sense, the following hold for matrices, A, B, C and a, b scalars.

| A (aB + bC) = a (AB) + b (AC) (B + C) A = BA + CA A (BC) = (AB) C | (5.13) (5.14) (5.15)

Proof: Using Definition 5.1.14,

Thus A (B + C) = AB + AC as claimed. Formula (5.14) is entirely similar.

Formula (5.15) is the associative law of multiplication. Using Definition 5.1.14,

This proves (5.15).

5.1.5 The Transpose

Another important operation on matrices is that of taking the transpose. The following example shows what is meant by this operation, denoted by placing a T as an exponent on the matrix.

What happened? The first column became the first row and the second column became the second row. Thus the 3 × 2 matrix became a 2 × 3 matrix. The number 3 was in the second row and the first column and it ended up in the first row and second column. Here is the definition.

Definition 5.1.21 Let A be an m × n matrix. Then AT denotes the n × m matrix which is defined as follows.

(AT)ij = Aji

Example 5.1.22

The transpose of a matrix has the following important properties.

Lemma 5.1.23 Let A be an m x n matrix and let B be a n x p matrix. then

| (AB)T = BTAT | (5.16)

and if α and β are scalars,

| (αA + βB)T = αAT + βBT  | (5.17)

Proof: From the definition,

The proof of Formula (5.17) is left as an exercise.

Definition 5.1.24 An n × n matrix A is said to be symmetric if A = AT. It is said to be skew symmetric if A = −AT.

Example 5.1.25 Let

Then A is symmetric.

Example 5.1.26 Let

Then A is skew symmetric.

5.1.6 The Identity And Inverses

There is a special matrix called I and referred to as the identity matrix. It is always a square matrix, meaning the number of rows equals the number of columns and it has the property that there are ones down the main diagonal and zeroes elsewhere. Here are some identity matrices of various sizes.

The first is the 1 × 1 identity matrix, the second is the 2 × 2 identity matrix, the third is the 3 × 3 identity matrix, and the fourth is the 4 × 4 identity matrix. By extension, you can likely see what the n × n identity matrix would be. It is so important that there is a special symbol to denote the ijth entry of the identity matrix

Iij = δij

where δij is the Kronecker symbol defined by

It is called the identity matrix because it is a multiplicative identity in the following sense.

Lemma 5.1.27 Suppose A is an m × n matrix and In is the n × n identity matrix. Then AIn = A. If Im is the m × m identity matrix, it also follows that ImA = A.

Proof:

and so AIn = A. The other case is left as an exercise for you.

Definition 5.1.28 An n × n matrix A has an inverse, A⁻¹ if and only if AA⁻¹ = A⁻¹A = I. Such a matrix is called invertible.

It is very important to observe that the inverse of a matrix, if it exists, is unique. Another way to think of this is that if it acts like the inverse, then it is the inverse.

Theorem 5.1.29 Suppose A⁻¹ exists and AB = BA = I. Then B = A⁻¹.

Proof:

A⁻¹ = A⁻¹I = A⁻¹ (AB) = (A⁻¹A)B = IB = B.

Unlike ordinary multiplication of numbers, it can happen that A ≠ 0 but A may fail to have an inverse. This is illustrated in the following example.

Example 5.1.30 Let A = Does A have an inverse?

One might think A would have an inverse because it does not equal zero. However,

and if A⁻¹ existed, this could not happen because you could write

a contradiction. Thus the answer is that A does not have an inverse. 

Example 5.1.31 Let A = . Show is the inverse of A.

To check this, multiply

and

showing that this matrix is indeed the inverse of A.

5.1.7 Finding The Inverse Of A Matrix

In the last example, how would you find A⁻¹? You wish to find a matrix  such that

This requires the solution of the systems of equations,

x + y = 1, x + 2y = 0

and

z + w = 0, z + 2w = 1.

Writing the augmented matrix for these two systems gives

| [picture] | (5.18)

for the first system and

| [picture] | (5.19)

for the second. Lets solve the first system. Take (−1) times the first row and add to the second to get

Now take (−1) times the second row and add to the first to get

Putting in the variables, this says x = 2 and y = −1.

Now solve the second system, (5.19) to find z and w. Take (−1) times the first row and add to the second to get

Now take (−1) times the second row and add to the first to get

Putting in the variables, this says z = −1 and w = 1. Therefore, the inverse is

Didn’t the above seem rather repetitive? Note that exactly the same row operations were used in both systems. In each case, the end result was something of the form (I|v) where I is the identity and v gave a column of the inverse. In the above, , the first column of the inverse was obtained first and then the second column

To simplify this procedure, you could have written

and row reduced till you obtained

and read off the inverse as the 2 × 2 matrix on the right side.

This is the reason for the following simple procedure for finding the inverse of a matrix. This procedure is called the Gauss-Jordan procedure.

Procedure 5.1.32 Suppose A is an n × n matrix. To find A⁻¹ if it exists, form the augmented n × 2n matrix

(A|I)

and then, if possible do row operations until you obtain an n× 2n matrix of the form

| (I|B) . | (5.20)

When this has been done, B = A⁻¹. If it is impossible to row reduce to a matrix of the form (I|B) , then A has no inverse.

Actually, all this shows is how to find a right inverse if it exists. Later, I will show that this right inverse is the inverse. See Corollary 7.2.15 or Theorem 8.2.11 presented later.

Example 5.1.33 Find A⁻¹ if it exists.

Set up the augmented matrix (A|I)

Next take (−1) times the first row and add to the second followed by (−3) times the first row added to the last. This yields

Then take 5 times the second row and add to -2 times the last row.

Next take the last row and add to (−7) times the top row. This yields

Now take (−7/5) times the second row and add to the top.

Finally divide the top row by -7, the second row by -10 and the bottom row by 14 which yields 

Therefore, the inverse is

Example 5.1.34 Find A⁻¹ if it exists.

Write the augmented matrix (A|I)

and proceed to do row operations attempting to obtain (I|A⁻¹). Take (−1) times the top row and add to the second. Then take (−2) times the top row and add to the bottom.

Next add (−1) times the second row to the bottom row.

At this point, you can see there will be no inverse because you have obtained a row of zeros in the left half of the augmented matrix (A|I) . Thus there will be no way to obtain I on the left.

Example 5.1.35 Find A⁻¹ if it exists.

Form the augmented matrix

Now do row operations until the n × n matrix on the left becomes the identity matrix. This yields after some computations,

and so the inverse of A is the matrix on the right, 

Checking the answer is easy. Just multiply the matrices and see if it works.

Always check your answer because if you are like some of us, you will usually have made a mistake.

Example 5.1.36 In this example, it is shown how to use the inverse of a matrix to find the solution to a system of equations. Consider the following system of equations. Use the inverse of a suitable matrix to give the solutions to this system.

The system of equations can be written in terms of matrices as

| [picture] | (5.21)

More simply, this is of the form Ax = b. Suppose you find the inverse of the matrix A⁻¹. Then you could multiply both sides of this equation by A⁻¹ to obtain

This gives the solution as x = A⁻¹b. Note that once you have found the inverse, you can easily get the solution for different right hand sides without any effort. It is always just A⁻¹b. In the given example, the inverse of the matrix is

This was shown in Example 5.1.35. Therefore, from what was just explained, the solution to the given system is

What if the right side of (5.21) had been

What would be the solution to

By the above discussion, it is just

This illustrates why once you have found the inverse of a given matrix, you can use it to solve many different systems easily.

5.2 Exercises

Here are some matrices:

Find if possible −3A, 3B  − A, AC, CB, AE, EA. If it is not possible explain why.

Here are some matrices:

Find if possible −3A, 3B  − A, AC, CA, AE, EA, BE, DE. If it is not possible explain why.

Here are some matrices:

Find if possible −3AT, 3B −AT, AC, CA, AE, ETB, BE, DE, EET, ETE. If it is not possible explain why.

Here are some matrices:

Find the following if possible and explain why it is not possible if this is the case.

AD, DA, DTB, DTBE, ETD, DET.

Let and

Find if possible.

(a) AB
(b) BA
(c) AC
(d) CA
(e) CB
(f) BC

Suppose A and B are square matrices of the same size. Which of the following are correct?

(a) (A − B)² = A² − 2AB + B²
(b) (AB)² = A²B²
(c) (A + B)² = A² + 2AB + B²
(d) (A + B)² = A² + AB + BA + B²
(e) A²B² = A (AB) B
(f) (A + B)³ = A³ + 3A²B + 3AB² + B³
(g) (A + B) (A − B) = A² − B²

Let Find 2 x 2 matrices, B such that AB = 0.

Let x = (−1,−1, 1) and y = (0, 1, 2) . Find xTy and xyT if possible.

Let Is it possible to choose k such that AB = BA? If so, what should k equal?

Let Is it possible to choose k such that AB = BA? If so, what should k equal?

In (5.1) - (5.8) describe −A and 0.

Let A be an n × n matrix. Show A equals the sum of a symmetric and a skew symmetric matrix. (M is skew symmetric if M = −MT. M is symmetric if MT = M.) Hint: Show that 1/2 (AT +  A) is symmetric and then consider using this as one of the matrices.

Show every skew symmetric matrix has all zeros down the main diagonal. The main diagonal consists of every entry of the matrix which is of the form aii. It runs from the upper left down to the lower right.

Suppose M is a 3 × 3 skew symmetric matrix. Show there exists a vector Ω such that for all u ∈ R³

M u = Ω × u

Hint: Explain why, since M is skew symmetric it is of the form

where the ωi are numbers. Then consider ω₁i + ω₂j + ω₃k.

Using only the properties (5.1) - (5.8) show −A is unique.

Using only the properties (5.1) - (5.8) show 0 is unique.

Using only the properties (5.1) - (5.8) show 0A = 0. Here the 0 on the left is the scalar 0 and the 0 on the right is the zero for m × n matrices.

Using only the properties (5.1) - (5.8) and previous problems show ( −1) A = −A.

Prove (5.17).

Prove that ImA = A where A is an m × n matrix.

Give an example of matrices, A, B, C such that B ≠ C, A ≠ 0, and yet AB = AC.

Suppose AB = AC and A is an invertible n × n matrix. Does it follow that B = C? Explain why or why not. What if A were a non invertible n × n matrix?

Find your own examples:

(a) 2 × 2 matrices, A and B such that A ≠ 0, B ≠ 0 with AB ≠ BA.
(b) 2 × 2 matrices, A and B such that A ≠ 0, B ≠ 0, but AB = 0.
(c) 2 × 2 matrices, A, D, and C such that A ≠ 0, C ≠ D, but AC = AD.

Explain why if AB = AC and A⁻¹ exists, then B = C.

Give an example of a matrix A such that A² = I and yet A ≠ I and A ≠ −I.

Give an example of matrices, A, B such that neither A nor B equals zero and yet AB = 0.

Give another example other than the one given in this section of two square matrices, A and B such that AB ≠ BA.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let A be a 2 × 2 matrix which has an inverse. Say   Find a formula for A⁻¹ in terms of a, b, c, d.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. If A⁻¹ does not exist, determine why.

Let

Find A⁻¹ if possible. A⁻¹ does not exist, determine why.

Write in the form where A is an appropriate matrix.

Write in the form where A is an appropriate matrix.

Write in the form where A is an appropriate matrix.

Using the inverse of the matrix, find the solution to the systems

Now give the solution in terms of a, b, and c to

Using the inverse of the matrix, find the solution to the systems

 Now give the solution in terms of a, b, and c to

Using the inverse of the matrix, find the solution to the system

Show that if A is an n × n invertible matrix and x is a n × 1 matrix such that A x = b for b an n × 1 matrix, then x = A⁻¹b.

Prove that if A⁻¹ exists and Ax = 0 then x = 0.

Show that if A⁻¹ exists for an n × n matrix, then it is unique. That is, if BA = I and AB = I, then B = A⁻¹.

Show that if A is an invertible n × n matrix, then so is AT and (AT)−1 = (A⁻¹)T.

Show (AB)⁻¹ = B⁻¹A⁻¹ by verifying that AB (B⁻¹A⁻¹) = I and

B⁻¹A⁻¹ (AB) = I. 

Hint: Use Problem 45.

Show that (ABC)⁻¹ = C⁻¹B⁻¹A⁻¹ by verifying that

(ABC)(C⁻¹B⁻¹A⁻¹) = I

and (C⁻¹B⁻¹A⁻¹) (ABC) = I. Hint: Use Problem 45.

If A is invertible, show (A²)⁻¹ = (A⁻¹)². Hint: Use Problem 45.

If A is invertible, show (A⁻¹)⁻¹ = A. Hint: Use Problem 45.

Let A and be a real m × n matrix and let x ∈ Rn and y ∈ Rm. Show (Ax, y)Rm = (x,ATy)Rn where (·,·)Rk denotes the dot product in Rk. In the notation above, Ax·y = x·ATy. Use the definition of matrix multiplication to do this.

Use the result of Problem 51 to verify directly that (AB)T = BTAT without making any reference to subscripts.

A matrix A is called a projection if A² = A. Here is a matrix.

Show that this is a projection. Show that a vector in the column space of a projection matrix is left unchanged by multiplication by A.

Determinants

6.1 Basic Techniques And Properties

6.1.1 Cofactors And 2 × 2 Determinants

Let A be an n × n matrix. The determinant of A, denoted as det (A) is a number. If the matrix is a 2 × 2 matrix, this number is very easy to find.

Deﬁnition 6.1.1 Let Then

det (A) ≡ ad − cb.

The determinant is also often denoted by enclosing the matrix with two vertical lines. Thus

Example 6.1.2 Find det

From the deﬁnition this is just (2) (6)− (−1) (4) = 16.

Having deﬁned what is meant by the determinant of a 2 × 2 matrix, what about a 3 × 3 matrix?

Deﬁnition 6.1.3 Suppose A is a 3 × 3 matrix. The ijth minor, denoted as minor(A)ij , is the determinant of the 2 × 2 matrix which results from deleting the ith row and the jth column.

Example 6.1.4 Consider the matrix

The (1, 2) minor is the determinant of the 2 × 2 matrix which results when you delete the first row and the second column. This minor is therefore

The (2, 3) minor is the determinant of the 2 × 2 matrix which results when you delete the second row and the third column. This minor is therefore

Deﬁnition 6.1.5 Suppose A is a 3 × 3 matrix. The ijth cofactor is deﬁned to be (−1)i+j × (ijth minor). In words, you multiply (−1)i + j times the ijth minor to get the ijth cofactor. The cofactors of a matrix are so important that special notation is appropriate when referring to them. The ijth cofactor of a matrix A will be denoted by cof (A)ij. It is also convenient to refer to the cofactor of an entry of a matrix as follows. For aij an entry of the matrix, its cofactor is just cof (A)ij. Thus the cofactor of the ijth entry is just the ijth cofactor.

Example 6.1.6 Consider the matrix

The (1, 2) minor is the determinant of the 2 × 2 matrix which results when you delete the first row and the second column. This minor is therefore

It follows

The (2, 3) minor is the determinant of the 2 × 2 matrix which results when you delete the second row and the third column. This minor is therefore

Therefore,

Similarly,

Deﬁnition 6.1.7 The determinant of a 3 × 3 matrix A, is obtained by picking a row (column) and taking the product of each entry in that row (column) with its cofactor and adding these up. This process when applied to the ith row (column) is known as expanding the determinant along the ith row (column).

Example 6.1.8 Find the determinant of

Here is how it is done by “expanding along the first column”.

You see, we just followed the rule in the above deﬁnition. We took the 1 in the first column and multiplied it by its cofactor, the 4 in the first column and multiplied it by its cofactor, and the 3 in the first column and multiplied it by its cofactor. Then we added these numbers together.

You could also expand the determinant along the second row as follows.

Observe this gives the same number. You should try expanding along other rows and columns. If you don’t make any mistakes, you will always get the same answer.

What about a 4 × 4 matrix? You know now how to find the determinant of a 3 × 3 matrix. The pattern is the same.

Deﬁnition 6.1.9 Suppose A is a 4 × 4 matrix. The ijth minor is the determinant of the 3 × 3 matrix you obtain when you delete the ith row and the jth column. The ijth cofactor, cof (A)ij is deﬁned to be (−1)i+j × (ijth minor). In words, you multiply (−1)i+j times the ijth minor to get the ijth cofactor.

Deﬁnition 6.1.10 The determinant of a 4 × 4 matrix A, is obtained by picking a row (column) and taking the product of each entry in that row (column) with its cofactor and adding these together. This process when applied to the ith row (column) is known as expanding the determinant along the ith row (column).

Example 6.1.11 Find det (A) where

As in the case of a 3 × 3 matrix, you can expand this along any row or column. Lets pick the third column. det (A) =

Now you know how to expand each of these 3 × 3 matrices along a row or a column. If you do so, you will get −12 assuming you make no mistakes. You could expand this matrix along any row or any column and assuming you make no mistakes, you will always get the same thing which is deﬁned to be the determinant of the matrix A. This method of evaluating a determinant by expanding along a row or a column is called the method of Laplace expansion.

Note that each of the four terms above involves three terms consisting of determinants of 2 × 2 matrices and each of these will need 2 terms. Therefore, there will be 4 × 3 × 2 = 24 terms to evaluate in order to find the determinant using the method of Laplace expansion. Suppose now you have a 10 × 10 matrix and you follow the above pattern for evaluating determinants. By analogy to the above, there will be 10! = 3, 628 , 800 terms involved in the evaluation of such a determinant by Laplace expansion along a row or column. This is a lot of terms.

In addition to the difficulties just discussed, you should regard the above claim that you always get the same answer by picking any row or column with considerable skepticism. It is incredible and not at all obvious. However, it requires a little effort to establish it. This is done in the section on the theory of the determinant.

Deﬁnition 6.1.12 Let A = (aij) be an n × n matrix and suppose the determinant of a (n − 1) × (n − 1) matrix has been deﬁned. Then a new matrix called the cofactor matrix, cof (A) is deﬁned by cof (A) = (cij) where to obtain cij delete the ith row and the jth column of A, take the determinant of the (n − 1) × (n − 1) matrix which results, (This is called the ijth minor of A.) and then multiply this number by (−1)i+j. Thus (−1)i+j × (the ijth minor) equals the ijth cofactor. To make the formulas easier to remember, cof (A)ij will denote the ijth entry of the cofactor matrix.

With this deﬁnition of the cofactor matrix, here is how to deﬁne the determinant of an n × n matrix.

Deﬁnition 6.1.13 Let A be an n × n matrix where n ≥ 2 and suppose the determinant of an (n− 1) × (n− 1) has been deﬁned. Then

| [picture] | (6.1)

The first formula consists of expanding the determinant along the ith row and the second expands the determinant along the jth column.

Theorem 6.1.14 Expanding the n × n matrix along any row or column always gives the same answer so the above deﬁnition is a good deﬁnition.

6.1.2 The Determinant Of A Triangular Matrix

Notwithstanding the difficulties involved in using the method of Laplace expansion, certain types of matrices are very easy to deal with.

Deﬁnition 6.1.15 A matrix M, is upper triangular if Mij = 0 whenever i > j. Thus such a matrix equals zero below the main diagonal, the entries of the form Mii, as shown.

A lower triangular matrix is deﬁned similarly as a matrix for which all entries above the main diagonal are equal to zero.

You should verify the following using the above theorem on Laplace expansion.

Corollary 6.1.16 Let M be an upper (lower) triangular matrix. Then det (M) is obtained by taking the product of the entries on the main diagonal.

Example 6.1.17 Let

Find det (A).

From the above corollary, it suffices to take the product of the diagonal elements. Thus det (A) = 1 × 2 × 3 × (−1) = −6. Without using the corollary, you could expand along the first column. This gives

and the only nonzero term in the expansion is

Now expand this along the first column to obtain

Next expand this last determinant along the first column to obtain the above equals

which is just the product of the entries down the main diagonal of the original matrix.

6.1.3 Properties Of Determinants

There are many properties satisﬁed by determinants. Some of these properties have to do with row operations. Recall the row operations.

Deﬁnition 6.1.18 The row operations consist of the following

Switch two rows.

Multiply a row by a nonzero number.

Replace a row by a multiple of another row added to itself.

Theorem 6.1.19 Let A be an n × n matrix and let A₁ be a matrix which results from multiplying some row of A by a scalar c. Then c det (A) = det (A₁).

Example 6.1.20 Let

Theorem 6.1.21 Let A be an n × n matrix and let A₁ be a matrix which results from switching two rows of A. Then det (A) = −det (A₁) . Also, if one row of A is a multiple of another row of A, then det (A) = 0.

Example 6.1.22 Let

Theorem 6.1.23 Let A be an n × n matrix and let A₁ be a matrix which results from applying row operation 3. That is you replace some row by a multiple of another row added to itself. Then det (A) = det (A₁).

Example 6.1.24 Let and let Thus the second row of A₁ is one times the first row added to the second row. det (A) = −2 and det (A₁) = −2.

Theorem 6.1.25 In Theorems 6.1.19 - 6.1.23 you can replace the word, “row” with the word “column”.

There are two other major properties of determinants which do not involve row operations.

Theorem 6.1.26 Let A and B be two n × n matrices. Then

Also,

Example 6.1.27 Compare det (AB) and det (A) det (B) for

First

and so

Now

and

Thus det (A) det (B) = 8 × (−5) = −40.

6.1.4 Finding Determinants Using Row Operations

Theorems 6.1.23 - 6.1.25 can be used to find determinants using row operations. As pointed out above, the method of Laplace expansion will not be practical for any matrix of large size. Here is an example in which all the row operations are used.

Example 6.1.28 Find the determinant of the matrix

Replace the second row by ( −5) times the first row added to it. Then replace the third row by (−4) times the first row added to it. Finally, replace the fourth row by (−2) times the first row added to it. This yields the matrix

and from Theorem 6.1.23, it has the same determinant as A. Now using other row operations, det (B) = (⁻¹⁄₃) det (C) where

The second row was replaced by (−3) times the third row added to the second row. By Theorem 6.1.23 this didn’t change the value of the determinant. Then the last row was multiplied by (−3) . By Theorem 6.1.19 the resulting matrix has a determinant which is (−3) times the determinant of the un-multiplied matrix. Therefore, we multiplied by −1/3 to retain the correct value. Now replace the last row with 2 times the third added to it. This does not change the value of the determinant by Theorem 6.1.23. Finally switch the third and second rows. This causes the determinant to be multiplied by (−1) . Thus det (C) = −det (D) where

You could do more row operations or you could note that this can be easily expanded along the first column followed by expanding the 3 × 3 matrix which results along its first column. Thus

and so det (C) = −1485 and det (A) = det (B) =  (−1485) = 495.

Example 6.1.29 Find the determinant of the matrix

Replace the second row by (−1) times the first row added to it. Next take −2 times the first row and add to the third and finally take −3 times the first row and add to the last row. This yields

By Theorem 6.1.23 this matrix has the same determinant as the original matrix. Remember you can work with the columns also. Take −5 times the last column and add to the second column. This yields

By Theorem 6.1.25 this matrix has the same determinant as the original matrix. Now take (−1) times the third row and add to the top row. This gives.

which by Theorem 6.1.23 has the same determinant as the original matrix. Lets expand it now along the first column. This yields the following for the determinant of the original matrix.

which equals

We suggest you do not try to be fancy in using row operations. That is, stick mostly to the one which replaces a row or column with a multiple of another row or column added to it. Also note there is no way to check your answer other than working the problem more than one way. To be sure you have gotten it right you must do this.

6.2 Applications

6.2.1 A Formula For The Inverse

The deﬁnition of the determinant in terms of Laplace expansion along a row or column also provides a way to give a formula for the inverse of a matrix. Recall the deﬁnition of the inverse of a matrix in Deﬁnition 5.1.28 on Page 87. Also recall the deﬁnition of the cofactor matrix given in Deﬁnition 6.1.12 on Page 102. This cofactor matrix was just the matrix which results from replacing the ijth entry of the matrix with the ijth cofactor.

The following theorem says that to find the inverse, take the transpose of the cofactor matrix and divide by the determinant. The transpose of the cofactor matrix is called the adjugate or sometimes the classical adjoint of the matrix A. In other words, A⁻¹ is equal to one divided by the determinant of A times the adjugate matrix of A. This is what the following theorem says with more precision.

Theorem 6.2.1 A⁻¹ exists if and only if det (A) ≠ 0. If det(A) ≠ 0, then A⁻¹ = (aij⁻¹) where

aij⁻¹ = det(A)⁻¹ cof (A)ji

for cof (A)ij the ijth cofactor of A.

Example 6.2.2 Find the inverse of the matrix

First find the determinant of this matrix. Using Theorems 6.1.23 - 6.1.25 on Page 104, the determinant of this matrix equals the determinant of the matrix

which equals 12. The cofactor matrix of A is

Each entry of A was replaced by its cofactor. Therefore, from the above theorem, the inverse of A should equal

Does it work? You should check to see if it does. When the matrices are multiplied

and so it is correct.

Example 6.2.3 Find the inverse of the matrix

First find its determinant. This determinant is  The inverse is therefore equal to

Expanding all the 2 × 2 determinants this yields

Always check your work.

and so we got it right. If the result of multiplying these matrices had been something other than the identity matrix, you would know there was an error. When this happens, you need to search for the mistake if you are interested in getting the right answer. A common mistake is to forget to take the transpose of the cofactor matrix.

Proof of Theorem 6.2.1: From the deﬁnition of the determinant in terms of expansion along a column, and letting (air) = A, if det (A) ≠ 0,

Now consider

when k ≠ r. Replace the kth column with the rth column to obtain a matrix Bk whose determinant equals zero by Theorem 6.1.21. However, expanding this matrix Bk along the kth column yields

Summarizing,

Now

which is the krth entry of cof (A)TA. Therefore,

| [picture] | (6.2)

Using the other formula in Deﬁnition 6.1.13, and similar reasoning,

Now

which is the rkth entry of A cof (A)T. Therefore,

| [picture] | (6.3)

and it follows from (6.2) and (6.3) that A⁻¹ = (aij⁻¹), where

aij⁻¹ = cof (A)ji det (A)⁻¹.

In other words,

Now suppose A⁻¹ exists. Then by Theorem 6.1.26,

1 = det (I) = det (AA⁻¹) = det (A) det (A⁻¹)

so det (A) ≠ 0.

This way of finding inverses is especially useful in the case where it is desired to find the inverse of a matrix whose entries are functions.

Example 6.2.4 Suppose

Show that A (t)⁻¹ exists and then find it.

First note det (A (t)) = et ≠ 0 A (t)−1 exists. The cofactor matrix is

and so the inverse is

6.2.2 Cramer’s Rule

This formula for the inverse also implies a famous procedure known as Cramer’s rule. Cramer’s rule gives a formula for the solutions, x, to a system of equations, Ax = y in the special case that A is a square matrix. Note this rule does not apply if you have a system of equations in which there is a different number of equations than variables.

In case you are solving a system of equations, Ax = y for x, it follows that if A⁻¹ exists,

x =(A⁻¹A) x = A⁻¹ (Ax) = A⁻¹y

thus solving the system. Now in the case that A⁻¹ exists, there is a formula for A⁻¹ given above. Using this formula,

By the formula for the expansion of a determinant along a column,

where here the ith column of A is replaced with the column vector (y₁· · · ·, yn)T, and the determinant of this modiﬁed matrix is taken and divided by det (A). This formula is known as Cramer’s rule.

Procedure 6.2.5 Suppose A is an n × n matrix and it is desired to solve the system Ax = y, y = (y₁,· · ·, yn)T for x = (x₁,· · ·, xn)T. Then Cramer’s rule says

where Ai is obtained from A by replacing the ith column of A with the column

(y₁,· · ·, yn)T.

Example 6.2.6 Find x, y if

From Cramer’s rule,

Now to find y,

You see the pattern. For large systems Cramer’s rule is less than useful if you want to find an answer. This is because to use it you must evaluate determinants. However, you have no practical way to evaluate determinants for large matrices other than row operations and if you are using row operations, you might just as well use them to solve the system to begin with. It will be a lot less trouble. Nevertheless, there are situations in which Cramer’s rule is useful.

Example 6.2.7 Solve for z if

You could do it by row operations but it might be easier in this case to use Cramer’s rule because the matrix of coefficients does not consist of numbers but of functions. Thus

You end up doing this sort of thing sometimes in ordinary differential equations in the method of variation of parameters.

6.3 Exercises

Find the determinants of the following matrices.

Find the following determinant by expanding along the first row and second column.

Find the following determinant by expanding along the first column and third row.

Find the following determinant by expanding along the second row and first column.

Compute the determinant by cofactor expansion. Pick the easiest row or column to use.

Find the determinant using row operations.

Find the determinant using row operations.

Find the determinant using row operations.

Find the determinant using row operations.

Verify an example of each property of determinants found in Theorems 6.1.23 - 6.1.25 for 2 × 2 matrices.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.

Let A be an r × r matrix and suppose there are r − 1 rows (columns) such that all rows (columns) are linear combinations of these r − 1 rows (columns). Show det (A) = 0.

Show det (aA) = an det (A) where here A is an n × n matrix and a is a scalar.

Illustrate with an example of 2 × 2 matrices that the determinant of a product equals the product of the determinants.

Is it true that det (A + B) = det (A) + det (B)? If this is so, explain why it is so and if it is not so, give a counter example.

An n × n matrix is called nilpotent if for some positive integer, k it follows Ak = 0. If A is a nilpotent matrix and k is the smallest possible integer such that Ak = 0, what are the possible values of det (A)?

A matrix is said to be orthogonal if ATA = I. Thus the inverse of an orthogonal matrix is just its transpose. What are the possible values of det (A) if A is an orthogonal matrix?

Fill in the missing entries to make the matrix orthogonal as in Problem 21.

Let A and B be two n × n matrices. A ∼ B (A is similar to B) means there exists an invertible matrix S such that A = S⁻¹BS. Show that if A ∼ B, then B ∼ A. Show also that A ∼ A and that if A ∼ B and B ∼ C, then A ∼ C.

In the context of Problem 23 show that if A ∼ B, then det (A) = det (B) .

Two n × n matrices, A and B, are similar if B = S⁻¹AS for some invertible n × n matrix S. Show that if two matrices are similar, they have the same characteristic polynomials. The characteristic polynomial of an n × n matrix M is the polynomial, det (λI − M).

Tell whether the statement is true or false.

(a)   If A is a 3 × 3 matrix with a zero determinant, then one column must be a multiple of some other column.

(b)   If any two columns of a square matrix are equal, then the determinant of the matrix equals zero.

(c)   For A and B two n × n matrices, det (A + B) = det (A) + det (B).

(d)   For A an n × n matrix, det (3A) = 3 det (A)

(e)   If A⁻¹ exists then det (A⁻¹) = det (A)⁻¹.

(f)    If B is obtained by multiplying a single row of A by 4 then det (B) = 4 det (A).

(g)   For A an n × n matrix, det (−A) = (−1)n det (A).

(h)   If A is a real n × n matrix, then det (ATA) ≥ 0.

(i)   Cramer’s rule is useful for finding solutions to systems of linear equations in which there is an inﬁnite set of solutions.

(j)    If Ak = 0 for some positive integer, k, then det (A) = 0.

(k)   If Ax = 0 for some x ≠ 0, then det (A) = 0.

Use Cramer’s rule to find the solution to

Use Cramer’s rule to find the solution to

Here is a matrix,

Determine whether the matrix has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Here is a matrix,

Determine whether the matrix has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Here is a matrix,

Determine whether the matrix has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Here is a matrix,

Determine whether the matrix has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Here is a matrix,

Determine whether the matrix has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Use the formula for the inverse in terms of the cofactor matrix to find if possible the inverses of the matrices

If the inverse does not exist, explain why.

Here is a matrix,

Does there exist a value of t for which this matrix fails to have an inverse? Explain.

Here is a matrix,

Does there exist a value of t for which this matrix fails to have an inverse? Explain.

Here is a matrix,

Does there exist a value of t for which this matrix fails to have an inverse? Explain.

Show that if det (A) ≠ for A an n × n matrix, it follows that if Ax = 0, then x = 0.

Suppose A, B are n × n matrices and that AB = I. Show that then BA = I. Hint: You might do something like this: First explain why det (A) , det (B) are both nonzero. Then (AB) A = A and then show BA (BA−I) = 0. From this use what is given to conclude A (BA−I) = 0. Then use Problem 38.

Use the formula for the inverse in terms of the cofactor matrix to find the inverse of the matrix

Find the inverse if it exists of the matrix

Here is a matrix,

Does there exist a value of t for which this matrix fails to have an inverse? Explain.

Suppose A is an upper triangular matrix. Show that A⁻¹ exists if and only if all elements of the main diagonal are non zero. Is it true that A⁻¹ will also be upper triangular? Explain. Is everything the same for lower triangular matrices?

If A, B, and C are each n × n matrices and ABC is invertible, why are each of A, B, and C invertible.

Let Verify

Now suppose

Use Laplace expansion and the first part to verify F′ (t) =

Conjecture a general result valid for n × n matrices and explain why it will be true. Can a similar thing be done with the columns?

Let Ly = y(n) + an₋₁ (x) y(n−1) + · · · + a₁ (x) y′ + a₀ (x) y where the ai are given continuous functions deﬁned on a closed interval, (a, b) and y is some function which has n derivatives so it makes sense to write Ly. Suppose Lyk = 0 for k = 1, 2,· · ·, n. The Wronskian of these functions, yi is deﬁned as

Show that for W (x) = W (y₁,· · ·, yn) (x) to save space,

Now use the differential equation, Ly = 0 which is satisﬁed by each of these functions, yi and properties of determinants presented above to verify that W′ + an₋₁ (x) W = 0. Give an explicit solution of this linear differential equation, Abel’s formula, and use your answer to verify that the Wronskian of these solutions to the equation, Ly = 0 either vanishes identically on (a, b) or never. Hint: To solve the differential equation, let A′ (x) = an₋₁ (x) and multiply both sides of the differential equation by eA(x) and then argue the left side is the derivative of something.

Find the following determinants.

The Mathematical Theory Of Determinants∗

7.1   The Function sgnn

This material is deﬁnitely not for the faint of heart. It is only for people who want to see everything proved. It is a fairly complete and unusually elementary treatment of the subject. There will be some repetition between this section and the earlier section on determinants. The main purpose is to give all the missing proofs. Two books which give a good introduction to determinants are Apostol [1] and Rudin [13]. A recent book which also has a good introduction is Baker [2]. Most linear algebra books do not do an honest job presenting this topic.

It is easiest to give a different deﬁnition of the determinant which is clearly well deﬁned and then prove the earlier one in terms of Laplace expansion. Let (i₁,· · ·, in) be an ordered list of numbers from {1,· · ·, n}. This means the order is important so (1, 2, 3) and (2, 1, 3) are different.

The following Lemma will be essential in the deﬁnition of the determinant.

Lemma 7.1.1 There exists a unique function, sgnn which maps each list of numbers from {1,· · ·, n} to one of the three numbers, 0, 1, or −1 which also has the following properties.

| sgnn (1,· · ·, n) = 1 | (7.1)

| sgnn (i₁,· · ·, p,· · ·, q,· · ·, in) = −sgnn (i₁,· · ·, q,· · ·, p,· · ·, in) | (7.2)

In words, the second property states that if two of the numbers are switched, the value of the function is multiplied by −1. Also, in the case where n > 1 and {i₁,· · ·, in} = {1,· · ·, n} so that every number from {1,· · ·, n} appears in the ordered list, (i₁,· · ·, in) ,

sgnn (i₁,· · ·, i−1, n, i+1,· · ·, in) ≡

| (−1)n− sgnn−1 (i₁,· · ·, i−1, i+1,· · ·, in) | (7.3)

where n = i in the ordered list, (i₁,· · ·, in) .

Proof: To begin with, it is necessary to show the existence of such a function. This is clearly true if n = 1. Deﬁne sgn₁ (1) ≡ 1 and observe that it works. No switching is possible. In the case where n = 2, it is also clearly true. Let sgn₂ (1, 2) = 1 and sgn₂ (2, 1) = 0 while sgn₂ (2, 2) = sgn₂ (1, 1) = 0 and verify it works. Assuming such a function exists for n, sgnn+1 will be deﬁned in terms of sgnn. If there are any repeated numbers in (i₁,· · ·, in+1) , sgnn+1 (i₁,· · ·, in+1) ≡ 0. If there are no repeats, then n + 1 appears somewhere in the ordered list. Let be the position of the number n + 1 in the list. Thus, the list is of the form (i₁,· · ·, i−1, n + 1, i+1,· · ·, in+1) . From (7.3) it must be that

sgnn+1 (i₁,· · ·, i−1, n + 1, i+1,· · ·, in+1) ≡

(−1)n+1− sgnn (i₁,· · ·, i−1, i+1,· · ·, in+1) .

It is necessary to verify this satisﬁes (7.1) and (7.2) with n replaced with n + 1. The first of these is obviously true because

sgnn+1 (1,· · ·, n, n + 1) ≡ (−1)n+1−(n+1) sgnn (1,· · ·, n) = 1.

If there are repeated numbers in ( i₁,· · ·, in+1) , then it is obvious (7.2) holds because both sides would equal zero from the above deﬁnition. It remains to verify (7.2) in the case where there are no numbers repeated in (i₁,· · ·, in+1) . Consider

sgnn+1(i₁,· · ·, rp,· · ·, sq,· · ·, in+1) ,

where the r above the p indicates the number, p is in the rth position and the s above the q indicates that the number q is in the sth position. Suppose first that r < < s. Then

while

and so, by induction, a switch of p and q introduces a minus sign in the result. Similarly, if > s or if < r it also follows that (7.2) holds. The interesting case is when = r or = s. Consider the case where = r and note the other case is entirely similar.

| [picture] | (7.4)

while

| [picture] | (7.5)

By making s−1−r switches, move the q which is in the s−1th position in (7.4) to the rth position in (7.5). By induction, each of these switches introduces a factor of −1 and so

Therefore,

This proves the existence of the desired function.

To see this function is unique, note that you can obtain any ordered list of distinct numbers from a sequence of switches. If there exist two functions, f and g both satisfying (7.1) and (7.2), you could start with f (1,· · ·, n) = g (1,· · ·, n) and applying the same sequence of switches, eventually arrive at f (i₁,· · ·, in) = g (i₁,· · ·, in) . If any numbers are repeated, then (7.2) gives both functions are equal to zero for that ordered list.

In what follows sgn will often be used rather than sgnn because the context supplies the appropriate n.

7.2 The Determinant

Deﬁnition 7.2.1 Let f be a function which has the set of ordered lists of numbers from {1,· · ·, n} as its domain. Deﬁne

to be the sum of all the f (k₁· · ·kn) for all possible choices of ordered lists (k₁,· · ·, kn) of numbers of {1,· · ·, n}. For example,

7.2.1 The Deﬁnition

Deﬁnition 7.2.2 Let (aij) = A denote an n × n matrix. The determinant of A, denoted by det (A) is deﬁned by

where the sum is taken over all ordered lists of numbers from {1,· · ·, n}. Note it suffices to take the sum over only those ordered lists in which there are no repeats because if there are, sgn (k₁,· · ·, kn) = 0 and so that term contributes 0 to the sum.

7.2.2 Permuting Rows Or Columns

Let A be an n × n matrix, A = (aij) and let (r₁,· · ·, rn) denote an ordered list of n numbers from {1,· · ·, n}. Let A (r₁,· · ·, rn) denote the matrix whose kth row is the rk row of the matrix A. Thus

| [picture] | (7.6)

and 

A (1,· · ·, n) = A.

Proposition 7.2.3 Let 

(r₁,· · ·, rn) 

be an ordered list of numbers from {1,· · ·, n}. Then 

sgn (r₁,· · ·, rn) det (A) 

| [picture] | (7.7) (7.8)

Proof: Let (1,· · ·, n) = (1,· · ·, r,· · ·s,· · ·, n) so r < s.

| det (A (1,· · ·, r,· · ·, s,· · ·, n)) = | (7.9)

and renaming the variables, calling ks, kr and kr, ks, this equals

| [picture] | (7.10)

Consequently,

det (A (1,· · ·, s,· · ·, r,· · ·, n)) =

-det (A (1,· · ·, r,· · ·, s,· · ·, n)) = −det (A)

Now letting A (1,· · ·, s,· · ·, r,· · ·, n) play the role of A, and continuing in this way, switching pairs of numbers,

det (A(r₁, · · · , rn)) = (−1)p det (A)

where it took p switches to obtain(r₁,· · ·, rn) from (1,· · ·, n). By Lemma 7.1.1, this implies

det (A (r₁, · · · , rn)) = (-1)p det (A) = sgn (r₁,· · ·, rn) det (A)

and proves the proposition in the case when there are no repeated numbers in the ordered list, (r₁,· · ·, rn). However, if there is a repeat, say the rth row equals the sth row, then the reasoning of (7.9) -(7.10) shows that A (r₁,· · ·, rn) = 0 and also sgn (r₁,· · ·, rn) = 0 so the formula holds in this case also.

Observation 7.2.4  There are n! ordered lists of distinct numbers from {1,· · ·, n}.

To see this, consider n slots placed in order. There are n choices for the first slot. For each of these choices, there are n − 1 choices for the second. Thus there are n (n − 1) ways to fill the first two slots. Then for each of these ways there are n − 2 choices left for the third slot. Continuing this way, there are n! ordered lists of distinct numbers from {1,· · ·, n} as stated in the observation.

7.2.3 A Symmetric Deﬁnition

With the above, it is possible to give a more symmetric description of the determinant from which it will follow that det (A) = det (AT).

Corollary 7.2.5 The following formula for det (A) is valid.

| [picture] | (7.11)

And also det (AT) = det (A) where AT is the transpose of A. (Recall that for )

Proof: From Proposition 7.2.3, if the ri are distinct,

Summing over all ordered lists, (r₁,· · ·, rn) where the ri are distinct, (If the ri are not distinct, sgn (r₁,· · ·, rn) = 0 and so there is no contribution to the sum.)

This proves the corollary since the formula gives the same number for A as it does for AT.

7.2.4 The Alternating Property Of The Determinant

Corollary 7.2.6  If two rows or two columns in an n × n matrix A, are switched, the determinant of the resulting matrix equals (−1) times the determinant of the original matrix. If A is an n × n matrix in which two rows are equal or two columns are equal then det (A) = 0. Suppose the ith row of A equals (xa₁ + yb₁,· · ·, xan + ybn). Then

det (A) = x det (A₁) + y det (A₂)

where the ith row of A₁ is (a₁,· · ·, an) and the ith row of A₂ is (b₁,· · ·, bn) , all other rows of A₁ and A₂ coinciding with those of A. In other words, det is a linear function of each row A. The same is true with the word “row” replaced with the word “column”.

Proof: By Proposition 7.2.3 when two rows are switched, the determinant of the resulting matrix is (−1) times the determinant of the original matrix. By Corollary 7.2.5 the same holds for columns because the columns of the matrix equal the rows of the transposed matrix. Thus if A₁ is the matrix obtained from A by switching two columns,

det (A) = det (AT) = - (AT₁) = - det (A₁).

If A has two equal columns or two equal rows, then switching them results in the same matrix. Therefore, det (A) = −det (A) and so det (A) = 0. 

It remains to verify the last assertion.

The same is true of columns because det (AT) = det (A) and the rows of AT are the columns of A.

7.2.5 Linear Combinations And Determinants

Deﬁnition 7.2.7 A vector  w, is a linear combination of the vectors {v₁,· · ·, vr} if there exists scalars, c₁,· · ·cr such that  This is the same as saying

w ∈ span{v₁,· · ·, vr}.

The following corollary is also of great use.

Corollary 7.2.8 Suppose A is an n × n matrix and some column (row) is a linear combination of r other columns (rows). Then det (A) = 0.

Proof: Let A = ( a₁ · · · an ) be the columns of A and suppose the condition that one column is a linear combination of r of the others is satisﬁed. Then by using Corollary 7.2.6 you may rearrange the columns to have the nth column a linear combination of the first r columns. Thus  and so

det (A) = det ( a₁ · · · ar · · · an−1  ) .

By Corollary 7.2.6

The case for rows follows from the fact that det (A) = det (AT).

7.2.6 The Determinant Of A Product

Recall the following deﬁnition of matrix multiplication.

Deﬁnition 7.2.9 If A and B are n × n matrices, A = (aij) and B = (bij), AB = (cij) where

One of the most important rules about determinants is that the determinant of a product equals the product of the determinants.

Theorem 7.2.10 Let A and B be n × n matrices. Then

det (AB) = det (A) det (B) .

Proof: Let cij be the ijth entry of AB. Then by Proposition 7.2.3,

det (AB) =

7.2.7 Cofactor Expansions

Lemma 7.2.11 Suppose a matrix is of the form

| [picture] | (7.12)

or

| [picture] | (7.13)

where a is a number and A is an (n − 1) × (n − 1) matrix and ∗ denotes either a column or a row having length n − 1 and the 0 denotes either a column or a row of length n − 1 consisting entirely of zeros. Then

det (M) = a det (A) . 

Proof: Denote M by (mij) . Thus in the first case, mnn = a and mni = 0 if i ≠ n while in the second case, mnn = a and min = 0 if i ≠ n. From the deﬁnition of the determinant,

Letting denote the position of n in the ordered list, (k₁,· · ·, kn) then using the earlier conventions used to prove Lemma 7.1.1, det (M) equals

Now suppose (7.13). Then if kn ≠ n, the term involving mnkn in the above expression equals zero. Therefore, the only terms which survive are those for which = n or in other words, those for which kn = n. Therefore, the above expression reduces to

To get the assertion in the situation of (7.12) use Corollary 7.2.5 and (7.13) to write

In terms of the theory of determinants, arguably the most important idea is that of Laplace expansion along a row or a column. This will follow from the above deﬁnition of a determinant.

Deﬁnition 7.2.12 Let A = (aij) be an n × n matrix. Then a new matrix called the cofactor matrix, cof (A) is deﬁned by cof (A) = (cij) where to obtain cij delete the ith row and the jth column of A, take the determinant of the (n − 1) × (n − 1) matrix which results, (This is called the ijth minor of A. ) and then multiply this number by (−1)i+j. To make the formulas easier to remember, cof (A)ij will denote the ijth entry of the cofactor matrix.

The following is the main result. Earlier this was given as a deﬁnition and the outrageous totally unjustiﬁed assertion was made that the same number would be obtained by expanding the determinant along any row or column. The following theorem proves this assertion.

Theorem 7.2.13 Let A be an n × n matrix where n ≥ 2. Then

| [picture] | (7.14)

The first formula consists of expanding the determinant along the ith row and the second expands the determinant along the jth column.

Proof: Let (ai1,· · ·, ain) be the ith row of A. Let Bj be the matrix obtained from A by leaving every row the same except the ith row which in Bj equals

(0,· · ·, 0, aij, 0,· · ·, 0) .

Then by Corollary 7.2.6,

Denote by Aij the (n − 1) x (n − 1) matrix obtained by deleting the ith row and the jth column of A. Thus cof (A)ij ≡ (−1)i+j det (Aij). At this point, recall that from Proposition 7.2.3, when two rows or two columns in a matrix M, are switched, this results in multiplying the determinant of the old matrix by −1 to get the determinant of the new matrix. Therefore, by Lemma 7.2.11,

Therefore,

which is the formula for expanding det (A) along the ith row. Also,

which is the formula for expanding det (A) along the ith column.

7.2.8 Formula For The Inverse

Note that this gives an easy way to write a formula for the inverse of an n × n matrix. 

Theorem 7.2.14  A⁻¹ exists if and only if det(A) ≠ 0. If det(A) ≠ 0, then A⁻¹ =(aij⁻¹) where

aij⁻¹= det (A)−1 cof(A)ji

for cof (A)ij the ijth cofactor of A. 

Proof: By Theorem 7.2.13 and letting (air) = A, if det (A) ≠ 0,

Now consider

when k ≠ r. Replace the kth column with the rth column to obtain a matrix Bk whose determinant equals zero by Corollary 7.2.6. However, expanding this matrix along the kth column yields

Summarizing,

Using the other formula in Theorem 7.2.13, and similar reasoning,

This proves that if det (A) ≠ 0, then A⁻¹ exists with A⁻¹ = (aij⁻¹), where

aij⁻¹ = cof (A)ji det (A)⁻¹.

Now suppose A⁻¹ exists. Then by Theorem 7.2.10,

 1 = det (I)  = det(AA⁻¹⁾ = det (A) det (A⁻¹)

so det (A)  ≠ 0.

The next corollary points out that if an n × n matrix A has a right or a left inverse, then it has an inverse.

Corollary 7.2.15 Let A be an n × n matrix and suppose there exists an n × n matrix B such that BA = I. Then A⁻¹ exists and A⁻¹ = B. Also, if there exists C an n × n matrix such that AC = I, then A⁻¹ exists and A⁻¹ = C.

Proof: Since BA = I, Theorem 7.2.10 implies 

det B det A = 1 

and so det A  ≠ 0. Therefore from Theorem 7.2.14, A⁻¹ exists. Therefore, 

A⁻¹ = (BA) A⁻¹ = B(AA⁻¹) = BI = B. 

The case where CA = I is handled similarly.

The conclusion of this corollary is that left inverses, right inverses and inverses are all the same in the context of n × n matrices.

Theorem 7.2.14 says that to find the inverse, take the transpose of the cofactor matrix and divide by the determinant. The transpose of the cofactor matrix is called the adjugate or sometimes the classical adjoint of the matrix A. It is an abomination to call it the adjoint although you do sometimes see it referred to in this way. In words, A⁻¹ is equal to one over the determinant of A times the adjugate matrix of A.

7.2.9    Cramer’s Rule

In case you are solving a system of equations, Ax = y for x, it follows that if A⁻¹ exists,

 x =(A⁻¹A)x = A⁻¹ (Ax) = A⁻¹y

thus solving the system. Now in the case that A⁻¹ exists, there is a formula for A⁻¹ given above. Using this formula,

By the formula for the expansion of a determinant along a column,

where here the ith column of A is replaced with the column vector (y₁· · · ·, yn)T, and the determinant of this modiﬁed matrix is taken and divided by det (A). This formula is known as Cramer’s rule.

7.2.10 Upper Triangular Matrices

Deﬁnition 7.2.16 A matrix M, is upper triangular if Mij = 0 whenever i > j. Thus such a matrix equals zero below the main diagonal, the entries of the form Mii as shown.

A lower triangular matrix is deﬁned similarly as a matrix for which all entries above the main diagonal are equal to zero.

With this deﬁnition, here is a simple corollary of Theorem 7.2.13.

Corollary 7.2.17  Let M be an upper (lower) triangular matrix. Then det (M) is obtained by taking the product of the entries on the main diagonal.

7.3 The Cayley Hamilton Theorem∗

Deﬁnition 7.3.1 Let A be an n×n matrix. The characteristic polynomial is deﬁned as

pA (t) ≡ det (tI −A) 

and the solutions to pA (t) = 0 are called eigenvalues. For A a matrix and p (t) = tn + an−1tn−1 + · · · + a₁t + a₀, denote by p (A) the matrix deﬁned by

p (A) ≡ An + an−1An−1 +· · · + a₁A + a₀I.

The explanation for the last term is that A⁰ is interpreted as I, the identity matrix.

The Cayley Hamilton theorem states that every matrix satisﬁes its characteristic equation, that equation deﬁned by PA (t) = 0. It is one of the most important theorems in linear algebra¹. The following lemma will help with its proof.

Lemma 7.3.2 Suppose for all |λ| large enough,

A₀ + A₁λ +· · · + Amλm = 0,

where the Ai are n × n matrices. Then each Ai = 0.

Proof: Multiply by λ−m to obtain

A₀λ−m + A₁λ−m+1 + · · · + Am−1λ⁻¹ + Am = 0.

Now let |λ| → ∞ to obtain Am = 0. With this, multiply by λ to obtain

A₀λ−m+1 + A₁λ−m+2 + · · · + Am−1 = 0.

Now let |λ| → ∞ to obtain Am−1 = 0. Continue multiplying by λ and letting λ → ∞ to obtain that all the Ai = 0.

With the lemma, here is a simple corollary.

Corollary 7.3.3 Let Ai and Bi be n×n matrices and suppose

A₀ + A₁λ +· · · + Amλm = B₀ + B₁λ +· · · + Bmλm

for all |λ| large enough. Then Ai = Bi for all i. Consequently if λ is replaced by any n × n matrix, the two sides will be equal. That is, for C any n × n matrix,

A₀ + A₁C +· · · + AmCm = B₀ + B₁C +· · · + BmCm.

Proof: Subtract and use the result of the lemma.

With this preparation, here is a relatively easy proof of the Cayley Hamilton theorem.

Theorem 7.3.4 Let A be an n × n matrix and let p (λ) ≡ det (λI − A) be the characteristic polynomial. Then p (A) = 0.

Proof: Let C (λ) equal the transpose of the cofactor matrix of (λI − A) for |λ| large. (If |λ| is large enough, then λ cannot be in the finite list of eigenvalues of A and so for such λ, (λI − A)⁻¹ exists.) Therefore, by Theorem 7.2.14

C (λ) = p (λ) (λI −A)⁻¹.

Note that each entry in C (λ) is a polynomial in λ having degree no more than n − 1. Therefore, collecting the terms,

C (λ) = C₀ + C₁λ +· · · + Cn−1λn−1 

for Cj some n × n matrix. It follows that for all |λ| large enough,

(λI −A)(C₀ + C₁λ +· · · + Cn−1λn−1) = p (λ) I

and so Corollary 7.3.3 may be used. It follows the matrix coeﬃcients corresponding to equal powers of λ are equal on both sides of this equation. Therefore, if λ is replaced with A, the two sides will be equal. Thus

0 = (A − A) (C₀ + C₁A +· · · + Cn−1An−1) = p (A) I = p (A). 

This proves the Cayley Hamilton theorem.

¹ A special case was first proved by Hamilton in 1853. The general case was announced by Cayley some time later and a proof was given by Frobenius in 1878.

Rank Of A Matrix

8.1 Elementary Matrices

The elementary matrices result from doing a row operation to the identity matrix.

Deﬁnition 8.1.1 Row operations consist of the following

Switch two rows.

Multiply a row by a nonzero number.

Replace a row by a multiple of another row added to it.

The elementary matrices are n x n matricies that result in the following deﬁnition.

Deﬁnition 8.1.2 Elementary matrices consist of those matrices which result by applying a row operation to an identity matrix. Those which involve switching rows of the identity are called permutation matrices 1.

As an example of why these elementary matrices are interesting, consider the following.

A 3 × 4 matrix was multiplied on the left by an elementary matrix which was obtained from row operation 1 applied to the identity matrix. This resulted in applying the operation 1 to the given matrix. This is what happens in general.

Now consider what these elementary matrices look like. First consider the one which involves switching row i and row j where i < j. This matrix is of the form

The two exceptional rows are shown. The ith row was the jth and the jth row was the ith in the identity matrix. Now consider what this does to a column vector.

Now denote by Pij the elementary matrix which comes from the identity from switching rows i and j. From what was just explained consider multiplication on the left by this elementary matrix.

From the way you multiply matrices this is a matrix which has the indicated columns.  

This has established the following lemma.

Lemma 8.1.3 Let Pij denote the elementary matrix which involves switching the ith and the jth rows. Then

PijA = B

where B is obtained from A by switching the ith and the jth rows.

Next consider the row operation which involves multiplying the ith row by a nonzero constant, c. The elementary matrix which results from applying this operation to the ith row of the identity matrix is of the form

Now consider what this does to a column vector.

Denote by E (c, i) this elementary matrix which multiplies the ith row of the identity by the nonzero constant, c. Then from what was just discussed and the way matrices are multiplied,

equals a matrix having the columns indicated below.

This proves the following lemma.

Lemma 8.1.4 Let E (c, i) denote the elementary matrix corresponding to the row operation in which the ith row is multiplied by the nonzero constant, c. Thus E (c, i) involves multiplying the ith row of the identity matrix by c. Then

E (c, i) A = B 

where B is obtained from A by multiplying the ith row of A by c.

Finally consider the third of these row operations. Denote by E (c × i + j) the elementary matrix which replaces the jth row with itself added to c times the ith row added to it. In case i < j this will be of the form

Now consider what this does to a column vector.

Now from this and the way matrices are multiplied,

equals a matrix of the following form having the indicated columns.

The case where i > j is handled similarly. This proves the following lemma.

Lemma 8.1.5 Let E (c × i + j) denote the elementary matrix obtained from I by replacing the jth row with c times the ith row added to it. Then

E (c×i + j) A = B

where B is obtained from A by replacing the jth row of A with itself added to c times the ith row of A.

The next theorem is the main result.

Theorem 8.1.6 To perform any of the three row operations on a matrix A it suﬃces to do the row operation on the identity matrix obtaining an elementary matrix E and then take the product, EA. Furthermore, each elementary matrix is invertible and its inverse is an elementary matrix.

Proof: The first part of this theorem has been proved in Lemmas 8.1.3 - 8.1.5. It only remains to verify the claim about the inverses. Consider first the elementary matrices corresponding to row operation of type three.

E (−c × i + j) E (c × i + j) = I

This follows because the first matrix takes c times row i in the identity and adds it to row j. When multiplied on the left by E (−c × i + j) it follows from the first part of this theorem that you take the ith row of E (c × i + j) which coincides with the ith row of I since that row was not changed, multiply it by −c and add to the jth row of E (c × i + j) which was the jth row of I added to c times the ith row of I. Thus E (−c × i + j) multiplied on the left, undoes the row operation which resulted in E (c × i + j). The same argument applied to the product

E (c × i + j) E (−c × i + j)

replacing c with−c in the argument yields that this product is also equal to I. Therefore, E (c × i + j)⁻¹ = E (−c × i + j) .

Similar reasoning shows that for E (c, i) the elementary matrix which comes from multiplying the ith row by the nonzero constant, c,

E (c, i)⁻¹ = E(c⁻¹, i).

Finally, consider Pij which involves switching the ith and the jth rows. 

PijPij = I

because by the first part of this theorem, multiplying on the left by Pij switches the ith and jth rows of Pij which was obtained from switching the ith and jth rows of the identity. First you switch them to get Pij and then you multiply on the left by Pij which switches these rows again and restores the identity matrix. Thus (Pij)−1 = Pij.

8.2 The Row Reduced Echelon Form Of A Matrix

Recall that putting a matrix in row reduced echelon form involves doing row operations as described on Page 66. In this section we review the description of the row reduced echelon form and prove the row reduced echelon form for a given matrix is unique. That is, every matrix can be row reduced to a unique row reduced echelon form. Of course this is not true of the echelon form. The signiﬁcance of this is that it becomes possible to use the deﬁnite article in referring to the row reduced echelon form and hence important conclusions about the original matrix may be logically deduced from an examination of its unique row reduced echelon form. First we need the following deﬁnition of some terminology.

Deﬁnition 8.2.1 Let v₁,· · ·, vk, u be vectors. Then u is said to be a linear combination of the vectors {v₁,· · ·, vk} if there exist scalars, c₁,· · ·, ck such that

The collection of all linear combinations of the vectors, {v₁,· · ·, vk} is known as the span of these vectors and is written as span (v₁,· · ·, vk).

Another way to say the same thing as expressed in the earlier deﬁnition of row reduced echelon form found on Page 65 is the following which is a more useful description when proving the major assertions about the row reduced echelon form.

Deﬁnition 8.2.2 Let ei denote the column vector which has all zero entries except for the ith slot which is one. An m × n matrix is said to be in row reduced echelon form if, in viewing successive columns from left to right, the first nonzero column encountered is e₁ and if you have encountered e₁, e₂,· · ·, ek, the next column is either ek+1 or is a linear combination of the vectors, e₁, e₂,· · ·, ek.

Theorem 8.2.3 Let A be an m × n matrix. Then A has a row reduced echelon form determined by a simple process.

Proof: Viewing the columns of A from left to right take the first nonzero column. Pick a nonzero entry in this column and switch the row containing this entry with the top row of A. Now divide this new top row by the value of this nonzero entry to get a 1 in this position and then use row operations to make all entries below this equal to zero. Thus the first nonzero column is now e₁. Denote the resulting matrix by A₁. Consider the sub-matrix of A₁ to the right of this column and below the first row. Do exactly the same thing for this sub-matrix that was done for A. This time the e₁ will refer to Fm−1. Use the first 1 obtained by the above process which is in the top row of this sub-matrix and row operations to zero out every entry above it in the rows of A₁. Call the resulting matrix A₂. Thus A₂ satisﬁes the conditions of the above deﬁnition up to the column just encountered. Continue this way till every column has been dealt with and the result must be in row reduced echelon form.

The following diagram illustrates the above procedure. Say the matrix looked something like the following.

First step would yield something like

For the second step you look at the lower right corner as described,

and if the first column consists of all zeros but the next one is not all zeros, you would get something like this.

Thus, after zeroing out the term in the top row above the 1, you get the following for the next step in the computation of the row reduced echelon form for the original matrix.

Next you look at the lower right matrix below the top two rows and to the right of the first four columns and repeat the process.

Recall the following deﬁnition which was discussed earlier.

Deﬁnition 8.2.4 The first pivot column of A is the first nonzero column of A. The next pivot column is the first column after this which becomes e₂ in the row reduced echelon form. The third is the next column which becomes e₃ in the row reduced echelon form and so forth.

There are three choices for row operations at each step in the above theorem. A natural question is whether the same row reduced echelon matrix always results in the end from following the above algorithm applied in any way. The next corollary says this is the case but first, here is a fundamental lemma.

In rough terms, the following lemma states that linear relationships between columns in a matrix are preserved by row operations. This simple lemma is the main result in understanding all the major questions related to the row reduced echelon form as well as many other topics.

Lemma 8.2.5 Let A and B be two m × n matrices and suppose B results from a row operation applied to A. Then the kth column of B is a linear combination of the i₁,· · ·, ir columns of B if and only if the kth column of A is a linear combination of the i₁,· · ·, ir columns of A. Furthermore, the scalars in the linear combination are the same. (The linear relationship between the kth column of A and the i₁,· · ·, ir columns of A is the same as the linear relationship between the kth column of B and the i₁,· · ·, ir columns of B.)

Proof: Let A equal the following matrix in which the ak are the columns 

( a₁ a₂ · · · an )

and let B equal the following matrix in which the columns are given by the bk 

( b₁ b₂ · · · bn)

Then by Theorem 8.1.6 on this Page bk = Eak where E is an elementary matrix. Suppose then that one of the columns of A is a linear combination of some other columns of A. Say

Then multiplying by E,

Deﬁnition 8.2.6 Two matrices are said to be row equivalent if one can be obtained from the other by a sequence of row operations.

It has been shown above that every matrix is row equivalent to one which is in row reduced echelon form. Note

so to say two column vectors are equal is to say they are the same linear combination of the special vectors ej.

Corollary 8.2.7 The row reduced echelon form is unique. That is if B, C are two matrices in row reduced echelon form and both are row equivalent to A, then B = C.

Proof: Suppose B and C are both row reduced echelon forms for the matrix A. Then they clearly have the same zero columns since row operations leave zero columns unchanged. If B has the sequence e₁, e₂,· · ·, er occurring for the first time in the positions, i₁, i₂,· · ·, ir, the description of the row reduced echelon form means that each of these columns is not a linear combination of the preceding columns. Therefore, by Lemma 8.2.5, the same is true of the columns in positions i₁, i₂,· · ·, ir for C. It follows from the description of the row reduced echelon form that e₁,· · ·, er occur respectively for the first time in columns i₁, i₂,· · ·, ir for C. Therefore, both B and C have the sequence e₁, e₂,· · ·, er occurring for the first time in the positions, i₁, i₂,· · ·, ir. By Lemma 8.2.5, the columns between the ik and ik+1 position in the two matrices are linear combinations involving the same scalars of the columns in the i₁,· · ·, ik position. Also the columns after the ir position are linear combinations of the columns in the i₁,· · ·, ir positions involving the same scalars in both matrices. This is equivalent to the assertion that each of these columns is identical and this proves the corollary.

The above corollary shows that you can determine whether two matrices are row equivalent by simply checking their row reduced echelon forms. The matrices are row equivalent if and only if they have the same row reduced echelon form.

With Corollary 8.2.7 we make an observation concerning matricies with more columns than rows.

Corollary 8.2.8 Suppose A is an m × n matrix and that m < n. That is, the number of rows is less than the number of columns. Then one of the columns of A is a linear combination of the preceding columns of A. Also, there exists a nonzero solution x to the equation Ax = 0.

Proof: Since m < n, not all the columns of A can be pivot columns. In reading from left to right, pick the first one which is not a pivot column. Then from the description of the row reduced echelon form, this column is a linear combination of the preceding columns. Denote the jth column of A by aj. Thus for some k > 1,

Let x = (x₁, . . . , xk-1, -1, 0, . . . , 0)T. Then Ax = 0.

Example 8.2.9 Find the row reduced echelon form of the matrix

The first nonzero column is the second in the matrix. We switch the third and first rows to obtain

Now we multiply the top row by −2 and add to the second.

Next, add the second row to the bottom and then divide the bottom row by −6

Next use the bottom row to obtain zeros in the last column above the 1 and divide the second row by −2

Finally, add −1 times the middle row to the top.

This is in row reduced echelon form.

Example 8.2.10 Find the row reduced echelon form for the matrix

You should verify that the row reduced echelon form is 

Having developed the row reduced echelon form, it is now easy to verify that the right inverse found earlier using the Gauss Jordan procedure is the inverse.

Theorem 8.2.11 Suppose A, B are n × n matrices and AB = I. Then it follows that BA = I also, and so B = A⁻¹. For n × n matrices, the left inverse and the right inverse are equal.

Proof. If AB = I, there exists a unique solution x to the equation

Bx = y 

for any choice of y. In fact,

x = A (Bx) = Ay.

This means the row reduced echelon form of B must be I. Thus every column is a pivot column. Otherwise, there exists a free variable and the solution, if it exists, would not be unique, contrary to what was just shown must happen if AB = I. It follows that a right inverse B⁻¹ for B exists. The Gauss Jordan procedure for finding the inverse yields

( B I ) →( I B−1 ) .

Now multiply both sides of the equation AB = I on the right by B⁻¹. Then

A = A(BB⁻¹) = (AB) B⁻¹ = B⁻¹.

Thus A is the right inverse of B, and so BA = I. This shows that if AB = I, then BA = I also. Exchanging roles of A and B, we see that if BA = I, then AB = I.

8.3 The Rank Of A Matrix 

8.3.1 The Deﬁnition Of Rank

To begin, here is a deﬁnition to introduce some terminology.

Deﬁnition 8.3.1 Let A be an m × n matrix. The column space of A is the span of the columns. The row space is the span of the rows.

There are three deﬁnitions of the rank of a matrix which are useful. These are given in the following deﬁnition. It turns out that the concept of determinant rank is often important but is virtually impossible to find directly. The other two concepts of rank are very easily determined and it is a happy fact that all three yield the same number. This is shown later.

Deﬁnition 8.3.2 A sub-matrix of a matrix A is a rectangular array of numbers obtained by deleting some rows and columns of A. Let A be an m×n matrix. The determinant rank of the matrix equals r where r is the largest number such that some r × r sub-matrix of A has a non zero determinant. The row space of a matrix is the span of the rows and the column space of a matrix is the span of the columns. The row rank of a matrix is the number of nonzero rows in the row reduced echelon form and the column rank is the number columns in the row reduced echelon form which are one of the ek vectors. Thus the column rank equals the number of pivot columns. It follows the row rank equals the column rank. This is also called the rank of the matrix. The rank of a matrix A is denoted by rank (A) .

Example 8.3.3 Consider the matrix

What is its rank?

You could look at all the 2 × 2 submatrices

Each has determinant equal to 0. Therefore, the rank is less than 2. Now look at the 1 × 1 submatrices. There exists one of these which has nonzero determinant. For example (1) has determinant equal to 1 and so the rank of this matrix equals 1.

Of course this example was pretty easy but what if you had a 4 × 7 matrix? You would have to consider all the 4 × 4 submatrices and then all the 3 × 3 submatrices and then all the 2 × 2 matrices and finally all the 1 × 1 matrices in order to compute the rank. Clearly this is not practical. The following theorem will remove the diﬃculties just indicated.

The following theorem is proved later.

Theorem 8.3.4 Let A be an m × n matrix. Then the row rank, column rank and determinant rank are all the same.

Example 8.3.5 Find the rank of the matrix

From the above deﬁnition, all you have to do is find the row reduced echelon form and then count up the number of nonzero rows. But the row reduced echelon form of this matrix is

and so the rank of this matrix is 4.

Find the rank of the matrix

The row reduced echelon form is

and so this time the rank is 3.

8.3.2 Finding The Row And Column Space Of A Matrix

The row reduced echelon form also can be used to obtain an eﬃcient description of the row and column space of a matrix. Of course you can get the column space by simply saying that it equals the span of all the columns but often you can get the column space as the span of fewer columns than this. This is what we mean by an “eﬃcient description”. This is illustrated in the next example.

Example 8.3.6 Find the rank of the following matrix and describe the column and row spaces efficiently.

The row reduced echelon form is

Therefore, the rank of this matrix equals 2. All columns of this row reduced echelon form are in

For example,

By Lemma 8.2.5, all columns of the original matrix, are similarly contained in the span of the first two columns of that matrix. For example, consider the third column of the original matrix. 

How did I know to use −9 and 5 for the coeﬃcients? This is what Lemma 8.2.5 says! It says linear relationships are all preserved. Therefore, the column space of the original matrix equals the span of the first two columns. This is the desired eﬃcient description of the column space.

What about an eﬃcient description of the row space? When row operations are used, the resulting vectors remain in the row space. Thus the rows in the row reduced echelon form are in the row space of the original matrix. Furthermore, by reversing the row operations, each row of the original matrix can be obtained as a linear combination of the rows in the row reduced echelon form. It follows that the span of the nonzero rows in the row reduced echelon matrix equals the span of the original rows. In the above example, the row space equals the span of the two vectors,( 1 0 −9 9 2) and ( 0 1 5 −3 0 ).

Example 8.3.7 Find the rank of the following matrix and describe the column and row spaces efficiently.

The row reduced echelon form is

and so the rank is 3, the row space is the span of the vectors,

and the column space is the span of the first three columns in the original matrix,

Example 8.3.8 Find the rank of the following matrix and describe the column and row spaces eﬃciently.

The row reduced echelon form is,

It follows the rank is three and the column space is the span of the first, second and fourth columns of the original matrix.

while the row space is the span of the vectors

Procedure 8.3.9 To find the rank of a matrix, obtain the row reduced echelon form for the matrix. Then count the number of nonzero rows or equivalently the number of pivot columns. This is the rank. The row space is the span of the nonzero rows in the row reduced echelon form and the column space is the span of the pivot columns of the original matrix.

8.4 Linear Independence And Bases 

8.4.1 Linear Independence And Dependence

First we consider the concept of linear independence. We deﬁne what it means for vectors in Fn to be linearly independent and then give equivalent descriptions. In the following deﬁnition, the symbol,

( v₁  v₂ · · · vk )

denotes the matrix which has the vector v₁ as the first column, v₂ as the second column and so forth until vk is the kth column.

Deﬁnition 8.4.1 Let {v₁,· · ·, vk} be vectors in Fn. Then this collection of vectors is said to be linearly independent if each of the columns of the n × k matrix

( v₁  v₂ · · · vk )

is a pivot column. Thus the row reduced echelon form for this matrix is

( e1  e₂ · · · ek ) .

The question whether any vector in the first k columns in a matrix is a pivot column is independent of the presence of later columns. Thus each of {v₁,· · ·, vk} is a pivot column in

( v₁  v₂ · · · vk )

if and only if these vectors are each pivot columns in

( v1  v₂ · · · vk  w₁ · · · wr )

Here is what the linear independence means in terms of linear relationships.

Corollary 8.4.2 The collection of vectors, {v₁,· · ·, vk} is linearly independent if and only if none of these vectors is a linear combination of the others.

Proof: If {v₁,· · ·, vk} is linearly independent, then every column in 

( v₁  v₂ · · · vk )

is a pivot column which requires that the row reduced echelon form is 

( e1  e₂ · · · ek ) .

Now none of the ei vectors is a linear combination of the others. By Lemma 8.2.5 on Page 138 none of the vi is a linear combination of the others. Recall this lemma says linear relationships between the columns are preserved under row operations.

Next suppose none of the vectors {v₁,· · ·, vk} is a linear combination of the others. Then none of the columns in

( v₁ v₂ · · · vk )

is a linear combination of the others. By Lemma 8.2.5 the same is true of the row reduced echelon form for this matrix. From the description of the row reduced echelon form, it follows that the ith column of the row reduced echelon form must be ei since otherwise, it would be a linear combination of the first i − 1 vectors e₁,· · ·, ei−1 and by Lemma 8.2.5, it follows vi would be the same linear combination of v₁,· · ·, vi−1 contrary to the assumption that none of the columns in ( v₁ v₂ · · · vk ) is a linear combination of the others. Therefore, each of the k columns in ( v₁ v₂ · · · vk is a pivot column and so {v₁,· · ·, vk} is linearly independent.

Corollary 8.4.3 The collection of vectors, {v₁,· · ·, vk} is linearly independent if and only if whenever

it follows each ci = 0.

Proof: Suppose first {v₁,· · ·, vk} is linearly independent. Then by Corollary 8.4.2, none of the vectors is a linear combination of the others. Now suppose

and not all ci = 0. Then pick ci which is not zero, divide by it and solve for vi in terms of the other vj, contradicting the fact that none of the vi equals a linear combination of the others.

Now suppose the condition about the sum holds. If vi is a linear combination of the other vectors in the list, then you could obtain an equation of the form

and so

contradicting the condition about the sum.

Sometimes we refer to this last condition about sums as follows: The set of vectors, {v₁,· · ·, vk} is linearly independent if and only if there is no nontrivial linear combination which equals zero. (A nontrivial linear combination is one in which not all the scalars equal zero.)

We give the following equivalent deﬁnition of linear independence which follows from the above corollaries.

Deﬁnition 8.4.4 A set of vectors, {v₁,· · ·, vk} is linearly independent if and only if none of the vectors is a linear combination of the others or equivalently if there is no nontrivial linear combination of the vectors which equals 0. It is said to be linearly dependent if at least one of the vectors is a linear combination of the others or equivalently there exists a nontrivial linear combination which equals zero.

Note the meaning of the words. To say a set of vectors is linearly dependent means at least one is a linear combination of the others. In other words, it is in a sense “dependent” on these other vectors.

The following corollary follows right away from the row reduced echelon form. It concerns a matrix with more columns than rows.

Corollary 8.4.5 Let {v₁,· · ·, vk} be a set of vectors in Fn. Then if k > n, it must be the case that {v₁,· · ·, vk} is not linearly independent. In other words, if k > n, then {v₁,· · ·, vk} is dependent.

Proof: If k > n, then the columns of ( v₁ v₂ · · · vk ) cannot each be a pivot column because there are at most n pivot columns due to the fact the matrix has only n rows. In reading from left to right, pick the first column which is not a pivot column. Then from the description of row reduced echelon form, this column is a linear combination of the preceding columns and so the given vectors are dependent by Corollary 8.4.2.

Example 8.4.6Determine whether the vectors

are linearly independent. If they are linearly dependent, exhibit one of the vectors as a linear combination of the others.

Form the matrix mentioned above.

Then the row reduced echelon form of this matrix is

Thus not all the columns are pivot columns and so the vectors are not linear independent. Note the fourth column is of the form

From Lemma 8.2.5, the same linear relationship exists between the columns of the original matrix. Thus

Note the usefulness of the row reduced echelon form in discovering hidden linear relationships in collections of vectors.

Example 8.4.7 Determine whether the vectors  

are linearly independent. If they are linearly dependent, exhibit one of the vectors as a linear combination of the others.

The matrix used to find this is

The row reduced echelon form is

and so every column is a pivot column. Therefore, these vectors are linearly independent and there is no way to obtain one of the vectors as a linear combination of the others.

8.4.2 Subspaces

A subspace is a set of vectors with the property that linear combinations of these vectors remain in the set. Geometrically, subspaces are like lines and planes which contain the origin. More precisely, the following deﬁnition is the right way to think of this.

Deﬁnition 8.4.8 Let V be a nonempty collection of vectors in Fn. Then V is called a subspace if whenever α, β are scalars and u, v are vectors in V, the linear combination αu + βv is also in V .

It turns out that every subspace equals the span of some vectors. This is the content of the next theorem.

Theorem 8.4.9 V is a subspace of Fn if and only if there exist vectors of V

{u₁,· · ·, uk}

 such that V = span (u₁,· · ·, uk) .

Proof: Pick a vector of V, u₁. If V = span {u₁}, then stop. You have found your list of vectors. If V ≠  span (u₁) , then there exists u₂ a vector of V which is not a vector in span (u₁) . Consider span (u₁, u₂) . If V = span (u₁, u₂) , stop. Otherwise, pick u₃ ∉ span (u₁, u₂) . Continue this way. Note that since V is a subspace, these spans are each contained in V . The process must stop with uk for some k ≤ n since otherwise, the matrix

(u₁ · · · uk )

having these vectors as columns would have n rows and k > n columns. Consequently, it can have no more than n pivot columns and so the first column which is not a pivot column would be a linear combination of the preceding columns contrary to the construction.

For the other half, suppose V = span (u₁, . . . , uk and let ∑ki=1 ciui and ∑ki=1 diui be two vectors in V. Now let α and β be two scalars. Then

which is one of the things in span (u₁,· · ·, uk) showing that span (u₁,· · ·, uk) has the properties of a subspace.

The following corollary also follows easily.

Corollary 8.4.10 If V is a subspace of Fn, then there exist vectors of V,{u₁,· · ·, uk} such that V = span (u₁,· · ·, uk) and {u₁,· · ·, uk} is linearly independent.

Proof: Let V = span (u₁,· · ·, uk) . Then let the vectors {u₁,· · ·, uk} be the columns of the following matrix.

( u₁ · · · uk )

Retain only the pivot columns. That is, determine the pivot columns from the row reduced echelon form and these are a basis for span (u₁,· · ·, uk).

The message is that subspaces of Fn consist of spans of finite, linearly independent collections of vectors of Fn.

The following fundamental lemma is very useful.

Lemma 8.4.11 Suppose {x₁,· · ·, xr} is linearly independent and each xk is contained in span (y₁,· · ·, ys) . Then s ≥r. In words, spanning sets have at least as many vectors as linearly independent sets.

Proof: Since {y₁,· · ·, ys} is a spanning set, there exist scalars aij such that

Suppose s < r. Then the matrix A whose ijth entry is aij has fewer rows, s than columns, r. By Corollary 8.2.8 there exists d such that d ≠ 0 but Ad = 0. In other words,

Therefore,

which contradicts {x₁,· · ·, xr} is linearly independent, because not all the dj = 0. Thus s ≥ r.

Note how this lemma was totally dependent on algebraic considerations and was independent of context. This will be considered more later in the chapter on abstract vector spaces. I didn’t need to know what the xk, yk were, only that the {x₁,· · ·, xr} were independent and contained in the span of the yk.

8.4.3 Basis Of A Subspace

It was just shown in Corollary 8.4.10 that every subspace of Fn is equal to the span of a linearly independent collection of vectors of Fn. Such a collection of vectors is called a basis.

Deﬁnition 8.4.12 Let V be a subspace of Fn. Then {u₁,· · ·, uk} is a basis for V if the following two conditions hold.

1. span (u₁,· · ·, uk) = V.

2. {u₁,· · ·, uk} is linearly independent.

The plural of basis is bases.

The main theorem about bases is the following.

Theorem 8.4.13 Let V be a subspace of Fn and suppose {u₁,· · ·, uk}, {v₁,· · ·, vm} are two bases for V . Then k = m.

Proof: This follows right away from Lemma 8.4.11. {u₁,· · ·, uk} is a spanning set while {v₁,· · ·, vm} is linearly independent so k ≥ m. Also {v₁,· · ·, vm} is a spanning set while{u₁,· · ·, uk} is linearly independent so m ≥ k.

Now here is another proof. Suppose k < m. Then since {u₁,· · ·, uk} is a basis for V, each vi is a linear combination of the vectors of {u₁,· · ·, uk}. Consider the matrix

( u₁ · · · uk  v₁ · · · vm ) 

in which each of the ui is a pivot column because the {u₁,· · ·, uk} are linearly independent. Therefore, the row reduced echelon form of this matrix is

| ( e₁ · · · ek  w₁ · · · wm) | (8.3)

where each wj has zeroes below the kth row. This is because of Lemma 8.2.5 which implies each wi is a linear combination of the e₁,· · ·, ek. Discarding the bottom n − k rows of zeroes in the above, yields the matrix

( e′₁ · · · e′k w′₁ · · · w′m )

in which all vectors are in Fk. Since m > k, it follows from Corollary 8.4.5 that the vectors, {w′₁,· · ·, w′m } are dependent. Therefore, some w′j is a linear combination of the other w′. Therefore, wj is a linear combination of the other wi in (8.3). By Lemma 8.2.5 again, the same linear relationship exists between the {v₁,· · ·, vm} showing that {v₁,· · ·, vm} is not linearly independent and contradicting the assumption that {v₁,· · ·, vm} is a basis. It follows m ≤ k. Similarly, k ≤ m.

This is a very important theorem so here is yet another proof of it.

Theorem 8.4.14 Let V be a subspace and suppose {u₁,· · ·, uk} and {v₁,· · ·, vm} are two bases for V . Then k = m.

Proof: Suppose k > m. Then since the vectors, {u₁,· · ·, uk} span V, there exist scalars, cij such that

Therefore,

if and only if

Now since {v₁,· · ·, vn} is independent, this happens if and only if

However, this is a system of m equations in k variables, d₁,· · ·, dk and m < k. Therefore, there exists a solution to this system of equations in which not all the dj are equal to zero. Recall why this is so. The augmented matrix for the system is of the form ( C 0 ) where C is a matrix which has more columns than rows. Therefore, there are free variables and hence nonzero solutions to the system of equations. However, this contradicts the linear independence of {u₁,· · ·, uk} because, as explained above, ∑kj=1 djuj = 0. Similarly it cannot happen that m > k.

The following deﬁnition can now be stated.

Deﬁnition 8.4.15 Let V be a subspace of Fn. Then the dimension of V is deﬁned to be the number of vectors in a basis.

Corollary 8.4.16 The dimension of Fn is n.

Proof: You only need to exhibit a basis for Fn which has n vectors. Such a basis is {e₁,· · ·, en}.

Corollary 8.4.17 Suppose {v₁,· · ·, vn} is linearly independent and each vi is a vector in Fn. Then {v₁,· · ·, vn} is a basis for Fn. Suppose {v₁,· · ·, vm} spans Fn. Then m ≥n. If {v₁,· · ·, vn} spans Fn, then {v₁,· · ·, vn} is linearly independent.

Proof: Let u be a vector of Fn and consider the matrix

( v₁ · · · vn  u ) .

Since each vi is a pivot column, the row reduced echelon form is

( e₁ · · · en  w )

and so, since w is in span (e₁,· · ·, en) , it follows from Lemma 8.2.5 that u is one of the vectors in span (v₁,· · ·, vn) . Therefore, {v₁,· · ·, vn} is a basis as claimed.

To establish the second claim, suppose that m < n. Then letting vi1,· · ·, vik be the pivot columns of the matrix

( v₁ · · · vm )

it follows k ≤ m < n and these k pivot columns would be a basis for Fn having fewer than n vectors, contrary to Theorem 8.4.13 which states every two bases have the same number of vectors in them. 

Finally consider the third claim. If {v₁,· · ·, vn} is not linearly independent, then replace this list with {vi1,· · ·, vik} where these are the pivot columns of the matrix

( v₁ · · · vn )

Then {vi1,· · ·, vik} spans Fn and is linearly independent so it is a basis having less than n vectors contrary to Theorem 8.4.13 which states every two bases have the same number of vectors in them.

Example 8.4.18 Find the rank of the following matrix. If the rank is r, identify r columns in the original matrix which have the property that every other column may be written as a linear combination of these. Also find a basis for the row and column spaces of the matrices.

The row reduced echelon form is

and so the rank of the matrix is 3. A basis for the column space is the first three columns of the original matrix. I know they span because the first three columns of the row reduced echelon form above span the column space of that matrix. They are linearly independent because the first three columns of the row reduced echelon form are linearly independent. By Lemma 8.2.5 all linear relationships are preserved and so these first three vectors form a basis for the column space. The four rows of the row reduced echelon form form a basis for the row space of the original matrix.

Example 8.4.19 Find the rank of the following matrix. If the rank is r, identify r columns in the original matrix which have the property that every other column may be written as a linear combination of these. Also find a basis for the row and column spaces of the matrices.

The row reduced echelon form is

A basis for the column space of this row reduced echelon form is the first second and fourth columns. Therefore, a basis for the column space in the original matrix is the first second and fourth columns. The rank of the matrix is 3. A basis for the row space of the original matrix is the columns of the row reduced echelon form.

8.4.4 Extending An Independent Set To Form A Basis

Suppose {v₁,· · ·, vm} is a linearly independent set of vectors in Fn. It turns out there is a larger set of vectors, {v₁,· · ·, vm, vm+1,· · ·, vn} which is a basis for Fn. It is easy to do this using the row reduced echelon form. Consider the following matrix having rank n in which the columns are shown.

( v₁ · · · vm e₁ e₂ · · · en ) .

Since the {v₁,· · ·, vm} are linearly independent, the row reduced echelon form of this matrix is of the form

 ( e1 · · · em u1 u2 · · · un )

Now the pivot columns can be identiﬁed and this leads to a basis for the column space of the original matrix which is of the form

{v₁,· · ·, vm, ei1,· · ·, ein−m} .

This proves the following theorem.

Theorem 8.4.20 Let {v₁,· · ·, vm} be a linearly independent set of vectors in Fn. Then there is a larger set of vectors, {v₁,· · ·, vm, vm+1,· · ·, vn} which is a basis for Fn.

Example 8.4.21 The vectors, are linearly independent.

Enlarge this set of vectors to form a basis for R⁴.

Using the above technique, consider the following matrix.

whose row reduced echelon form is

The pivot columns are numbers 1,2,3, and 6. Therefore, a basis is

8.4.5 Finding The Null Space Or Kernel Of A Matrix

Let A be an m × n matrix.

Deﬁnition 8.4.22 ker (A), also referred to as the null space of A is deﬁned as follows.

ker (A) = {x : Ax = 0}

and to find ker (A) one must solve the system of equations Ax = 0.

This is not new! There is just some new terminology being used. To repeat, ker (A) is the solution to the system Ax = 0.

Example 8.4.23 Let

Find ker (A).

You need to solve the equation Ax = 0. To do this you write the augmented matrix and then obtain the row reduced echelon form and the solution. The augmented matrix is

Next place this matrix in row reduced echelon form,

Note that x₁ and x₂ are basic variables while x₃ is a free variable. Therefore, the solution to this system of equations, Ax = 0 is given by

Example 8.4.2.24 Let

Find the null space of A.

You need to solve the equation, Ax = 0. The augmented matrix is

Its row reduced echelon form is

It follows x₁ and x₂ are basic variables and x₃, x₄, x₅ are free variables. Therefore, ker (A) is given by

We write this in the form

In other words, the null space of this matrix equals the span of the three vectors above. Thus

This is the same as

Notice also that the three vectors above are linearly independent and so the dimension of ker (A) is 3. This is generally the way it works. The number of free variables equals the dimension of the null space while the number of basic variables equals the number of pivot columns which equals the rank. We state this in the following theorem.

Deﬁnition 8.4.25 The dimension of the null space of a matrix is called the nullity² and written as null (A).

Theorem 8.4.26 Let A be an m × n matrix. Then rank (A) + null (A) = n.

8.4.6 Rank And Existence Of Solutions To Linear Systems

Consider the linear system of equations,

| Ax = b | (8.4)

where A is an m × n matrix, x is a n × 1 column vector, and b is an m × 1 column vector. Suppose

A =( a₁ · · · an )

where the ak denote the columns of A. Then x = (x₁,· · ·, xn)T is a solution of the system (8.4), if and only if

x₁a₁ +· · · + xnan = b

which says that b is a vector in span (a₁,· · ·, an) . This shows that there exists a solution to the system, (8.4) if and only if b is contained in span (a₁,· · ·, an) . In words, there is a solution to (8.4) if and only if b is in the column space of A. In terms of rank, the following proposition describes the situation.

Proposition 8.4.27 Let A be an m×n matrix and let b be an m × 1 column vector. Then there exists a solution to (8.4) if and only if

| rank (A | b )= rank (A) | (8.5)

Proof: Place ( A | b ) and A in row reduced echelon form, respectively B and C. If the above condition on rank is true, then both B and C have the same number of nonzero rows. In particular, you cannot have a row of the form

( 0 · · · 0 )

where  ≠ 0 in B. Therefore, there will exist a solution to the system (8.4).

Conversely, suppose there exists a solution. This means there cannot be such a row in B described above. Therefore, B and C must have the same number of zero rows and so they have the same number of nonzero rows. Therefore, the rank of the two matrices in (8.5) is the same.

8.5 Fredholm Alternative

There is a very useful version of Proposition 8.4.27 known as the Fredholm alternative. I will only present this for the case of real matrices here. Later a much more elegant and general approach is presented which allows for the general case of complex matrices.

The following deﬁnition is used to state the Fredholm alternative.

Deﬁnition 8.5.1 Let S ⊆ Rm. Then S⊥ ≡ {z ∈ Rm : z·s = 0 for every s ∈ S}. The funny exponent, ⊥ is called “perp”.

Now note

Lemma 8.5.2 Let A be a real m × n matrix, let x ∈ Rn and y ∈ Rm. Then

Proof: This follows right away from the deﬁnition of the dot product and matrix multiplication.

Now it is time to state the Fredholm alternative. The first version of this is the following theorem. 

Theorem 8.5.3 Let A be a real m × n matrix and let b ∈ Rm. There exists a solution, x to the equation Ax = b if and only if b ∈ ker (AT)⊥.

Proof: First suppose b(AT)⊥. Then this says that if ATx = 0, it follows that b·x = 0. In other words, taking the transpose, if

xT A = 0, then b· x· x = 0.

In other words, letting x = (x₁, . . . ,xm)T, it follows that if

then it follows

In other words, if you get a row of zeros in row reduced echelon form for A then you the same row operations produce a zero in the m × 1 matrix b.

Consequently

rank ( A | b ) = rank (A)

and so by Proposition 8.4.27, there exists a solution, x to the system Ax = b. It remains to go the other direction.

Let z ∈ ker (AT)and suppose Ax = b. I need to verify b · z = 0. By Lemma 8.5.2, 

b·z = Ax·z = x·ATz = x·0 = 0

This implies the following corollary which is also called the Fredholm alternative. The “alternative” becomes more clear in this corollary.

Corollary 8.5.4 Let A be an m × n matrix. Then A maps Rn onto Rm if and only if the only solution to ATx = 0 is x = 0.

Proof: If the only solution to ATx = 0 is x = 0, then ker (AT) = {0} and so ker (AT)⊥ =Rm because every b ∈ Rm has the property that b · 0 = 0. Therefore, Ax = b has a solution for any b ∈ Rm because the b for which there is a solution are those in ker (AT)⊥ by Theorem 8.5.3. In other words, A maps Rn onto Rm.

Conversely if A is onto, then by Theorem 8.5.3 every b ∈ Rm is in ker (AT)⊥ and so if ATx = 0, then b · x = 0 for every b. In particular, this holds for b = x. Hence if ATx = 0, then x = 0.

Here is an amusing example.

Example 8.5.5 Let A be an m × n matrix in which m > n. Then A cannot map onto Rm.

The reason for this is that AT is an n × m where m > n and so in the augmented matrix

(AT|0)

there must be some free variables. Thus there exists a nonzero vector x such that ATx = 0.

8.5.1 Row, Column, And Determinant Rank

I will now present a review of earlier topics and prove Theorem 8.3.4.

Deﬁnition 8.5.6 A sub-matrix of a matrix A is the rectangular array of numbers obtained by deleting some rows and columns of A. Let A be an m × n matrix. The determinant rank of the matrix equals r where r is the largest number such that some r × r sub-matrix of A has a non zero determinant. The row rank is deﬁned to be the dimension of the span of the rows. The column rank is deﬁned to be the dimension of the span of the columns.

Theorem 8.5.7 If A, an m × n matrix has determinant rank, r, then there exist r rows of the matrix such that every other row is a linear combination of these r rows.

Proof: Suppose the determinant rank of A = (aij) equals r. Thus some r×r submatrix has non zero determinant and there is no larger square submatrix which has non zero determinant. Suppose such a submatrix is determined by the r columns whose indices are

j₁< · · ·< jr

and the r rows whose indices are

 i₁< · · ·< ir 

I want to show that every row is a linear combination of these rows. Consider the lth row and let p be an index between 1 and n. Form the following (r + 1)× (r + 1) matrix

Of course you can assume l ∉ {i₁,· · ·, ir} because there is nothing to prove if the lth row is one of the chosen ones. The above matrix has determinant 0. This is because if p ∉ {j₁,· · ·, jr} then the above would be a submatrix of A which is too large to have non zero determinant. On the other hand, if p ∈ {j₁,· · ·, jr} then the above matrix has two columns which are equal so its determinant is still 0.

Expand the determinant of the above matrix along the last column. Let Ck denote the cofactor associated with the entry aikp. This is not dependent on the choice of p. Remember, you delete the column and the row the entry is in and take the determinant of what is left and multiply by −1 raised to an appropriate power. Let C denote the cofactor associated with alp. This is given to be nonzero, it being the determinant of the matrix

Thus

which implies

Since this is true for every p and since mk does not depend on p, this has shown the lth row is a linear combination of the i₁, i₂,· · ·, ir rows.

Corollary 8.5.8 The determinant rank equals the row rank.

Proof: From Theorem 8.5.7, the row rank is no larger than the determinant rank. Could the row rank be smaller than the determinant rank? If so, there exist p rows for p < r such that the span of these p rows equals the row space. But this implies that the r × r sub-matrix whose determinant is nonzero also has row rank no larger than p which is impossible if its determinant is to be nonzero because at least one row is a linear combination of the others. 

Corollary 8.5.9 If A has determinant rank, r, then there exist r columns of the matrix such that every other column is a linear combination of these r columns. Also the column rank equals the determinant rank.

Proof: This follows from the above by considering AT. The rows of AT are the columns of A and the determinant rank of AT and A are the same. Therefore, from Corollary 8.5.8, column rank of A = row rank of AT = determinant rank of AT = determinant rank of A.

The following theorem is of fundamental importance and ties together many of the ideas presented above.

Theorem 8.5.10 Let A be an n × n matrix. Then the following are equivalent.

1. det (A) = 0.

2. A, AT are not one to one.

3. A is not onto.

Proof: Suppose det (A) = 0. Then the determinant rank of A = r < n. Therefore, there exist r columns such that every other column is a linear combination of these columns by Theorem 8.5.7. In particular, it follows that for some m, the mth column is a linear combination of all the others. Thus letting A = ( a₁ · · · am · · · an ) where the columns are denoted by ai, there exists scalars, αi such that 

Now consider the column vector  Then

Since also A0 = 0, it follows A is not one to one. Similarly, AT is not one to one by the same argument applied to AT. This veriﬁes that 1.) implies 2.).

Now suppose 2.). Then since AT is not one to one, it follows there exists x 0 such that

ATx = 0.

Taking the transpose of both sides yields

xTA = 0

where the 0 is a 1 × n matrix or row vector. Now if Ay = x, then 

|x|² = xT (Ay) = (xTA) y = 0y = 0

contrary to x ≠ 0. Consequently there can be no y such that Ay = x and so A is not onto. This shows that 2.) implies 3.).

Finally, suppose 3 .). If 1.) does not hold, then det (A) ≠ 0 but then from Theorem 7.2.14 A⁻¹ exists and so for every y ∈ Fn there exists a unique x ∈ Fn such that Ax = y. In fact x = A⁻¹y. Thus A would be onto contrary to 3.). This shows 3.) implies 1.)

Corollary 8.5.11 Let A be an n × n matrix. Then the following are equivalent.

1. det (A) ≠ 0.

2. A and AT are one to one.
3. A is onto.

Proof: This follows immediately from the above theorem. 

Corollary 8.5.12 Let A be an invertible n×n matrix. Then A equals a finite product of elementary matrices.

Proof: Since A⁻¹ is given to exist, det (A) ≠ 0 and it follows A must have rank n and so the row reduced echelon form of A is I. Therefore, by Theorem 8.1.6 there is a sequence of elementary matrices, E₁,· · ·, Ep which accomplish successive row operations such that

(EpEp−1· · ·E₁) A = I.

But now multiply on the left on both sides by then by and then by until you get

and by Theorem 8.1.6 each of these in this product is an elementary matrix.

8.6 Exercises

Let {u₁,· · ·, un} be vectors in Rn. The parallelepiped determined by these vectors

P (u₁,· · ·, un)

is deﬁned as

Now let A be an n × n matrix. Show that

{Ax : x ∈P (u₁,· · ·, un)}

is also a parallelepiped.

In the context of Problem 1, draw P (e₁, e₂) where e₁, e₂ are the standard basis vectors for R². Thus e₁ = (1, 0) , e₂ = (0, 1) . Now suppose

where E is the elementary matrix which takes the third row and adds to the first. Draw

{Ex : x ∈P (e₁, e₂)}.

In other words, draw the result of doing E to the vectors in P (e₁, e₂). Next draw the results of doing the other elementary matrices to P (e₁, e₂).

In the context of Problem 1, either draw or describe the result of doing elementary matrices to P (e₁, e₂, e₃). Describe geometrically the conclusion of Corollary 8.5.12.

Determine which matrices are in row reduced echelon form.

Row reduce the following matrices to obtain the row reduced echelon form. List the pivot columns in the original matrix.

Find the rank of the following matrices. If the rank is r, identify r columns in the original matrix which have the property that every other column may be written as a linear combination of these. Also find a basis for the row and column spaces of the matrices.

Suppose A is an m × n matrix. Explain why the rank of A is always no larger than min (m, n).

Let H denote span . Find the

dimension of H and determine a basis.

Let H denote span

. Find the dimension of H and determine a basis.

Let H denote span

. Find the dimension of H and determine a basis.

Let M = {u = (u₁, u₂, u₃, u₄) ∈ R⁴ : u₃ = u₁ = 0}. Is M a subspace? Explain.

Let M ={u = (u₁, u₂, u₃, u₄) ∈ R⁴ : u₃ ≥ u₁. Is M a subspace? Explain.

Let w ∈ R⁴ and let M = {u = (u₁, u₂, u₃, u₄) ∈ R⁴ : w·u = 0}. Is M a subspace? Explain.

Let M = {u = (u₁, u₂, u₃, u₄) ∈ R⁴ : ui ≥ 0 for each i = 1, 2, 3, 4}. Is M a subspace? Explain.

Let w, w₁ be given vectors in R⁴ and deﬁne

M = {u = (u₁, u₂, u₃, u₄) ∈ R⁴ : w·u = 0 and w₁ · u = 0}.

Is M a subspace? Explain.

Let M ={u = (u₁, u₂, u₃, u₄) ∈ R⁴ : |u₁| ≤ 4}. Is M a subspace? Explain.

Let M = {u = (u₁, u₂, u₃, u₄) ∈ R⁴ : sin (u₁) = 1}. Is M a subspace? Explain.

Study the deﬁnition of span. Explain what is meant by the span of a set of vectors. Include pictures.

Suppose {x₁,· · ·, xk} is a set of vectors from Fn. Show that span (x₁,· · ·, xk) contains 0.

Study the deﬁnition of linear independence. Explain in your own words what is meant by linear independence and linear dependence. Illustrate with pictures.

Use Corollary 8.4.17 to prove the following theorem: If A, B are n × n matrices and if AB = I, then BA = I and B = A⁻¹. Hint: First note that if AB = I, then it must be the case that A is onto. Explain why this requires span (columns of A) = Fn. Now explain why, using the corollary that this requires A to be one to one. Next explain why A (BA−I) = 0 and why the fact that A is one to one implies BA = I.

Here are three vectors. Determine whether they are linearly independent or linearly dependent.

Here are three vectors. Determine whether they are linearly independent or linearly dependent.

Here are three vectors. Determine whether they are linearly independent or linearly dependent.

Here are four vectors. Determine whether they span R³. Are these vectors linearly independent?

Here are four vectors. Determine whether they span R³. Are these vectors linearly independent?

Determine whether the following vectors are a basis for R³. If they are, explain why they are and if they are not, give a reason and tell whether they span R³.

Determine whether the following vectors are a basis for R³. If they are, explain why they are and if they are not, give a reason and tell whether they span R³.

Determine whether the following vectors are a basis for R³. If they are, explain why they are and if they are not, give a reason and tell whether they span R³.

Determine whether the following vectors are a basis for R³. If they are, explain why they are and if they are not, give a reason and tell whether they span R³.

Consider the vectors of the form

Is this set of vectors a subspace of R³? If so, explain why, give a basis for the subspace and find its dimension.

Consider the vectors of the form

Is this set of vectors a subspace of R⁴? If so, explain why, give a basis for the subspace and find its dimension.

Consider the vectors of the form

Is this set of vectors a subspace of R⁴? If so, explain why, give a basis for the subspace and find its dimension.

If you have 5 vectors in F⁵ and the vectors are linearly independent, can it always be concluded they span F⁵? Explain.

If you have 6 vectors in F⁵, is it possible they are linearly independent? Explain.

Suppose A is an m × n matrix and {w₁,· · ·, wk} is a linearly independent set of vectors in A (Fn) ⊆ Fm. Now suppose A (zi) = wi. Show {z₁,· · ·, zk} is also independent.

Suppose V, W are subspaces ofn. Show V ∩ W deﬁned to be all vectors which are in both V and W is a subspace also.

Suppose V and W both have dimension equal to 7 and they are subspaces of F¹⁰. What are the possibilities for the dimension of V ∩ W ? Hint: Remember that a linear independent set can be extended to form a basis.

Suppose V has dimension p and W has dimension q and they are each contained in a subspace, U which has dimension equal to n where n > max (p, q) . What are the possibilities for the dimension of V ∩ W ? Hint: Remember that a linear independent set can be extended to form a basis.

If b ≠ 0, can the solution set of Ax = b be a plane through the origin? Explain.

Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain. 

Suppose a system of linear equations has a 2 × 4 augmented matrix and the last column is a pivot column. Could the system of linear equations be consistent? Explain.

Suppose the coeﬃcient matrix of a system of n equations with n variables has the property that every column is a pivot column. Does it follow that the system of equations must have a solution? If so, must the solution be unique? Explain.

Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix.

State whether each of the following sets of data are possible for the matrix equation Ax = b. If possible, describe the solution set. That is, tell whether there exists a unique solution no solution or inﬁnitely many solutions.

(a) A is a 5 × 6 matrix, rank (A) = 4 and rank (A|b) = 4. Hint: This says b is in the span of four of the columns. Thus the columns are not independent.

(b) A is a 3 × 4 matrix, rank (A) = 3 and rank (A|b) = 2.

(c) A is a 4 × 2 matrix, rank (A) = 4 and rank (A|b) = 4. Hint: This says b is in the span of the columns and the columns must be independent.

(d) A is a 5 × 5 matrix, rank (A) = 4 and rank (A|b) = 5. Hint: This says b is not in the span of the columns.

(e) A is a 4× 2 matrix, rank (A) = 2 and rank (A|b) = 2.

Suppose A is an m × n matrix in which m ≤ n. Suppose also that the rank of A equals m. Show that A maps Fn onto Fm. Hint: The vectors e₁,· · ·, em occur as columns in the row reduced echelon form for A.

Suppose A is an m × n matrix in which m ≥ n. Suppose also that the rank of A equals n. Show that A is one to one. Hint: If not, there exists a vector x such that Ax = 0, and this implies at least one column of A is a linear combination of the others. Show this would require the column rank to be less than n.

Explain why an n × n matrix A is both one to one and onto if and only if its rank is n. 

Suppose A is an m × n matrix and B is an n × p matrix. Show that

dim (ker (AB)) ≤ dim (ker (A)) + dim (ker (B)).

Hint: Consider the subspace, B (Fp) ∩ ker (A) and suppose a basis for this subspace is

{w₁,· · ·, wk}.

Now suppose {u₁,· · ·, ur} is a basis for ker (B) . Let {z₁,· · ·, zk} be such that Bzi = wi and argue that

ker (AB) ⊆ span (u₁,· · ·, ur, z₁,· · ·, zk) .

Here is how you do this. Suppose ABx = 0. Then Bx ∈ ker (A)∩B (Fp) and so showing that

Explain why Ax = 0 always has a solution even when A⁻¹ does not exist.

(a) What can you conclude about A if the solution is unique?
(b) What can you conclude about A if the solution is not unique?

Suppose det (A−λI) = 0. Show using Theorem 9.2.9 there exists x ≠ 0 such that

(A−λI) x = 0.

Let A be an n × n matrix and let x be a nonzero vector such that Ax = λx for some scalar λ. When this occurs, the vector x is called an eigenvector and the scalar λ is called an eigenvalue. It turns out that not every number is an eigenvalue. Only certain ones are. Why? Hint: Show that if Ax = λx, then (A−λI) x = 0. Explain why this shows that (A−λI)is not one to one and not onto. Now use Theorem) 9.2.9 to argue det (A−λI) = 0. What sort of equation is this? How many solutions does it have?

Let m < n and let A be an m × n matrix. Show that A is not one to one. Hint: Consider the n × n matrix A₁ which is of the form

where the 0 denotes an (n−m) × n matrix of zeros. Thus det A₁ = 0 and so A₁ is not one to one. Now observe that A₁x is the vector

which equals zero if and only if Ax = 0. Do this using the Fredholm alternative.

Let A be an m × n real matrix and let b ∈ Rm. Show there exists a solution, x to the system 

ATAx = ATb

Next show that if x, x₁ are two solutions, then Ax = Ax₁. Hint: First show that (ATA)T = ATA. Next show if x ∈ ker (ATA), then Ax = 0. Finally apply the Fredholm alternative. This will give existence of a solution.

Show that in the context of Problem 54 that if x is the solution there, then |b−Ax| ≤ |b−Ay| for every y. Thus Ax is the point of A (Rn) which is closest to b of every point in A (Rn). 

Let A be an n × n matrix and consider the matrices {I, A, A²,· · ·, An²}. Explain why there exist scalars, ci not all zero such that

Then argue there exists a polynomial, p (λ) of the form

λm + dm−1λm−1 +· · · + d₁λ + d₀

such that p (A) = 0 and if q (λ) is another polynomial such that q (A) = 0, then q (λ) is of the form p (λ) l (λ) for some polynomial, l (λ) . This extra special polynomial, p (λ) is called the minimal polynomial. Hint: You might consider an n × n matrix as a vector in Fn².

¹More generally, a permutation matrix is a matrix which comes by permuting the rows of the identity matrix, not just switching two rows.

² Isn’t it amazing how many different words are available for use in linear algebra?

Linear Transformations

9.1 Linear Transformations

An m × n matrix can be used to transform vectors in Fn to vectors in Fm through the use of matrix multiplication.

Example 9.1.1 Consider the matrix  Think of it as a function which takes vectors in F³ and makes them in to vectors in F² as follows. For  a vector in F³, multiply on the left by the given matrix to obtain the vector in F². Here are some numerical examples.

More generally,

The idea is to deﬁne a function which takes vectors in F³ and delivers new vectors in F².

This is an example of something called a linear transformation.

Deﬁnition 9.1.2 Let T : Fn Fm be a function. Thus for each x ∈ Fn, T x ∈ Fm. Then T is a linear transformation if whenever α, β are scalars and x₁ and x₂ are vectors in Fn,

T (αx₁ + βx₂) = α₁T x₁ + βT x₂.

A linear transformation is also called a homomorphism. In the case that T is in addition to this one to one and onto, it is sometimes called an isomorphism.

The last two terms are typically used more in abstract algebra than in linear algebra so in this book, such mappings will be referred to as linear transformations. In sloppy language, it distributes across vector addition and you can factor out the scalars.

In words, linear transformations distribute across + and allow you to factor out scalars. At this point, recall the properties of matrix multiplication. The pertinent property is (5.14) on Page 85. Recall it states that for a and b scalars,

A (aB + bC) = aAB + bAC

In particular, for A an m × n matrix and B and C, n × 1 matrices (column vectors) the above formula holds which is nothing more than the statement that matrix multiplication gives an example of a linear transformation.

The reason this concept is so important is there are many examples of things which are linear transformations. You might remember from calculus that the operator which consists of taking the derivative is a linear transformation. That is, if f, g are functions (vectors) and α, β are numbers (scalars)

Another example of a linear transformation is that of rotation through an angle. For example, I may want to rotate every vector through an angle of 45 degrees. Such a rotation would achieve something like the following if applied to each vector corresponding to points on the picture which is standing upright.

More generally, denote a rotation by T. Why is such a transformation linear? Consider the following picture which illustrates a rotation.

To get T (a + b) , you can add Ta and Tb. Here is why. If you add Ta to Tb you get the diagonal of the parallelogram determined by Ta and Tb. This diagonal also results from rotating the diagonal of the parallelogram determined by a and b. This is because the rotation preserves all angles between the vectors as well as their lengths. In particular, it preserves the shape of this parallelogram. Thus both Ta + Tb and T(a + b) give the same directed line segment. Thus T distributes across + where + refers to vector addition. Similarly, if k is a number Tka = kTa (draw a picture) and so you can factor out scalars also. Thus rotations are an example of a linear transformation.

Definition 9.1.3 A linear transformation is called one to one (often written as 1 − 1) if it never takes two dierent vectors to the same vector. Thus T is one to one if whenever x  y

Tx Ty.

Equivalently, if T ( x ) = T ( y ) , then x = y.

In the case that a linear transformation comes from matrix multiplication, it is common usage to refer to the matrix as a one to one matrix when the linear transformation it determines is one to one.

Deﬁnition 9.1.4 A linear transformation mapping Fn to Fm is called onto if whenever y ∈ Fm there exists x ∈ Fn such that T ( x ) = y.

Thus T is onto if everything in Fm gets hit. In the case that a linear transformation comes from matrix multiplication, it is common to refer to the matrix as onto when the linear transformation it determines is onto. Also it is common usage to write TFn, T (Fn) ,or Im (T ) as the set of vectors of Fm which are of the form Tx for some x ∈ Fn. In the case that T is obtained from multiplication by an m × n matrix A, it is standard to simply write A (Fn), AFn, or Im (A) to denote those vectors in Fm which are obtained in the form Ax for some x ∈ Fn.

9.2 Constructing The Matrix Of A Linear Transformation

It turns out that if T is any linear transformation which maps Fn to Fm, there is always an m × n matrix A with the property that

| Ax = Tx | (9.1)

for all x ∈ Fn. Here is why. Suppose T : Fn Fm is a linear transformation and you want to find the matrix deﬁned by this linear transformation as described in (9.1). Then if x ∈ Fn it follows

where ei is the vector which has zeros in every slot but the ith and a 1 in this slot. Then since T is linear,

and so you see that the matrix desired is obtained from letting the ith column equal T (ei) . We state this as the following theorem.

Theorem 9.2.1 Let T be a linear transformation from Fn to Fm. Then the matrix A satisfying (9.1) is given by

9.2.1 Rotations in R²

Sometimes you need to find a matrix which represents a given linear transformation which is described in geometrical terms. The idea is to produce a matrix which you can multiply a vector by to get the same thing as some geometrical description. A good example of this is the problem of rotation of vectors discussed above. Consider the problem of rotating through an angle of .

Example 9.2.2 Determine the matrix which represents the linear transformation defined by rotating every vector through an angle of .

Let e₁ These identify the geometric vectors which point along the positive x axis and positive y axis as shown.

From the above, you only need to find Te₁ and Te₂, the first being the first column of the desired matrix A and the second being the second column. From the deﬁnition of the cos, sin the coordinates of T (e₁) are as shown in the picture. The coordinates of T (e₂) also follow from simple trigonometry. Thus

Therefore, from Theorem 9.2.1,

For those who prefer a more algebraic approach, the deﬁnition of (cos () , sin ()) is as the x and y coordinates of the point (1, 0) . Now the point of the vector from (0, 0) to (0, 1), e₂ is exactly π/2 further along along the unit circle. Therefore, when it is rotated through an angle of the x and y coordinates are given by

Example 9.2.3 Find the matrix of the linear transformation which is obtained by first rotating all vectors through an angle of ϕ and then through an angle . Thus you want the linear transformation which rotates all angles through an angle of ϕ.

Let T+ϕ denote the linear transformation which rotates every vector through an angle of + ϕ. Then to get T+ϕ, you could first do Tϕ and then do T where Tϕ is the linear transformation which rotates through an angle of ϕ and T is the linear transformation which rotates through an angle of . Denoting the corresponding matrices by A+ϕ, Aϕ, and A, you must have for every x

A+ϕx = T+ϕx = TTϕx = AAϕx.

Consequently, you must have

You know how to multiply matrices. Do so to the pair on the right. This yields

Don’t these look familiar? They are the usual trig. identities for the sum of two angles derived here using linear algebra concepts.

You do not have to stop with two dimensions. You can consider rotations and other geometric concepts in any number of dimensions. This is one of the major advantages of linear algebra. You can break down a diﬃcult geometrical procedure into small steps, each corresponding to multiplication by an appropriate matrix. Then by multiplying the matrices, you can obtain a single matrix which can give you numerical information on the results of applying the given sequence of simple procedures. That which you could never visualize can still be understood to the extent of finding exact numerical answers. Another example follows.

Example 9.2.4 Find the matrix of the linear transformation which is obtained by first rotating all vectors through an angle of π/6 and then reﬂecting through the x axis.

As shown in Example 9.2.3, the matrix of the transformation which involves rotating through an angle of π/6 is

The matrix for the transformation which reﬂects all vectors through the x axis is

Therefore, the matrix of the linear transformation which first rotates through π/6 and then reﬂects through the x axis is

9.2.2 Rotations About A Particular Vector

The problem is to find the matrix of the linear transformation which rotates all vectors about a given unit vector u which is possibly not one of the coordinate vectors i, j, or k. Suppose for |c|1

First I will produce a matrix which maps u to k such that the right handed rotation about k corresponds to the right handed rotation about u. Then I will rotate about k and finally, I will multiply by the inverse of the first matrix to get the desired result.

To begin, find vectors w, v such that w × v = u. Let

This vector is clearly perpendicular to u. Then v = (a, b, c) × w ≡ u × w. Thus from the geometric description of the cross product, w × v = u. Computing the cross product gives

Now I want to have Tw = i, Tv = j, Tu = k. What does this? It is the inverse of the matrix which takes i to w, j to v, and k to u. This matrix is

Its inverse is

Therefore, the matrix which does the rotating is

This yields a matrix whose columns are

Using the assumption that u is a unit vector so that a²+b²+c² = 1, it follows the desired matrix is

This was done under the assumption that |c|1. However, if this condition does not hold, you can verify directly that the above still gives the correct answer.

9.2.3 Projections

In Physics it is important to consider the work done by a force field on an object. This involves the concept of projection onto a vector. Suppose you want to find the projection of a vector v onto the given vector u, denoted by proju (v) This is done using the dot product as follows.

Because of properties of the dot product, the map vproju (v) is linear,

Example 9.2.5 Let the projection map be deﬁned above and let u = (1, 2, 3)T. Does this linear transformation come from multiplication by a matrix? If so, what is the matrix?

You can find this matrix in the same way as in the previous example. Let ei denote the vector in Rn which has a 1 in the ith position and a zero everywhere else. Thus a typical vector x = (x₁,· · ·, xn)T can be written in a unique way as

From the way you multiply a matrix by a vector, it follows that proju (ei) gives the ith column of the desired matrix. Therefore, it is only necessary to find

For the given vector in the example, this implies the columns of the desired matrix are

Hence the matrix is

9.2.4 Matrices Which Are One To One Or Onto

Lemma 9.2.6 Let A be an m × n matrix. Then A (Fn) = span (a₁,· · ·, an) where a₁,· · ·, an denote the columns of A. In fact, for x = (x₁,· · ·, xn)T,

Proof: This follows from the deﬁnition of matrix multiplication in Deﬁnition 5.1.9 on Page 80. 

The following is a theorem of major signiﬁcance. First here is an interesting observation.

Observation 9.2.7 Let A be an m×n matrix. Then A is one to one if and only if Ax = 0 implies x = 0.

Here is why: A0 = A (0 + 0) = A0 + A0 and so A0 = 0.

Now suppose A is one to one and Ax = 0. Then since A0 = 0, it follows x = 0. Thus if A is one to one and Ax = 0, then x = 0.

Next suppose the condition that Ax = 0 implies x = 0 is valid. Then if Ax = Ay, then A (x − y) = 0 and so from the condition, x − y = 0 so that x = y. Thus A is one to one.

Theorem 9.2.8 Suppose A is an n × n matrix. Then A is one to one if and only if A is onto. Also, if B is an n × n matrix and AB = I, then it follows BA = I.

Proof: First suppose A is one to one. Consider the vectors, {Ae₁,· · ·, Aen} where ek is the column vector which is all zeros except for a 1 in the kth position. This set of vectors is linearly independent because if

then since A is linear,

and since A is one to one, it follows

which implies each ck = 0. Therefore, {Ae₁,· · ·, Aen} must be a basis for Fn by Corollary 8.4.17 on Page 150. It follows that for y ∈ Fn there exist constants, ci such that

showing that, since y was arbitrary, A is onto.

Next suppose A is onto. This implies the span of the columns of A equals Fn and by Corollary 8.4.17 this implies the columns of A are independent. If Ax = 0, then letting x = (x₁,· · ·, xn)T, it follows

and so each xi = 0. If Ax = Ay, then A (x − y) = 0 and so x = y. This shows A is one to one.

Now suppose AB = I. Why is BA = I? Since AB = I it follows B is one to one since otherwise, there would exist, x ≠ 0 such that Bx = 0 and then ABx = A0 = 0 ≠ Ix. Therefore, from what was just shown, B is also onto. In addition to this, A must be one to one because if Ay = 0, then y = Bx for some x and then x = ABx = Ay = 0 showing y = 0. Now from what is given to be so, it follows (AB) A = A and so using the associative law for matrix multiplication,

A (BA) − A = A (BA − I) = 0.

But this means (BA − I) x = 0 for all x since otherwise, A would not be one to one. Hence BA = I as claimed.

This theorem shows that if an n × n matrix B acts like an inverse when multiplied on one side of A it follows that B = A⁻¹and it will act like an inverse on both sides of A.

The conclusion of this theorem pertains to square matrices only. For example, let

| [picture] | (9.2)

Then

but

There is also an important characterization in terms of determinants. This is proved completely in the section on the mathematical theory of the determinant.

Theorem 9.2.9 Let A be an n × n matrix and let TA denote the linear transformation determined by A. Then the following are equivalent.

1. TA is one to one.

2. TA is onto.

3. det (A) ≠ 0.

9.2.5 The General Solution Of A Linear System

Recall the following deﬁnition which was discussed above.

Deﬁnition 9.2.10 T is a linear transformation if whenever x, y are vectors and a, b scalars,

| T (ax + by) = aTx + bTy. | (9.3)

Thus linear transformations distribute across addition and pass scalars to the outside. A linear system is one which is of the form

Tx = b.

If Txp = b, then xp is called a particular solution to the linear system. 

For example, if A is an m×n matrix and TA is determined by

TA (x) = Ax,

then from the properties of matrix multiplication, TA is a linear transformation. In this setting, we will usually write A for the linear transformation as well as the matrix. There are many other examples of linear transformations other than this. In differential equations, you will encounter linear transformations which act on functions to give new functions. In this case, the functions are considered as vectors. Don’t worry too much about this at this time. It will happen later. The fundamental idea is that something is linear if (9.3) holds and if whenever a, b are scalars and x, y are vectors ax + by is a vector. That is you can add vectors and multiply by scalars.

Deﬁnition 9.2.11 Let T be a linear transformation. Deﬁne

ker (T) ≡ {x : Tx = 0}.

In words, ker (T) is called the kernel of T . As just described, ker (T) consists of the set of all vectors which T sends to 0. This is also called the null space of T . It is also called the solution space of the equation Tx = 0.

The above deﬁnition states that ker (T) is the set of solutions to the equation,

Tx = 0.

In the case where T is really a matrix, you have been solving such equations for quite some time. However, sometimes linear transformations act on vectors which are not in Fn. There is more on this in Chapter 16 on Page 16 and this is discussed more carefully then. However, consider the following familiar example.

Example 9.2.12 Let denote the linear transformation deﬁned on X, the functions which are deﬁned on R and have a continuous derivative. Find ker

The example asks for functions, f which the property that As you know from calculus, these functions are the constant functions. Thus ker = constant functions.

When T is a linear transformation, systems of the form Tx = 0 are called homogeneous systems. Thus the solution to the homogeneous system is known as ker (T) .

Systems of the form T x = b where b ≠ 0 are called nonhomogeneous systems. It turns out there is a very interesting and important relation between the solutions to the homogeneous systems and the solutions to the nonhomogeneous systems.

Theorem 9.2.13 Suppose xp is a solution to the linear system,

Tx = b

Then if y is any other solution, there exists x ∈ ker (T) such that

y = xp + x.

Proof: Consider y − xp ≡ y+(−1) xp. Then T (y − xp) = Ty − Txp = b − b = 0. Let 

x ≡ y − xp.

Sometimes people remember the above theorem in the following form. The solutions to the nonhomogeneous system, Tx = b are given by xp + ker (T) where xp is a particular solution to Tx = b.

I have been vague about what T is and what x is on purpose. This theorem is completely algebraic in nature and will work whenever you have linear transformations. In particular, it will be important in differential equations. For now, here is a familiar example.

Example 9.2.14 Let

Find ker (A). Equivalently, find the solution space to the system of equations Ax = 0.

This asks you to find {x : Ax = 0}. In other words you are asked to solve the system, Ax = 0. Let x = (x, y, z, w)T. Then this amounts to solving

This is the linear system

x + 2y + 3z = 0
2x + y + z + 2w = 0
4x + 5y + 7z + 2w = 0

and you know how to solve this using row operations, (Gauss Elimination). Set up the augmented matrix

Then row reduce to obtain the row reduced echelon form,

This yields . Thus ker (A) consists of vectors of the form,

Example 9.2.15 The general solution of a linear system of equations is just the set of all solutions. Find the general solution to the linear system,

given that ( 1 1 2 1 )T = ( x y z w )T is one solution.

Note the matrix on the left is the same as the matrix in Example 9.2.14. Therefore, from Theorem 9.2.13, you will obtain all solutions to the above linear system in the form

9.3 Exercises 

Study the deﬁnition of a linear transformation. State it from memory.

Show the map T : Rn → Rm deﬁned by T (x) = Ax where A is an m × n matrix and x is an m × 1 column vector is a linear transformation.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/3.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/4.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of −π/3.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of 2π/3.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/12. Hint: Note that π/12 = π/3−π/4.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of 2π/3 and then reﬂects across the x axis.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/3 and then reﬂects across the x axis.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/4 and then reﬂects across the x axis.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of π/6 and then reﬂects across the x axis followed by a reﬂection across the y axis.

Find the matrix for the linear transformation which reﬂects every vector in R² across the x axis and then rotates every vector through an angle of π/4.

Find the matrix for the linear transformation which reﬂects every vector in R² across the y axis and then rotates every vector through an angle of π/4.

Find the matrix for the linear transformation which reﬂects every vector in R² across the x axis and then rotates every vector through an angle of π/6.

Find the matrix for the linear transformation which reﬂects every vector in R² across the y axis and then rotates every vector through an angle of π/6.

Find the matrix for the linear transformation which rotates every vector in R² through an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4.

Find the matrix of the linear transformation which rotates every vector in R³ counter clockwise about the z axis when viewed from the positive z axis through an angle of 30◦ and then reflects through the xy plane.

Find the matrix for proju ( v ) where u = (1,−2, 3)T.

Find the matrix for proju ( v ) where u = (1, 5, 3)T.

Find the matrix for proju ( v ) where u = (1, 0, 3)T.

Show that the function Tu deﬁned by Tu ( v ) ≡ v− proju ( v ) is also a linear transformation.

Show that

(v − proju (v) , u) ≡ (v − proju(v) , u) ≡ (v− proju (v)) · u = 0

and conclude every vector in Rn can be written as the sum of two vectors, one which is perpendicular and one which is parallel to the given vector.

Here are some descriptions of functions mappingn to Rn.

(a) T multiplies the jth component of x by a nonzero number b.

(b) T replaces the ith component of x with b times the jth component added to the ith component.

(c) T switches two components.

Show these functions are linear and describe their matrices.

In Problem 23, sketch the effects of the linear transformations on the unit square in R². Give a geometric description of an arbitrary invertible matrix in terms of products of matrices of these special matrices in Problem 23.

Let u = (a, b) be a unit vector in R². Find the matrix which reﬂects all vectors across this vector.

Hint: You might want to notice that (a, b) = (cos , sin ) for some . First rotate through −. Next reﬂect through the x axis which is easy. Finally rotate through .

Let u be a unit vector. Show the linear transformation of the matrix I − 2uuT preserves all distances and satisﬁes

(I — 2uuT)T = (I − 2uuT) I.

This matrix is called a Householder reﬂection. More generally, any matrix Q which satisﬁes QTQ = QQT is called an orthogonal matrix. Show the linear transformation determined by an orthogonal matrix always preserves the length of a vector in Rn. Hint: First either recall, depending on whether you have done Problem 51 on Page 97 or show that for any matrix A,

Suppose |x| = |y| for x, y ∈ Rn. The problem is to find an orthogonal transformation Q, (see Problem 26) which has the property that Qx = y and Qy = x. Show

does what is desired.

Let a be a fixed vector. The function Ta deﬁned by Tav = a + v has the effect of translating all vectors by adding a. Show this is not a linear transformation. Explain why it is not possible to realize Ta in R³ by multiplying by a 3 × 3 matrix.

In spite of Problem 28 we can represent both translations and rotations by matrix multiplication at the expense of using higher dimensions. This is done by the homogeneous coordinates.

I will illustrate in R³ where most interest in this is found. For each vector v = (v₁, v₂, v₃)T, consider the vector in R⁴ (v₁, v₂, v₃, 1)T. What happens when you do

Describe how to consider both rotations and translations all at once by forming appropriate 4 × 4 matrices.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 30 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 32 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 34 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 36 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 38 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 40 find the general solution to the following linear system.

Give an example of a 3 × 2 matrix with the property that the linear transformation determined by this matrix is one to one but not onto.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 43 find the general solution to the following linear system.

Write the solution set of the following system as the span of vectors and find a basis for the solution space of the following system.

Using Problem 45 find the general solution to the following linear system.

Find Ker (A) for

Recall ker (A) is just the set of solutions to Ax = 0. It is the solution space to the system Ax = 0.

Using Problem 47, find the general solution to the following linear system.

Using Problem 47, find the general solution to the following linear system.

Suppose Ax = b has a solution. Explain why the solution is unique precisely when Ax = 0 has only the trivial (zero) solution.

Show that if A is an m × n matrix, then ker (A) is a subspace.

Verify the linear transformation determined by the matrix of (9.2) maps R³ onto R² but the linear transformation determined by this matrix is not one to one.

The LU Factorization

10.1 Deﬁnition Of An LU factorization

An LU factorization of a matrix involves writing the given matrix as the product of a lower triangular matrix which has the main diagonal consisting entirely of ones L, and an upper triangular matrix U in the indicated order. This is the version discussed here but it is sometimes the case that the L has numbers other than 1 down the main diagonal. It is still a useful concept. The L goes with “lower” and the U with “upper”. It turns out many matrices can be written in this way and when this is possible, people get excited about slick ways of solving the system of equations, Ax = y. It is for this reason that you want to study the LU factorization. It allows you to work only with triangular matrices. It turns out that it takes about half as many operations to obtain an LU factorization as it does to find the row reduced echelon form.

First it should be noted not all matrices have an LU factorization and so we will emphasize the techniques for achieving it rather than formal proofs.

Example 10.1.1 Can you write in the form LU as just described?

To do so you would need

Therefore, b = 1 and a = 0. Also, from the bottom rows, xa = 1 which can’t happen and have a = 0. Therefore, you can’t write this matrix in the form LU. It has no LU factorization. This is what we mean above by saying the method lacks generality.

10.2 Finding An LU Factorization By Inspection

Which matrices have an LU factorization? It turns out it is those whose row reduced echelon form can be achieved without switching rows and which only involve row operations of type 3 in which row j is replaced with a multiple of row i added to row j for i < j.

Example 10.2.1 Find an LU factorization of 

One way to find the LU factorization is to simply look for it directly. You need

Then multiplying these you get

and so you can now tell what the various quantities equal. From the first column, you need a = 1, x = 1, y = 2. Now go to the second column. You need d = 2, xd + b = 3 so b = 1, yd + zb = 3 so z = −1. From the third column, h = 0, e = 2, c = 6. Now from the fourth column, j = 2, i = −1, f = −5. Therefore, an LU factorization is

You can check whether you got it right by simply multiplying these two.

10.3 Using Multipliers To Find An LU Factorization

There is also a convenient procedure for finding an LU factorization. It turns out that it is only necessary to keep track of the multipliers which are used to row reduce to upper triangular form. This procedure is described in the following examples.

Example 10.3.1 Find an LU factorization for

Write the matrix next to the identity matrix as shown.

The process involves doing row operations to the matrix on the right while simultaneously updating successive columns of the matrix on the left. First take −2 times the first row and add to the second in the matrix on the right.

Note the way we updated the matrix on the left. We put a 2 in the second entry of the first column because we used −2 times the first row added to the second row. Now replace the third row in the matrix on the right by −1 times the first row added to the third. Notice that the product of the two matrices is unchanged and equals the original matrix. This is because a row operation was done on the original matrix to get the matrix on the right and then on the left, it was multiplied by an elementary matrix which “undid” the row operation which was done.

The next step is

Again, the product is unchanged because we just did and then undid a row operation. Finally, we will add the second row to the bottom row and make the following changes

At this point, we stop because the matrix on the right is upper triangular. An LU factorization is the above.

The justiﬁcation for this gimmick will be given later in a more general context.

Example 10.3.2 Find an LU factorization for

We will use the same procedure as above. However, this time we will do everything for one column at a time. First multiply the first row by (−1) and then add to the last row. Next take (−2) times the first and add to the second and then (−2) times the first and add to the third.

This finishes the first column of L and the first column of U. As in the above, what happened was this. Lots of row operations were done and then these were undone by multiplying by the matrix on the left. Thus the above product equals the original matrix. Now take −(1/4) times the second row in the matrix on the right and add to the third followed by −(1/2) times the second added to the last.

This finishes the second column of L as well as the second column of U. Since the matrix on the right is upper triangular, stop. The LU factorization has now been obtained. This technique is called Dolittle’s method.

This process is entirely typical of the general case. The matrix U is just the first upper triangular matrix you come to in your quest for the row reduced echelon form using only the row operation which involves replacing a row by itself added to a multiple of another row. The matrix L is what you get by updating the identity matrix as illustrated above.

You should note that for a square matrix, the number of row operations necessary to reduce to LU form is about half the number needed to place the matrix in row reduced echelon form. This is why an LU factorization is of interest in solving systems of equations.

10.4 Solving Systems Using The LU Factorization

One reason people care about the LU factorization is it allows the quick solution of systems of equations. Here is an example.

Example 10.4.1 Suppose you want to find the solutions to

Of course one way is to write the augmented matrix and grind away. However, this involves more row operations than the computation of the LU factorization and it turns out that the LU factorization can give the solution quickly. Here is how. The following is an LU factorization for the matrix.

Let Ux = y and consider Ly = b where in this case, b = (1, 2, 3)T . Thus

which yields very quickly that Now you can find x by solving Ux = y. Thus in this case,

which yields

10.5 Justiﬁcation For The Multiplier Method

Why does the multiplier method work for finding the LU factorization? Suppose A is a matrix which has the property that the row reduced echelon form for A may be achieved using only the row operations which involve replacing a row with itself added to a multiple of another row. It is not ever necessary to switch rows. Thus every row which is replaced using this row operation in obtaining the echelon form may be modiﬁed by using a row which is above it.

Lemma 10.5.1 Let L be a lower (upper) triangular matrix m × m which has ones down the main diagonal. Then L⁻¹ also is a lower (upper) triangular matrix which has ones down the main diagonal. Also L⁻¹ is obtained from L by simply multiplying each entry below the main diagonal in L with −1.

Proof: Consider the usual setup for finding the inverse ( L I ). Then each row operation done to L to reduce to row reduced echelon form results in changing only the entries in I below the main diagonal and also the resulting entry on the right of the above m × 2m matrix below the main diagonal is just −1 times the corresponding entry in L. 

Now let A be an m × n matrix, say

and assume A can be row reduced to an upper triangular form using only row operation 3. Thus, in particular, a₁₁≠ 0. Multiply on the left by E₁ =

This is the product of elementary matrices which make modiﬁcations in the first column only. It is equivalent to taking −a₂₁/a₁₁ times the first row and adding to the second. Then taking −a₃₁/a₁₁ times the first row and adding to the third and so forth. The quotients in the first column of the above matrix are the multipliers. Thus the result is of the form

By assumption, a′₂₂ ≠ 0 and so it is possible to use this entry to zero out all the entries below it in the matrix on the right by multiplication by a matrix of the form where E is an (m − 1) × (m − 1) matrix of the form

Again, the entries in the first column below the 1 are the multipliers. Continuing this way, zeroing out the entries below the diagonal entries, finally leads to

Em−1En−2· · ·E₁A = U

where U is upper triangular. Each Ej has all ones down the main diagonal and is lower triangular. Now multiply both sides by the inverses of the Ej in the reverse order. This yields

By Lemma 10.5.1, this implies that the product of those Ej⁻¹ is a lower triangular matrix having all ones down the main diagonal.

The above discussion and lemma gives the justiﬁcation for the multiplier method. The expressions

−a₂₁/a₁₁,−a₃₁/a₁₁,· · · −am1/a₁₁

denoted respectively by m₂₁,· · ·, mm1 to save notation which were obtained in building E₁ are the multipliers. . Then according to the lemma, to find E₁⁻¹ you simply write

Similar considerations apply to the other Ej⁻¹ Thus L is of the form

It follows from Theorem 8.1.6 about the effect of multiplying on the left by an elementary matrix that the above product is of the form

In words, beginning at the left column and moving toward the right, you simply insert, into the corresponding position in the identity matrix, −1 times the multiplier which was used to zero out an entry in that position below the main diagonal in A, while retaining the main diagonal which consists entirely of ones. This is L.

10.6 The PLU Factorization

As indicated above, some matrices don’t have an LU factorization. Here is an example.

| [picture] | (10.1)

In this case, there is another factorization which is useful called a PLU factorization. Here P is a permutation matrix.

Example 10.6.1 Find a PLU factorization for the above matrix in (10.1).

Proceed as before trying to find the row echelon form of the matrix. First add −1 times the first row to the second row and then add −4 times the first to the third. This yields

There is no way to do only row operations involving replacing a row with itself added to a multiple of another row to the matrix on the right in such a way as to obtain an upper triangular matrix.

Therefore, consider the original matrix with the bottom two rows switched.

Now try again with this matrix. First take −1 times the first row and add to the bottom row and then take −4 times the first row and add to the second row. This yields

The matrix on the right is upper triangular and so the LU factorization of the matrix M′ has been obtained above.

Thus M′ = PM = LU where L and U are given above. Notice that P² = I and therefore, M = P²M = PLU and so

This process can always be followed and so there always exists a PLU factorization of a given matrix even though there isn’t always an LU factorization.

Example 10.6.2 Use the PLU factorization of to solve the system Mx = b where b = (1, 2, 3)T.

Let Ux = y and consider PLy = b. In other words, solve,

Multiplying both sides by P gives

and so

Now Ux = y and so it only remains to solve

which yields

10.7 The QR Factorization

As pointed out above, the LU factorization is not a mathematically respectable thing because it does not always exist. There is another factorization which does always exist. Much more can be said about it than I will say here. I will only deal with real matrices and so the dot product will be the usual real dot product.

Deﬁnition 10.7.1 An n × n real matrix Q is called an orthogonal matrix if

QQT = QTQ = I.

Thus an orthogonal matrix is one whose inverse is equal to its transpose.

First note that if a matrix is orthogonal this says

Thus

This shows that orthogonal transformations preserve distances. You can show that if you have a matrix which does preserve distances, then it must be orthogonal also.

Example 10.7.2 One of the most important examples of an orthogonal matrix is the so called Householder matrix. You have v a unit vector and you form the matrix

I − 2vvT

This is an orthogonal matrix which is also symmetric. To see this, you use the rules of matrix operations.

so it is symmetric. Now to show it is orthogonal,

because vT v = v · v =|v|² = 1. Therefore, this is an example of an orthogonal matrix.

Consider the following problem.

Problem 10.7.3 Given two vectors x, y such that |x| = |y| ≠ 0 but x ≠ y and you want an orthogonal matrix Q such that Qx = y and Qy = x. The thing which works is the Householder matrix

Here is why this works.

Hence

Adding these equations, 2Qx = 2y and subtracting them yields 2Qy = 2x.

A picture of the geometric signiﬁcance follows.

The orthogonal matrix Q reﬂects across the dotted line taking x to y and y to x.

Deﬁnition 10.7.4 Let A be an m×n matrix. Then a QR factorization of A consists of two matrices, Q orthogonal and R upper triangular or in other words equal to zero below the main diagonal such that A = QR.

With the solution to this simple problem, here is how to obtain a QR factorization for any matrix A. Let

A = (a₁, a₂,· · ·, an)

where the ai are the columns. If a₁ = 0, let Q₁ = I. If a1 ≠ 0, let

and form the Householder matrix

As in the above problem Q₁a₁ = b and so

where A₂ is a m − 1 × n − 1 matrix. Now find in the same way as was just done a n − 1 × n − 1 matrix Q₂ such that

Let

Then

Continuing this way until the result is upper triangular, you get a sequence of orthogonal matrices QpQp−1· · ·Q₁ such that

| QpQp−1· · ·Q₁A = R | (10.2)

where R is upper triangular.

Now if Q₁ and Q₂ are orthogonal, then from properties of matrix multiplication,

and similarly

(Q₁Q₂)TQ₁Q₂ = I.

Thus the product of orthogonal matrices is orthogonal. Also the transpose of an orthogonal matrix is orthogonal directly from the deﬁnition. Therefore, from (10.2)

A = (QpQp−1· · ·Q₁)TR ≡QR,

where Q is orthogonal. This proves the following theorem.

Theorem 10.7.5 Let A be any real m × n matrix. Then there exists an orthogonal matrix Q and an upper triangular matrix R having nonnegative entries down the main diagonal such that

A = QR

and this factorization can be accomplished in a systematic manner.

10.8 Exercises

Find an LU factorization of

Find an LU factorization

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find an LU factorization of the matrix

Find the LU factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations.

x + 2y = 5

2x + 3y = 6

Find the LU factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations.

x + 2y + z = 1

y + 3z = 2

2x + 3y = 6

Find the LU factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations.

x + 2y + 3z = 5

2x + 3y + z = 6

x − y + z = 2

Find the LU factorization of the coefficient matrix using Dolittle’s method and use it to solve the system of equations.

x + 2y + 3z = 5

2x + 3y + z = 6

3x + 5y + 4z = 11

Is there only one LU factorization for a given matrix? Hint: Consider the equation

Look for all possible LU factorizations.

Find a PLU factorization of

Find a PLU factorization of

Find a PLU factorization of

Find a PLU factorization of  and use it to solve the systems

Find a PLU factorization of  and use it to solve the systems

Find a QR factorization for the matrix

Find a QR factorization for the matrix

If you had a QR factorization, A = QR, describe how you could use it to solve the equation Ax = b. This is not usually the way people solve this equation. However, the QR factorization is of great importance in certain other problems, especially in finding eigenvalues and eigenvectors.

Linear Programming

11.1 Simple Geometric Considerations

One of the most important uses of row operations is in solving linear program problems which involve maximizing a linear function subject to inequality constraints determined from linear equations. Here is an example. A certain hamburger store has 9000 hamburger patties to use in one week and a limitless supply of special sauce, lettuce, tomatoes, onions, and buns. They sell two types of hamburgers, the big stack and the basic burger. It has also been determined that the employees cannot prepare more than 9000 of either type in one week. The big stack, popular with the teenagers from the local high school, involves two patties, lots of delicious sauce, condiments galore, and a divider between the two patties. The basic burger, very popular with children, involves only one patty and some pickles and ketchup. Demand for the basic burger is twice what it is for the big stack. What is the maximum number of hamburgers which could be sold in one week given the above limitations?

Let x be the number of basic burgers and y the number of big stacks which could be sold in a week. Thus it is desired to maximize z = x + y subject to the above constraints. The total number of patties is 9000 and so the number of patty used is x + 2y. This number must satisfy x + 2y ≤ 9000 because there are only 9000 patty available. Because of the limitation on the number the employees can prepare and the demand, it follows 2x + y ≤ 9000. You never sell a negative number of hamburgers and so x, y ≥ 0. In simpler terms the problem reduces to maximizing z = x+y subject to the two constraints, x + 2y ≤ 9000 and 2x + y ≤ 9000. This problem is pretty easy to solve geometrically. Consider the following picture in which R labels the region described by the above inequalities and the line z = x + y is shown for a particular value of z.

As you make z larger this line moves away from the origin, always having the same slope and the desired solution would consist of a point in the region, R which makes z as large as possible or equivalently one for which the line is as far as possible from the origin. Clearly this point is the point of intersection of the two lines, (3000, 3000) and so the maximum value of the given function is 6000. Of course this type of procedure is fine for a situation in which there are only two variables but what about a similar problem in which there are very many variables. In reality, this hamburger store makes many more types of burgers than those two and there are many considerations other than demand and available patty. Each will likely give you a constraint which must be considered in order to solve a more realistic problem and the end result will likely be a problem in many dimensions, probably many more than three so your ability to draw a picture will get you nowhere for such a problem. Another method is needed. This method is the topic of this section. I will illustrate with this particular problem. Let x₁ = x and y = x₂. Also let x₃ and x₄ be nonnegative variables such that

x₁ + 2x₂ + x₃ = 9000, 2x₁ + x₂ + x₄ = 9000.

To say that x₃ and x₄ are nonnegative is the same as saying x₁ + 2x₂ ≤ 9000 and 2x₁ + x₂ ≤ 9000 and these variables are called slack variables at this point. They are called this because they “take up the slack”. I will discuss these more later. First a general situation is considered.

11.2 The Simplex Tableau

Here is some notation.

Deﬁnition 11.2.1 Let x, y be vectors in Rq. Then x ≤ y means for each i, xi ≤ yi.

The problem is as follows:

Let A be an m × (m + n) real matrix of rank m. It is desired to find x ∈ Rn+m such that x satisﬁes the constraints,

| x ≥ 0, Ax = b | (11.1)

and out of all such x,

is as large (or small) as possible. This is usually referred to as maximizing or minimizing z subject to the above constraints. First I will consider the constraints.

Let A = ( a₁ · · · an+m ). First you find a vector x⁰≥ 0, Ax⁰= b such that n of the components of this vector equal 0. Letting i₁,· · ·, in be the positions of x⁰ for which x⁰ = 0, suppose also that {aj₁ , · · · , ajm} is linearly independent for ji the other positions of x 0 i. Geometrically, this means that x⁰ is a corner of the feasible region, those x which satisfy the constraints. This is called a basic feasible solution. Also deﬁne

and

since The variables which are the components of the vector xB are called the basic variables and the variables which are the entries of xF are called the free variables. You set xF = 0. Now (x⁰, z⁰)T is a solution to

along with the constraints x ≥ 0. Writing the above in augmented matrix form yields

| [picture] | (11.2)

Permute the columns and variables on the left if necessary to write the above in the form

| [picture] | (11.3)

or equivalently in the augmented matrix form keeping track of the variables on the bottom as

| [picture] | (11.4)

Here B pertains to the variables xi₁,· · ·, xjm and is an m × m matrix with linearly independent columns, {aj₁,· · ·, ajm}, and F is an m × n matrix. Now it is assumed that

and since B is assumed to have rank m, it follows

| [picture] | (11.5)

This is very important to observe. B⁻¹b ≥ 0! This is by the assumption that x⁰ ≥ 0.

Do row operations on the top part of the matrix

| [picture] | (11.6)

and obtain its row reduced echelon form. Then after these row operations the above becomes

| [picture] | (11.7)

where B⁻¹b ≥ 0. Next do another row operation in order to get a 0 where you see a −cB. Thus

| [picture] | (11.8)

| [picture] | (11.9)

The reason there is a z⁰ on the bottom right corner is that is a solution of the system of equations represented by the above augmented matrix because it is a solution to the system of equations corresponding to the system of equations represented by (11.6) and row operations leave solution sets unchanged. Note how attractive this is. The z₀ is the value of z at the point x⁰. The augmented matrix of (11.9) is called the simplex tableau and it is the beginning point for the simplex algorithm to be described a little later. It is very convenient to express the simplex tableau in the above form in which the variables are possibly permuted in order to have on the left side. However, as far as the simplex algorithm is concerned it is not necessary to be permuting the variables in this manner. Starting with (11.9) you could permute the variables and columns to obtain an augmented matrix in which the variables are in their original order. What is really required for the simplex tableau?

It is an augmented m + 1× m + n + 2 matrix which represents a system of equations which has the same set of solutions, (x,z)T as the system whose augmented matrix is

(Possibly the variables for x are taken in another order.) There are m linearly independent columns in the first m + n columns for which there is only one nonzero entry, a 1 in one of the first m rows, the “simple columns”, the other first m + n columns being the “nonsimple columns”. As in the above, the variables corresponding to the simple columns are xB, the basic variables and those corresponding to the nonsimple columns are xF, the free variables. Also, the top m entries of the last column on the right are nonnegative. This is the description of a simplex tableau.

In a simplex tableau it is easy to spot a basic feasible solution. You can see one quickly by setting the variables, xF corresponding to the nonsimple columns equal to zero. Then the other variables, corresponding to the simple columns are each equal to a nonnegative entry in the far right column. Lets call this an “obvious basic feasible solution”. If a solution is obtained by setting the variables corresponding to the nonsimple columns equal to zero and the variables corresponding to the simple columns equal to zero this will be referred to as an “obvious” solution. Lets also call the first m + n entries in the bottom row the “bottom left row”. In a simplex tableau, the entry in the bottom right corner gives the value of the variable being maximized or minimized when the obvious basic feasible solution is chosen.

The following is a special case of the general theory presented above and shows how such a special case can be fit into the above framework. The following example is rather typical of the sorts of problems considered. It involves inequality constraints instead of Ax = b. This is handled by adding in “slack variables” as explained below.

The idea is to obtain an augmented matrix for the constraints such that obvious solutions are also feasible. Then there is an algorithm, to be presented later, which takes you from one obvious feasible solution to another until you obtain the maximum.

Example 11.2.2 Consider z = x₁ − x₂ subject to the constraints, x₁ + 2x₂ ≤ 10, x₁ + 2x₂ ≥ 2, and 2x₁ + x₂ ≤ 6, xi ≥ 0. Find a simplex tableau for a problem of the form x ≥ 0,Ax = b which is equivalent to the above problem.

You add in slack variables. These are positive variables, one for each of the first three constraints, which change the first three inequalities into equations. Thus the first three inequalities become x₁ + 2x₂ + x₃ = 10, x₁ + 2x₂ −x₄ = 2, and 2x₁ + x₂ + x₅ = 6, x₁, x₂, x₃, x₄, x₅ ≥ 0. Now it is necessary to find a basic feasible solution. You mainly need to find a positive solution to the equations,

x₁ + 2x₂ + x₃ = 10
x₁ + 2x₂ − x₄ = 2 .
2x₁ + x₂ + x₅ = 6

the solution set for the above system is given by

An easy way to get a basic feasible solution is to let x₄ = 8 and x₅ = 1. Then a feasible solution is

(x₁, x₂, x₃, x₄, x₅) = (0, 5, 0, 8, 1) .

It follows z⁰ = −5 and the matrix (11.2),with the variables kept track of on the bottom is

and the first thing to do is to permute the columns so that the list of variables on the bottom will have x₁ and x₃ at the end.

Next, as described above, take the row reduced echelon form of the top three lines of the above matrix. This yields

Now do row operations to

to finally obtain

and this is a simplex tableau. The variables are x₂, x₄, x₅, x₁, x₃, z.

It isn’t as hard as it may appear from the above. Lets not permute the variables and simply find an acceptable simplex tableau as described above.

Example 11.2.3 Consider z = x1 − x₂ subject to the constraints, x₁ + 2x₂ ≤ 10, x₁ + 2x₂ ≥ 2, and 2x₁ + x₂ ≤ 6, xi ≥ 0. Find a simplex tableau.

Adding in slack variables, an augmented matrix which is descriptive of the constraints is

The obvious solution is not feasible because of that -1 in the fourth column. When you let x₁, x₂ = 0, you end up having x₄ = −6 which is negative. Consider the second column and select the 2 as a pivot to zero out that which is above and below the 2.

This one is good. When you let x₁ = x₄ = 0, you find that x₂ = 3, x₃ = 4, x₅ = 3. The obvious solution is now feasible. You can now assemble the simplex tableau. The first step is to include a column and row for z. This yields

Now you need to get zeros in the right places so the simple columns will be preserved as simple columns in this larger matrix. This means you need to zero out the 1 in the third column on the bottom. A simplex tableau is now

Note it is not the same one obtained earlier. There is no reason a simplex tableau should be unique. In fact, it follows from the above general description that you have one for each basic feasible point of the region determined by the constraints.

11.3 The Simplex Algorithm

11.3.1 Maximums

The simplex algorithm takes you from one basic feasible solution to another while maximizing or minimizing the function you are trying to maximize or minimize. Algebraically, it takes you from one simplex tableau to another in which the lower right corner either increases in the case of maximization or decreases in the case of minimization.

I will continue writing the simplex tableau in such a way that the simple columns having only one entry nonzero are on the left. As explained above, this amounts to permuting the variables. I will do this because it is possible to describe what is going on without onerous notation. However, in the examples, I won’t worry so much about it. Thus, from a basic feasible solution, a simplex tableau of the following form has been obtained in which the columns for the basic variables, xB are listed first and b ≥ 0.

| [picture] | (11.10)

Then (x⁰, z⁰⁾ is a solution to the above system and since b ≥ 0 , it follows (x⁰, z⁰) is a basic feasible solution.

If ci < 0 for some i, and if Fji ≤ 0 so that a whole column of with the bottom entry < 0, then letting xi be the variable corresponding to that column, you could leave all the other entries of xF equal to zero but change xi to be positive. Let the new vector be denoted by x′F and letting x′B = b − F x′F it follows

Now this shows (x′B , x′F) is feasible whenever xi > 0 and so you could let xi become arbitrarily large and positive and conclude there is no maximum for z because

| z = (−ci) xi + z⁰ | (11.11)

If this happens in a simplex tableau, you can say there is no maximum and stop.

What if c ≥ 0? Then z = z⁰−cxF and to satisfy the constraints, you need xF ≥ 0. Therefore, in this case, z⁰ is the largest possible value of z and so the maximum has been found. You stop when this occurs. Next I explain what to do if neither of the above stopping conditions hold.

The only case which remains is that some ci < 0 and some Fji > 0. You pick a column in in which ci < 0, usually the one for which ci is the largest in absolute value. You pick Fji > 0 as a pivot element, divide the jth row by Fji and then use to obtain zeros above Fji and below Fji, thus obtaining a new simple column. This row operation also makes exactly one of the other simple columns into a nonsimple column. (In terms of variables, it is said that a free variable becomes a basic variable and a basic variable becomes a free variable.) Now permuting the columns and variables, yields

where z0′ ≥ z⁰ because . If b′ ≥ 0, you are in the same position you were at the beginning but now z⁰ is larger. Now here is the important thing. You don’t pick just any Fji when you do these row operations. You pick the positive one for which the row operation results in b′ ≥ 0. Otherwise the obvious basic feasible solution obtained by letting x′F = 0 will fail to satisfy the constraint that x ≥ 0.

How is this done? You need

| [picture] | (11.12)

for each k = 1,· · ·, m or equivalently,

| [picture] | (11.13)

Now if Fki ≤ 0 the above holds. Therefore, you only need to check Fpi for Fpi > 0. The pivot, Fji is the one which makes the quotients of the form

for all positive Fpi the smallest. This will work because for Fki> 0,

Having gotten a new simplex tableau, you do the same thing to it which was just done and continue. As long as b > 0, so you don’t encounter the degenerate case, the values for z associated with setting xF = 0 keep getting strictly larger every time the process is repeated. You keep going until you find c ≥ 0. Then you stop. You are at a maximum. Problems can occur in the process in the so called degenerate case when at some stage of the process some bj = 0. In this case you can cycle through different values for x with no improvement in z. This case will not be discussed here.

Example 11.3.1 Maximize 2x₁+3x₂ subject to the constraints x₁+x₂ ≥ 1, 2x₁+x₂ ≤ 6, x₁+2x₂ ≤ 6, x₁, x₂ ≥ 0.

The constraints are of the form

x₁ + x₂ − x₃ = 1
2x₁ + x₂ + x₄ = 6
x₁ + 2x₂ + x₅ = 6

where the x₃, x₄, x₅ are the slack variables. An augmented matrix for these equations is of the form

Obviously the obvious solution is not feasible. It results in x₃ < 0. We need to exchange basic variables. Lets just try something.

Now this one is all right because the obvious solution is feasible. Letting x₂ = x₃ = 0, it follows that the obvious solution is feasible. Now we add in the objective function as described above.

Then do row operations to leave the simple columns the same. Then

Now there are negative numbers on the bottom row to the left of the 1. Lets pick the first. (It would be more sensible to pick the second.) The ratios to look at are 5/1, 1/1 so pick for the pivot the 1 in the second column and first row. This will leave the right column above the lower right corner nonnegative. Thus the next tableau is

There is still a negative number there to the left of the 1 in the bottom row. The new ratios are 4/2, 5/1 so the new pivot is the 2 in the third column. Thus the next tableau is

Still, there is a negative number in the bottom row to the left of the 1 so the process does not stop yet. The ratios are 3/ (3/2) and 3/ (1/2) and so the new pivot is that 3/2 in the first column. Thus the new tableau is

Now stop. The maximum value is 10. This is an easy enough problem to do geometrically and so you can easily verify that this is the right answer. It occurs when x₄ = x₅ = 0, x₁ = 2, x₂ = 2, x₃ = 3.

11.3.2 Minimums

How does it differ if you are finding a minimum? From a basic feasible solution, a simplex tableau of the following form has been obtained in which the simple columns for the basic variables, xB are listed first and b ≥ 0.

| [picture] | (11.14)

Let for i = 1,· · ·, m and for i > m. Then (x⁰, z⁰) is a solution to the above system and since b ≥ 0, it follows (x⁰, z⁰) is a basic feasible solution. So far, there is no change.

Suppose first that some ci > 0 and Fji ≤ 0 for each j. Then let x′F consist of changing xi by making it positive but leaving the other entries of xF equal to 0. Then from the bottom row,

z = −cixi + z⁰

and you let x′B = b − Fx′F ≥ 0. Thus the constraints continue to hold when xi is made increasingly positive and it follows from the above equation that there is no minimum for z. You stop when this happens.

Next suppose c ≤ 0. Then in this case, z = z⁰ − cxF and from the constraints, xF ≥ 0 and so −cxF ≥ 0 and so z⁰ is the minimum value and you stop since this is what you are looking for.

What do you do in the case where some ci > 0 and some Fji > 0? In this case, you use the simplex algorithm as in the case of maximums to obtain a new simplex tableau in which z⁰′ is smaller. You choose Fji the same way to be the positive entry of the ith column such that bp/Fpi ≥ bj/Fji for all positive entries, Fpi and do the same row operations. Now this time,

As in the case of maximums no problem can occur and the process will converge unless you have the degenerate case in which some bj = 0. As in the earlier case, this is most unfortunate when it occurs. You see what happens of course. z⁰ does not change and the algorithm just delivers different values of the variables forever with no improvement.

To summarize the geometrical signiﬁcance of the simplex algorithm, it takes you from one corner of the feasible region to another. You go in one direction to find the maximum and in another to find the minimum. For the maximum you try to get rid of negative entries of c and for minimums you try to eliminate positive entries of c, where the method of elimination involves the auspicious use of an appropriate pivot element and row operations.

Now return to Example 11.2.2. It will be modiﬁed to be a maximization problem.

Example 11.3.2 Maximize z = x₁ − x₂ subject to the constraints,

x₁ + 2x₂ ≤ 10, x₁ + 2x₂ ≥ 2,

and 2x₁ + x₂ ≤ 6, xi ≥ 0.

Recall this is the same as maximizing z = x₁ − x₂ subject to

the variables, x₃, x₄, x₅ being slack variables. Recall the simplex tableau was

with the variables ordered as x₂, x₄, x₅, x₁, x₃ and so xB = (x₂, x₄, x₅) and

xF = (x₁, x₃) .

Apply the simplex algorithm to the fourth column because and this is the most negative entry in the bottom row. The pivot is 3/2 because 1/(3/2) = 2/3 < 5/ (1/2) . Dividing this row by 3/2 and then using this to zero out the other elements in that column, the new simplex tableau is

Now there is still a negative number in the bottom left row. Therefore, the process should be continued. This time the pivot is the 2/3 in the top of the column. Dividing the top row by 2/3 and then using this to zero out the entries below it,

Now all the numbers on the bottom left row are nonnegative so the process stops. Now recall the variables and columns were ordered as x₂, x₄, x₅, x₁, x₃. The solution in terms of x₁ and x₂ is x₂ = 0 and x₁ = 3 and z = 3. Note that in the above, I did not worry about permuting the columns to keep those which go with the basic variables on the left.

Here is a bucolic example.

Example 11.3.3 Consider the following table.

This information is available to a pig farmer and Fi denotes a particular feed. The numbers in the table contain the number of units of a particular nutrient contained in one pound of the given feed. Thus F₂ has 2 units of iron in one pound. Now suppose the cost of each feed in cents per pound is given in the following table.

A typical pig needs 5 units of iron, 8 of protein, 6 of folic acid, 7 of copper and 4 of calcium. (The units may change from nutrient to nutrient.) How many pounds of each feed per pig should the pig farmer use in order to minimize his cost?

His problem is to minimize C ≡ 2x₁ + 3x₂ + 2x₃ + 3x₄ subject to the constraints

x₁ + 2x₂ + x₃ + 3x₄ ≥ 5,
5x₁ + 3x₂ + 2x₃ + x₄ ≥ 8,
x₁ + 2x₂ + 2x₃ + x₄ ≥ 6,
2x₁ + x₂ + x₃ + x₄ ≥ 7,
x₁ + x₂ + x₃ + x₄ ≥ 4.

where each xi ≥ 0. Add in the slack variables,

x₁ + 2x₂ + x₃ + 3x₄ − x₅ = 5
5x₁ + 3x₂ + 2x₃ + x₄ − x₆ = 8
x₁ + 2x₂ + 2x₃ + x₄ − x₇ = 6
2x₁ + x₂ + x₃ + x₄ − x₈ = 7
x₁ + x₂ + x₃ + x₄ − x₉ = 4

The augmented matrix for this system is

How in the world can you find a basic feasible solution? Remember the simplex algorithm is designed to keep the entries in the right column nonnegative so you use this algorithm a few times till the obvious solution is a basic feasible solution.

Consider the first column. The pivot is the 5. Using the row operations described in the algorithm, you get

Now go to the second column. The pivot in this column is the 7 /5. This is in a different row than the pivot in the first column so I will use it to zero out everything below it. This will get rid of the zeros in the fifth column and introduce zeros in the second. This yields

Now consider another column, this time the fourth. I will pick this one because it has some negative numbers in it so there are fewer entries to check in looking for a pivot. Unfortunately, the pivot is the top 2 and I don’t want to pivot on this because it would destroy the zeros in the second column. Consider the fifth column. It is also not a good choice because the pivot is the second element from the top and this would destroy the zeros in the first column. Consider the sixth column. I can use either of the two bottom entries as the pivot. The matrix is

Next consider the third column. The pivot is the 1 in the third row. This yields

There are still 5 columns which consist entirely of zeros except for one entry. Four of them have that entry equal to 1 but one still has a -1 in it, the -1 being in the fourth column. I need to do the row operations on a nonsimple column which has the pivot in the fourth row. Such a column is the second to the last. The pivot is the 3. The new matrix is

| [picture] | (11.15)

Now the obvious basic solution is feasible. You let x₄ = 0 = x₅ = x₇ = x₈ and x₁ = 8/3, x₂ = 2/3, x₃ = 1, and x₆ = 28/3. You don’t need to worry too much about this. It is the above matrix which is desired. Now you can assemble the simplex tableau and begin the algorithm. Remember C ≡ 2x₁ + 3x₂ + 2x₃ + 3x₄. First add the row and column which deal with C. This yields

| [picture] | (11.16)

Now you do row operations to keep the simple columns of (11.15) simple in (11.16). Of course you could permute the columns if you wanted but this is not necessary.

This yields the following for a simplex tableau. Now it is a matter of getting rid of the positive entries in the bottom row because you are trying to minimize.

The most positive of them is the 2/3 and so I will apply the algorithm to this one first. The pivot is the 7/3. After doing the row operation the next tableau is

and you see that all the entries are negative and so the minimum is 64/7 and it occurs when x₁ = 18/7, x₂ = 0, x₃ = 11/7, x₄ = 2/7.

There is no maximum for the above problem. However, I will pretend I don’t know this and attempt to use the simplex algorithm. You set up the simiplex tableau the same way. Recall it is

Now to maximize, you try to get rid of the negative entries in the bottom left row. The most negative entry is the -1 in the fifth column. The pivot is the 1 in the third row of this column. The new tableau is

Consider the fourth column. The pivot is the top 1/3. The new tableau is

There is still a negative in the bottom, the -4. The pivot in that column is the 3. The algorithm yields

Note how z keeps getting larger. Consider the column having the −13/3 in it. The pivot is the single positive entry, 1/3. The next tableau is

There is a column consisting of all negative entries. There is therefore, no maximum. Note also how there is no way to pick the pivot in that column.

Example 11.3.4 Minimize z = x₁ − 3x₂ + x₃ subject to the constraints x₁ + x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 2, x₁ + x₂ + 3x₃ ≤ 8 and x₁ + 2x₂ + x₃ ≤ 7 with all variables nonnegative.

There exists an answer because the region deﬁned by the constraints is closed and bounded. Adding in slack variables you get the following augmented matrix corresponding to the constraints.

Of course there is a problem with the obvious solution obtained by setting to zero all variables corresponding to a nonsimple column because of the simple column which has the−1 in it. Therefore, I will use the simplex algorithm to make this column non simple. The third column has the 1 in the second row as the pivot so I will use this column. This yields

| [picture] | (11.17)

and the obvious solution is feasible. Now it is time to assemble the simplex tableau. First add in the bottom row and second to last column corresponding to the equation for z. This yields

Next you need to zero out the entries in the bottom row which are below one of the simple columns in (11.17). This yields the simplex tableau

The desire is to minimize this so you need to get rid of the positive entries in the left bottom row. There is only one such entry, the 4. In that column the pivot is the 1 in the second row of this column. Thus the next tableau is

There is still a positive number there, the 3. The pivot in this column is the 2. Apply the algorithm again. This yields

Now all the entries in the left bottom row are nonpositive so the process has stopped. The minimum is −21/2. It occurs when x₁ = 0, x₂ = 7/2, x₃ = 0.

Now consider the same problem but change the word, minimize to the word, maximize.

Example 11.3.5 Maximize z = x₁ − 3x₂ + x₃ subject to the constraints x₁ + x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 2, x₁ + x₂ + 3x₃ ≤ 8 and x₁ + 2x₂ + x₃ ≤ 7 with all variables nonnegative.

The first part of it is the same. You wind up with the same simplex tableau,

but this time, you apply the algorithm to get rid of the negative entries in the left bottom row. There is a −1. Use this column. The pivot is the 3. The next tableau is

There is still a negative entry, the −2/3. This will be the new pivot column. The pivot is the 2/3 on the fourth row. This yields

and the process stops. The maximum for z is 7 and it occurs when x₁ = 13/2, x₂ = 0, x₃ = 1/2.

11.4 Finding A Basic Feasible Solution

By now it should be fairly clear that finding a basic feasible solution can create considerable difficulty. Indeed, given a system of linear inequalities along with the requirement that each variable be nonnegative, do there even exist points satisfying all these inequalities? If you have many variables, you can’t answer this by drawing a picture. Is there some other way to do this which is more systematic than what was presented above? The answer is yes. It is called the method of artiﬁcial variables. I will illustrate this method with an example.

Example 11.4.1 Find a basic feasible solution to the system 2x₁ + x₂ − x₃ ≥ 3, x₁ + x₂ + x₃ ≥ 2, x₁ + x₂ + x₃ ≤ 7 and x ≥ 0.

If you write the appropriate augmented matrix with the slack variables,

| [picture] | (11.18)

The obvious solution is not feasible. This is why it would be hard to get started with the simplex method. What is the problem? It is those −1 entries in the fourth and fifth columns. To get around this, you add in artiﬁcial variables to get an augmented matrix of the form

| [picture] | (11.19)

Thus the variables are x₁, x₂, x₃, x₄, x₅, x₆, x₇, x₈. Suppose you can find a feasible solution to the system of equations represented by the above augmented matrix. Thus all variables are nonnegative. Suppose also that it can be done in such a way that x₈ and x₇ happen to be 0. Then it will follow that x₁,· · ·, x₆ is a feasible solution for (11.18). Conversely, if you can find a feasible solution for (11.18), then letting x₇ and x₈ both equal zero, you have obtained a feasible solution to (11.19). Since all variables are nonnegative, x₇ and x₈ both equalling zero is equivalent to saying the minimum of z = x7 + x₈ subject to the constraints represented by the above augmented matrix equals zero. This has proved the following simple observation.

Observation 11.4.2 There exists a feasible solution to the constraints represented by the augmented matrix of (11.18) and x ≥ 0 if and only if the minimum of x₇+x₈ subject to the constraints of (11.19) and x ≥ 0 exists and equals 0.

Of course a similar observation would hold in other similar situations. Now the point of all this is that it is trivial to see a feasible solution to (11.19), namely x₆ = 7, x₇ = 3, x₈ = 2 and all the other variables may be set to equal zero. Therefore, it is easy to find an initial simplex tableau for the minimization problem just described. First add the column and row for z

Next it is necessary to make the last two columns on the bottom left row into simple columns. Performing the row operation, this yields an initial simplex tableau,

Now the algorithm involves getting rid of the positive entries on the left bottom row. Begin with the first column. The pivot is the 2. An application of the simplex algorithm yields the new tableau

Now go to the third column. The pivot is the 3/2 in the second row. An application of the simplex algorithm yields

| [picture] | (11.20)

and you see there are only nonpositive numbers on the bottom left column so the process stops and yields 0 for the minimum of z = x₇+ x₈. As for the other variables, x₁ = 5/3, x₂ = 0, x₃ = 1/3, x₄ = 0, x₅ = 0, x₆ = 5. Now as explained in the above observation, this is a basic feasible solution for the original system (11.18).

Now consider a maximization problem associated with the above constraints.

Example 11.4.3 Maximize x₁−x₂+2x₃ subject to the constraints, 2x₁+x₂−x₃ ≥ 3, x₁+x₂+x₃ ≥ 2, x₁ + x₂ + x₃ ≤ 7 and x ≥ 0.

From (11.20) you can immediately assemble an initial simplex tableau. You begin with the first 6 columns and top 3 rows in (11.20). Then add in the column and row for z. This yields

and you first do row operations to make the first and third columns simple columns. Thus the next simplex tableau is

You are trying to get rid of negative entries in the bottom left row. There is only one, the −5/3. The pivot is the 1. The next simplex tableau is then

and so the maximum value of z is 32/3 and it occurs when x₁ = 10/3, x₂ = 0 and x₃ = 11/3.

11.5 Duality

You can solve minimization problems by solving maximization problems. You can also go the other direction and solve maximization problems by minimization problems. Sometimes this makes things much easier. To be more speciﬁc, the two problems to be considered are

A.) Minimize z = cx subject to x ≥ 0 and Ax ≥ b and

B.) Maximize w = yb such that y ≥ 0 and yA ≤ c,

(equivalently ATyT ≥ cT and w = bTyT).

In these problems it is assumed A is an m × p matrix.

I will show how a solution of the first yields a solution of the second and then show how a solution of the second yields a solution of the first. The problems, A.) and B.) are called dual problems.

Lemma 11.5.1 Let x be a solution of the inequalities of A.) and let y be a solution of the inequalities of B.). Then

cx ≥ yb.

and if equality holds in the above, then x is the solution to A.) and y is a solution to B.).

Proof: This follows immediately. Since c ≥ yA,

cx ≥ yAx ≥ yb.

It follows from this lemma that if y satisﬁes the inequalities of B.) and x satisﬁes the inequalities of A.) then if equality holds in the above lemma, it must be that x is a solution of A.) and y is a solution of B.).

Now recall that to solve either of these problems using the simplex method, you first add in slack variables. Denote by x′ and y′ the enlarged list of variables. Thus x′ has at least m entries and so does y′ and the inequalities involving A were replaced by equalities whose augmented matrices were of the form

( A −I b ), and ( AT I cT )

Then you included the row and column for z and w to obtain

| [picture] | (11.21)

Then the problems have basic feasible solutions if it is possible to permute the first p + m columns in the above two matrices and obtain matrices of the form

| [picture] | (11.22)

where B, B₁ are invertible m × m and p × p matrices and denoting the variables associated with these columns by xB, yB and those variables associated with F or F₁ by xF and yF, it follows that letting BxB = b and xF = 0, the resulting vector x′ is a solution to x′ ≥ 0 and ( A −I )x′ = b with similar constraints holding for y′. In other words, it is possible to obtain simplex tableaus,

| [picture] | (11.23)

Similar considerations apply to the second problem. Thus as just described, a basic feasible solution is one which determines a simplex tableau like the above in which you get a feasible solution by setting all but the first m variables equal to zero. The simplex algorithm takes you from one basic feasible solution to another till eventually, if there is no degeneracy, you obtain a basic feasible solution which yields the solution of the problem of interest.

Theorem 11.5.2 Suppose there exists a solution, x to A.) where x is a basic feasible solution of the inequalities of A.). Then there exists a solution, y to B.) and cx = by. It is also possible to find y from x using a simple formula.

Proof: Since the solution to A.) is basic and feasible, there exists a simplex tableau like (11.23) such that x′ can be split into xB and xF such that xF = 0 and xB = B⁻¹b. Now since it is a minimizer, it follows cBB⁻¹F − cF ≤ 0 and the minimum value for cx is cBB⁻¹b. Stating this again, cx = cBB⁻¹b. Is it possible you can take y = cBB⁻¹? From Lemma 11.5.1 this will be so if cBB⁻¹ solves the constraints of problem B.). Is cBB⁻¹ ≥ 0? Is cBB⁻¹A ≤ c? These two conditions are satisﬁed if and only if cBB−1 ( A −I ) ≤ ( c 0 ). Referring to the process of permuting the columns of the first augmented matrix of (11.21) to get (11.22) and doing the same permutations on the columns of ( A −I ) and ( c 0 ), the desired inequality holds if and only if cBB⁻¹ ( B F ) ≤ (cB cF) which is equivalent to saying (cB cBB⁻¹F ) ≤ ( cB cF ) and this is true because cBB⁻¹F −cF ≤ 0 due to the assumption that x is a minimizer. The simple formula is just

y = cBB⁻¹.

The proof of the following corollary is similar.

Corollary 11.5.3 Suppose there exists a solution, y to B.) where y is a basic feasible solution of the inequalities of B.). Then there exists a solution, x to A.) and cx = by. It is also possible to find x from y using a simple formula. In this case, and referring to (11.23), the simple formula is x = B₁−TbB1.

As an example, consider the pig farmers problem. The main difficulty in this problem was finding an initial simplex tableau. Now consider the following example and marvel at how all the diﬃculties disappear.

Example 11.5.4 minimize C ≡ 2x₁ + 3x₂ + 2x₃ + 3x₄ subject to the constraints

where each xi ≥ 0. 

Here the dual problem is to maximize w = 5y₁ + 8y₂ + 6y₃ + 7y₄ + 4y₅ subject to the constraints

Adding in slack variables, these inequalities are equivalent to the system of equations whose augmented matrix is

Now the obvious solution is feasible so there is no hunting for an initial obvious feasible solution required. Now add in the row and column for w. This yields

It is a maximization problem so you want to eliminate the negatives in the bottom left row. Pick the column having the one which is most negative, the −8. The pivot is the top 5. Then apply the simplex algorithm to obtain

There are still negative entries in the bottom left row. Do the simplex algorithm to the column which has the  The pivot is the This yields

and there are still negative numbers. Pick the column which has the −13/4. The pivot is the 3/8 in the top. This yields

which has only one negative entry on the bottom left. The pivot for this first column is the  The next tableau is

and all the entries in the left bottom row are nonnegative so the answer is 64 /7. This is the same as obtained before. So what values for x are needed? Here the basic variables are y₁, y₃, y₄, y₇. Consider the original augmented matrix, one step before the simplex tableau.

Permute the columns to put the columns associated with these basic variables first. Thus

The matrix B is

and so B−T equals

Also = ( 5 6 7 0 ) and so from Corollary 11.5.3,

which agrees with the original way of doing the problem.

Two good books which give more discussion of linear programming are Strang [15] and Nobel and Daniels [12]. Also listed in these books are other references which may prove useful if you are interested in seeing more on these topics. There is a great deal more which can be said about linear programming.

11.6 Exercises

Maximize and minimize z = x₁ −2x₂ + x₃ subject to the constraints x₁ + x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 2, and x₁ + 2x₂ + x₃ ≤ 7 if possible. All variables are nonnegative.

Maximize and minimize the following if possible. All variables are nonnegative.

(a) z = x₁ − 2x₂ subject to the constraints x₁ + x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 1, and x₁ + 2x₂ + x₃ ≤ 7

(b) z = x₁ − 2x₂ − 3x₃ subject to the constraints x₁ + x₂ + x₃ ≤ 8, x₁ + x₂ + 3x₃ ≥ 1, and x₁ + x₂ + x₃ ≤ 7

(c) z = 2x₁ + x₂ subject to the constraints x₁ −x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 1, and x₁ + 2x₂ + x₃ ≤ 7

(d) z = x₁ + 2x₂ subject to the constraints x₁ −x₂ + x₃ ≤ 10, x₁ + x₂ + x₃ ≥ 1, and x₁ + 2x₂ + x₃ ≤ 7

3. Consider contradictory constraints, x₁ + x₂ ≥ 12 and x₁ + 2x₂ ≤ 5, x₁ ≥ 0, x₂ ≥ 0. You know these two contradict but show they contradict using the simplex algorithm.

 4. Find a solution to the following inequalities for x, y ≥ 0 if it is possible to do so. If it is not possible, prove it is not possible.

5. Minimize z = x₁ + x₂ subject to x₁ + x₂ ≥ 2, x₁ + 3x₂ ≤ 20, x₁ + x₂ ≤ 18. Change to a maximization problem and solve as follows: Let yi = M − xi. Formulate in terms of y₁, y₂.

Spectral Theory

12.1 Eigenvalues And Eigenvectors Of A Matrix

Spectral Theory refers to the study of eigenvalues and eigenvectors of a matrix. It is of fundamental importance in many areas. Row operations will no longer be such a useful tool in this subject.

12.1.1 Denition Of Eigenvectors And Eigenvalues

In this section, F = C.

To illustrate the idea behind what will be discussed, consider the following example.

Example 12.1.1 Here is a matrix.

Multiply this matrix by the vector

and see what happens. Then multiply it by

and see what happens. Does this matrix act this way for some other vector?

First

Next

When you multiply the first vector by the given matrix, it stretched the vector, multiplying it by 10. When you multiplied the matrix by the second vector it sent it to the zero vector. Now consider

In this case, multiplication by the matrix did not result in merely multiplying the vector by a number.

In the above example, the first two vectors were called eigenvectors and the numbers, 10 and 0 are called eigenvalues. Not every number is an eigenvalue and not every vector is an eigenvector. When you have a nonzero vector which, when multiplied by a matrix results in another vector which is parallel to the first or equal to 0, this vector is called an eigenvector of the matrix. This is the meaning when the vectors are in Rn. Things are less apparent geometrically when the vectors are in Cn. The precise definition in all cases follows.

Definition 12.1.2  Let M be an n × n matrix and let x ∈ Cn be a nonzero vector for which

| Mx = λx | (12.1)

for some scalar λ. Then x is called an eigenvector and λ is called an eigenvalue (characteristic value) of the matrix M.

Note: Eigenvectors are never equal to zero!

The set of all eigenvalues of an n × n matrix M, is denoted by σ (M) and is referred to as the spectrum of M.

The eigenvectors of a matrix M are those vectors, x for which multiplication by M results in a vector in the same direction or opposite direction to x. Since the zero vector 0 has no direction this would make no sense for the zero vector. As noted above, 0 is never allowed to be an eigenvector. How can eigenvectors be identified? Suppose x satisfies (12.1). Then

(M − λI) x = 0

for some x ≠ 0. (Equivalently, you could write (λI −M) x = 0.) Sometimes we will use

(λI −M) x = 0

and sometimes (M − λI) x = 0. It makes absolutely no difference and you should use whichever you like better. Therefore, the matrix M − λI cannot have an inverse because if it did, the equation could be solved,

x = ( (M − λI)⁻¹ (M − λI) ) x = (M − λI)⁻¹ ((M − λI) x) = (M − λI)⁻¹ 0 = 0,

and this would require x = 0, contrary to the requirement that x ≠ 0. By Theorem 6.2.1 on Page 106,

| det (M − λI) = 0. | (12.2)

(Equivalently you could write det (λI −M) = 0.) The expression, det (λI − M) or equivalently, det (M − λI) is a polynomial called the characteristic polynomial and the above equation is called the characteristic equation. For M an n×n matrix, it follows from the theorem on expanding a matrix by its cofactor that det (M − λI) is a polynomial of degree n. As such, the equation (12.2) has a solution, λ ∈ C by the fundamental theorem of algebra. Is it actually an eigenvalue? The answer is yes, and this follows from Observation 9.2.7 on Page 171 along with Theorem 6.2.1 on Page 106. Since det (M − λI) = 0 the matrix det (M − λI) cannot be one to one and so there exists a nonzero vector x such that (M − λI) x = 0. This proves the following corollary.

Corollary 12.1.3 Let M be an n × n matrix and det (M − λI) = 0. Then there exists a nonzero vector x ∈ Cn such that (M − λI) x = 0.

12.1.2  Finding Eigenvectors And Eigenvalues

As an example, consider the following.

Example 12.1.4 Find the eigenvalues and eigenvectors for the matrix

You first need to identify the eigenvalues. Recall this requires the solution of the equation

det (A − λI) = 0.

In this case this equation is

When you expand this determinant and simplify, you find the equation you need to solve is

(λ − 5) ( λ2 − 20λ + 100 ) = 0

and so the eigenvalues are

5, 10, 10.

We have listed 10 twice because it is a zero of multiplicity two due to

λ² − 20λ + 100 = (λ − 10)² .

Having found the eigenvalues, it only remains to find the eigenvectors. First find the eigenvectors for λ = 5. As explained above, this requires you to solve the equation,

That is you need to find the solution to

By now this is an old problem. You set up the augmented matrix and row reduce to get the solution. Thus the matrix you must row reduce is

| [picture] | (12.3)

The row reduced echelon form is

and so the solution is any vector of the form

where t ∈ F. You would obtain the same collection of vectors if you replaced t with 4t. Thus a simpler description for the solutions to this system of equations whose augmented matrix is in (12.3) is

| [picture] | (12.4)

where t ∈ F. Now you need to remember that you can’t take t = 0 because this would result in the zero vector and

Eigenvectors are never equal to zero!

Other than this value, every other choice of z in (12.4) results in an eigenvector. It is a good idea to check your work! To do so, we will take the original matrix and multiply by this vector and see if we get 5 times this vector.

so it appears this is correct. Always check your work on these problems if you care about getting the answer right. 

The parameter, t is sometimes called a free variable. The set of vectors in (12.4) is called the eigenspace and it equals ker (A − λI) . You should observe that in this case the eigenspace has dimension 1 because the eigenspace is the span of a single vector. In general, you obtain the solution from the row echelon form and the number of different free variables gives you the dimension of the eigenspace. Just remember that not every vector in the eigenspace is an eigenvector. The vector 0 is not an eigenvector although it is in the eigenspace because

Eigenvectors are never equal to zero!

Next consider the eigenvectors for λ = 10. These vectors are solutions to the equation,

That is you must find the solutions to

which reduces to consideration of the augmented matrix

The row reduced echelon form for this matrix is

and so the eigenvectors are of the form

You can’t pick t and s both equal to zero because this would result in the zero vector and

Eigenvectors are never equal to zero!

However, every other choice of t and s does result in an eigenvector for the eigenvalue λ = 10. As in the case for λ = 5 you should check your work if you care about getting it right.

so it worked. The other vector will also work. Check it.

12.1.3     A Warning

The above example shows how to find eigenvectors and eigenvalues algebraically. You may have noticed it is a bit long. Sometimes students try to first row reduce the matrix before looking for eigenvalues. This is a because row operations destroy the eigenvalues. The eigenvalue problem is really not about row operations.

The general eigenvalue problem is the hardest problem in algebra and people still do research on ways to find eigenvalues and their eigenvectors. If you are doing anything which would yield a way to find eigenvalues and eigenvectors for general matrices without too much trouble, the thing you are doing will certainly be wrong. The problems you will see in this book are not too hard because they are cooked up to be easy. General methods to compute eigenvalues and eigenvectors numerically are presented later. These methods work even when the problem is not cooked up to be easy.

If you are so fortunate as to find the eigenvalues as in the above example, then finding the eigenvectors does reduce to row operations and this part of the problem is easy. However, finding the eigenvalues along with the eigenvectors is anything but easy because for an n × n matrix, it involves solving a polynomial equation of degree n. If you only find a good approximation to the eigenvalue, it won’t work. It either is or is not an eigenvalue and if it is not, the only solution to the equation, (M − λI) x = 0 will be the zero solution as explained above and

Eigenvectors are never equal to zero!

Here is another example

Example 12.1.5 Let

First find the eigenvalues.

This reduces to λ³ − 6λ² + 8λ = 0 and the solutions are 0, 2, and 4.

Now find the eigenvectors. For λ = 0 the augmented matrix for finding the solutions is

and the row reduced echelon form is

Therefore, the eigenvectors are of the form

where t ≠ 0.

Next find the eigenvectors for λ = 2. The augmented matrix for the system of equations needed to find these eigenvectors is

and the row reduced echelon form is

and so the eigenvectors are of the form

where t ≠ 0.

Finally find the eigenvectors for λ = 4. The augmented matrix for the system of equations needed to find these eigenvectors is

and the row reduced echelon form is

Therefore, the eigenvectors are of the form

where t ≠ 0.

12.1.4 Triangular Matrices

Although it is usually hard to solve the eigenvalue problem, there is a kind of matrix for which this is not the case. These are the upper or lower triangular matrices. I will illustrate by a examples.

Example 12.1.6 Let A = . Find its eigenvalues.

You need to solve

Thus the eigenvalues are just the diagonal entries of the original matrix. You can see it would work this way with any such matrix. These matrices are called upper triangular. Stated precisely, a matrix A is upper triangular if Aij = 0 for all i > j. Similarly, it is easy to find the eigenvalues for a lower triangular matrix, on which has all zeros above the main diagonal.

12.1.5 Defective And Nondefective Matrices

Deﬁnition 12.1.7 By the fundamental theorem of algebra, it is possible to write the characteristic equation in the form

(λ−λ₁)r1 (λ−λ₂)r2 · · ·(λ−λm)rm = 0

where ri is some integer no smaller than 1. Thus the eigenvalues are λ₁, λ₂,· · ·, λm. The algebraic multiplicity of λj is deﬁned to be rj.

Example 12.1.8 Consider the matrix

What is the algebraic multiplicity of the eigenvalue λ = 1?

In this case the characteristic equation is

det (A−λI) = (1−λ)³ = 0

or equivalently,

det (λI −A) = (λ− 1)³ = 0.

Therefore, λ is of algebraic multiplicity 3.

Deﬁnition 12.1.9 The geometric multiplicity of an eigenvalue is the dimension of the eigenspace,

ker (A−λI) .

Example 12.1.10 Find the geometric multiplicity of λ = 1 for the matrix in (12.5). 

We need to solve

The augmented matrix which must be row reduced to get this solution is therefore,

This requires z = y = 0 and x is arbitrary. Thus the eigenspace is

It follows the geometric multiplicity of λ = 1 is 1.

Deﬁnition 12.1.11 An n × n matrix is called defective if the geometric multiplicity is not equal to the algebraic multiplicity for some eigenvalue. Sometimes such an eigenvalue for which the geometric multiplicity is not equal to the algebraic multiplicity is called a defective eigenvalue. If the geometric multiplicity for an eigenvalue equals the algebraic multiplicity, the eigenvalue is sometimes referred to as nondefective.

Here is another more interesting example of a defective matrix.

Example 12.1.12 Let

Find the eigenvectors and eigenvalues.

In this case the eigenvalues are 3, 6, 6 where we have listed 6 twice because it is a zero of algebraic multiplicity two, the characteristic equation being

(λ− 3) (λ− 6)² = 0.

It remains to find the eigenvectors for these eigenvalues. First consider the eigenvectors for λ = 3. You must solve

The augmented matrix is

and the row reduced echelon form is

so the eigenvectors are nonzero vectors of the form

Next consider the eigenvectors for λ = 6. This requires you to solve

and the augmented matrix for this system of equations is

The row reduced echelon form is

and so the eigenvectors for λ = 6 are of the form

or written more simply,

where t ∈ F.

Note that in this example the eigenspace for the eigenvalue λ = 6 is of dimension 1 because there is only one parameter. However, this eigenvalue is of multiplicity two as a root to the characteristic equation. Thus this eigenvalue is a defective eigenvalue. However, the eigenvalue 3 is nondefective. The matrix is defective because it has a defective eigenvalue.

The word, defective, seems to suggest there is something wrong with the matrix. This is in fact the case. Defective matrices are a lot of trouble in applications and we may wish they never occurred. However, they do occur as the above example shows. When you study linear systems of differential equations, you will have to deal with the case of defective matrices and you will see how awful they are. The reason these matrices are so horrible to work with is that it is impossible to obtain a basis of eigenvectors. When you study differential equations, solutions to first order systems are expressed in terms of eigenvectors of a certain matrix times eλt where λ is an eigenvalue. In order to obtain a general solution of this sort, you must have a basis of eigenvectors. For a defective matrix, such a basis does not exist and so you have to go to something called generalized eigenvectors. Unfortunately, it is never explained in beginning differential equations courses why there are enough generalized eigenvectors and eigenvectors to represent the general solution. In fact, this reduces to a diﬃcult question in linear algebra equivalent to the existence of something called the Jordan Canonical form which is much more diﬃcult than everything discussed in the entire differential equations course. If you become interested in this, see Appendix A.

Ultimately, the algebraic issues which will occur in differential equations are a red herring anyway. The real issues relative to existence of solutions to systems of ordinary differential equations are analytical, having much more to do with calculus than with linear algebra although this will likely not be made clear when you take a beginning differential equations class.

In terms of algebra, this lack of a basis of eigenvectors says that it is impossible to obtain a diagonal matrix which is similar to the given matrix.

Although there may be repeated roots to the characteristic equation, (12.2) and it is not known whether the matrix is defective in this case, there is an important theorem which holds when considering eigenvectors which correspond to distinct eigenvalues.

Theorem 12.1.13 Suppose Mvi = λivi, i = 1,· · ·, r , vi≠ 0, and that if i≠ j, then λi≠ λj. Then the set of eigenvectors, {v₁,· · ·, vr} is linearly independent.

Proof. Suppose the claim of the lemma is not true. Then there exists a subset of this set of vectors

{w₁,· · ·, wr} ⊆ {v₁,· · ·, vk}

such that

where each cj≠ 0. Say Mwj = µjwj where

{µ₁,· · ·, µr} ⊆ {λ1,· · ·, λk},

the µj being distinct eigenvalues of M. Out of all such subsets, let this one be such that r is as small as possible. Then necessarily, r > 1 because otherwise, c₁w₁ = 0 which would imply w₁ = 0, which is not allowed for eigenvectors.

Now apply M to both sides of (12.6).

Next pick µk≠ 0 and multiply both sides of (12.6) by µk. Such a µk exists because r > 1. Thus

Subtract the sum in (12.8) from the sum in (12.7) to obtain

Now one of the constants cj (µk −µj) equals 0, when j = k. Therefore, r was not as small as possible after all.

Here is another proof in case you did not follow the above.

Theorem 12.1.14 Suppose Mvi = λivi, i = 1,· · ·, r , vi ≠ 0, and that if i≠ j, then λi≠ λj. Then the set of eigenvectors, {v₁,· · ·, vr} is linearly independent.

Proof: Suppose the conclusion is not true. Then in the matrix

(  v1   v2   ...   vr  )

not every column is a pivot column. Let the pivot columns be {w₁,· · ·, wk}, k < r. Then there exists v ∈ {v₁,· · ·, vr}, Mv =λvv, v ∉ {w₁,· · ·, wk}, such that

Then doing M to both sides yields

But also you could multiply both sides of (12.9) by λv to get

And now subtracting this from (12.10) yields

and by independence of the {w₁,· · ·, wk}, this requires ci (λv −λwi) = 0 for each i. Since the eigenvalues are distinct, λv−λwi ≠ 0 and so each ci = 0. > But from (12.9), this requires v = 0 which is impossible because v is an eigenvector and

12.1.6 Diagonalization

First of all, here is what it means for two matrices to be similar.

Deﬁnition 12.1.15 Let A, B be two n × n matrices. Then they are similar if and only if there exists an invertible matrix S such that

A = S⁻¹BS

Proposition 12.1.16 Deﬁne for n × n matrices A ~ B if A is similar to B. Then

A ~ A,

If A ~ B then B ~ A

If A ~ B and B ~ C then A ~ C

Proof: It is clear that A ~ A because you could just take S = I. If A ~ B, then for some S invertible,

A = S⁻¹ BS

and so

SAS⁻¹ = B

But then

(S⁻¹)⁻¹AS⁻¹ = B

which shows that B ~ A.

Now suppose A ~ B and B ~ C. Then there exist invertible matrices S, T such that

A = S⁻¹BS, B = T⁻¹CT.

Therefore,

A = S⁻¹T⁻¹CT S = (T S)⁻¹C (T S)

showing that A is similar to C

For your information, when ~ satisﬁes the above conditions, it is called a similarity relation. Similarity relations are very signiﬁcant in mathematics.

When a matrix is similar to a diagonal matrix, the matrix is said to be diagonalizable.

Deﬁnition 12.1.17 Let A be an n×n matrix. Then A is diagonalizable if there exists an invertible matrix S such that

S⁻¹AS = D

where D is a diagonal matrix. This means D has a zero as every entry except for the main diagonal. More precisely, Dij = 0 unless i = j. Such matrices look like the following.

where ∗ might not be zero.

The most important theorem about diagonalizability¹ is the following major result.

Theorem 12.1.18 An n x n matrix is diagonalizable if and only if F² has a basis of eigenvectors of A. Futhermorem, you can take the matrix S described above, to be given as

S = (v₁ v₂ · · · vn)

where here the vk are the eigenvectors in the basis for Fn. If A is diagonalizable, the eigenvalues of A are the diagonal entries of the diagonal matrix.

Proof: Suppose there exists a basis of eigenvectors {vk} where Avk = λkvk. Then let S be given as above. It follows S⁻¹ exists and is of the form

where Then

Next suppose A is diagonalizable so that S⁻¹AS = D. Let S = (v₁ v₂ · · · vn) where the columns are the vk and

then 

and so

( Av₁ Av₂ · · · Avn ) = ( λ₁ v₁ λ₂v₂ · · · λnvn)

showing the vi are eigenvectors of A and the λk are eigenvectors. Now the vk form a basis for Fn because the matrix S having these vectors as columns is given to be invertible.

Example 12.1.19 Let A = Find a matrix, S such that S⁻¹AS = D, a diagonal matrix.

Solving det (λI − A) = 0 yields the eigenvalues are 2 and 6 with 2 an eigenvalue of multiplicity two. Solving (2I −A) x = 0 to find the eigenvectors, you find that the eigenvectors are

where a, b are scalars. An eigenvector for λ = 6 is Let the matrix S be.

That is, the columns are the eigenvectors. Then

We know the result from the above theorem, but it is nice to see it work in a speciﬁc example just the same. You may wonder if there is a need to find S⁻¹. The following is an example of a situation where this is needed. It is one of the major applications of diagonalizability.

Example 12.1.20 Here is a matrix. A = Find A⁵⁰.

Sometimes this sort of problem can be made easy by using diagonalization. In this case there are eigenvectors,

the first two corresponding to λ = 1 and the last corresponding to λ = 2. Then let the eigenvectors be the columns of the matrix, S. Thus

Then also

and

Now it follows

Note that (SDS⁻¹)² = SDS⁻¹ SDS⁻¹ = SD²S⁻¹ and

(SDS⁻¹)³ = SDC⁻¹SDC⁻¹SDC⁻¹ = SD³S⁻¹,

etc. In general, you can see that

(SDS⁻¹)n = SDnS⁻¹

In other words, An = SDnS⁻¹. Therefore,

It is easy to raise a diagonal matrix to a power.

It follows

That isn’t too hard. However, this would have been horrendous if you had tried to multiply A⁵⁰ by hand.

This technique of diagonalization is also important in solving the differential equations resulting from vibrations. Sometimes you have systems of differential equation and when you diagonalize an appropriate matrix, you “decouple” the equations. This is very nice. It makes hard problems trivial.

The above example is entirely typical. If A = SDS⁻¹ then Am = SDmS⁻¹ and it is easy to compute Dm. More generally, you can deﬁne functions of the matrix using power series in this way.

12.1.7 The Matrix Exponential

The exponential of a matrix is a useful concept in differential equations and elsewhere in mathematics and physics. When the matrix A is diagonalizable, it is relatively straightforward to calculate eA. Here is how. You know:

S⁻¹AS = D

where D is a diagonal matrix. You also know that if D is of the form

| [picture] | (12.11)

then

and that

Am = SDmS⁻¹

as shown above. Recall why this was.

A = SDS⁻¹

and so

Now formally write the following power series for eA

If D is given above in (12.11), the above sum is of the form

and this last thing is the deﬁnition of what is meant by eA.

Example 12.1.21 Let

Find eA

The eigenvalues happen to be 1, 2, 3 and eigenvectors associated with these eigenvalues are

Then let

and so

and

Then the matrix exponential is

Isn’t that nice? You could also talk about sin ( A) or cos (A) etc. You would just have to use a different power series.

This matrix exponential is actually a useful idea when solving autonomous systems of first order linear differential equations. These are equations which are of the form

x′ = Ax

where x is a vector in Rn or Cn and A is an n×n matrix. Then it turns out that the solution to the above system of equations is x (t) = eAtc where c is a constant vector.

12.1.8 Complex Eigenvalues

Sometimes you have to consider eigenvalues which are complex numbers. This occurs in differential equations for example. You do these problems exactly the same way as you do the ones in which the eigenvalues are real. Here is an example.

Example 12.1.22 Find the eigenvalues and eigenvectors of the matrix

You need to find the eigenvalues. Solve

This reduces to (λ - 1)(λ² − 4λ + 5) = 0. The solutions are λ = 1, λ = 2 + i, λ = 2−i

There is nothing new about finding the eigenvectors for λ = 1 so consider the eigenvalue λ = 2+i. You need to solve

In other words, you must consider the augmented matrix

for the solution. Divide the top row by (1 + i) and then take −i times the second row and add to the bottom. This yields

Now multiply the second row by −i to obtain

Therefore, the eigenvectors are of the form

You should find the eigenvectors for λ = 2−i. These are

As usual, if you want to get it right you had better check it.

so it worked.

12.2 Some Applications Of Eigenvalues And Eigenvectors

12.2.1 Principal Directions

Recall that n × n matrices can be considered as linear transformations. If F is a 3 × 3 real matrix having positive determinant, it can be shown that F = RU where R is a rotation matrix and U is a symmetric real matrix having positive eigenvalues. An application of this wonderful result, known to mathematicians as the right polar factorization, is to continuum mechanics where a chunk of material is identiﬁed with a set of points in three dimensional space.

The linear transformation, F in this context is called the deformation gradient and it describes the local deformation of the material. Thus it is possible to consider this deformation in terms of two processes, one which distorts the material and the other which just rotates it. It is the matrix U which is responsible for stretching and compressing. This is why in elasticity, the stress is often taken to depend on U which is known in this context as the right Cauchy Green strain tensor. In this context, the eigenvalues will always be positive. The symmetry of U allows the proof of a theorem which says that if λM is the largest eigenvalue, then in every other direction other than the one corresponding to the eigenvector for λM the material is stretched less than λM and if λm is the smallest eigenvalue, then in every other direction other than the one corresponding to an eigenvector of λm the material is stretched more than λm. This process of writing a matrix as a product of two such matrices, one of which preserves distance and the other which distorts is also important in applications to geometric measure theory an interesting field of study in mathematics and to the study of quadratic forms which occur in many applications such as statistics. Here we are emphasizing the application to mechanics in which the eigenvectors of the symmetric matrix U determine the Principal directions, those directions in which the material is stretched the most or the least.

Example 12.2.1 Find the Principal directions determined by the matrix

The eigenvalues are 3, 1, and

It is nice to be given the eigenvalues. The largest eigenvalue is 3 which means that in the direction determined by the eigenvector associated with 3 the stretch is three times as large. The smallest eigenvalue is 1/2 and so in the direction determined by the eigenvector for 1/2 the material is stretched by a factor of 1/2, becoming locally half as long. It remains to find these directions. First consider the eigenvector for 3. It is necessary to solve

Thus the augmented matrix for this system of equations is

The row reduced echelon form is

and so the Principal direction for the eigenvalue, 3 in which the material is stretched to the maximum extent is

A direction vector (or unit vector) in this direction is

You should show that the direction in which the material is compressed the most is in the direction

Note this is meaningful information which you would have a hard time finding without the theory of eigenvectors and eigenvalues.

12.2.2 Migration Matrices

There are applications which are of great importance which feature only one eigenvalue.

Deﬁnition 12.2.2 Let n locations be denoted by the numbers 1, 2, · · ·, n. Also suppose it is the case that each year aij denotes the proportion of residents in location j which move to location i. Also suppose no one escapes or emigrates from without these n locations. This last assumption requires ∑i aij = 1. Such matrices in which the columns are nonnegative numbers which sum to one arei called Markov matrices. In this context describing migration, they are also called migration matrices.

Example 12.2.3 Here is an example of one of these matrices.

Thus if it is considered as a migration matrix, .4 is the proportion of residents in location 1 which stay in location one in a given time period while .6 is the proportion of residents in location 1 which move to location 2 and .2 is the proportion of residents in location 2 which move to location 1. Considered as a Markov matrix, these numbers are usually identiﬁed with probabilities.

If v = (xT where xi is the population of location i at a given instant, you obtain the₁,· · ·, xn) population of location i one year later by computing . Therefore, the population of location i after k years is (Akv)i. An obvious application of this would be to a situation in whichi you rent trailers which can go to various parts of a city and you observe through experiments the proportion of trailers which go from point i to point j in a single day. Then you might want to find how many trailers would be in all the locations after 8 days.

Proposition 12.2.4 Let A = (aij) be a migration matrix. Then 1 is always an eigenvalue for A.

Proof: Remember that det (BT) = det (B) . Therefore,

because IT = I. Thus the characteristic equation for A is the same as the characteristic equation for AT and so A and AT have the same eigenvalues. We will show that 1 is an eigenvalue for AT and then it will follow that 1 is an eigenvalue for A.

Remember that for a migration matrix,∑ aij = 1. Therefore, if AT = (bij) so bij = aji, it follows that

Therefore, from matrix multiplication,

which shows that is an eigenvector for AT corresponding to the eigenvalue, λ =

1. As explained above, this shows that λ = 1 is an eigenvalue for A because A and AT have the same eigenvalues.

Example 12.2.5 Consider the migration matrix for locations 1, 2,

and 3. Suppose initially there are 100 residents in location 1, 200 in location 2 and 400 in location 4. Find the population in the three locations after 10 units of time.

From the above, it suﬃces to consider

Of course you would need to round these numbers off.

A related problem asks for how many there will be in the various locations after a long time. It turns out that if some power of the migration matrix has all positive entries, then there is a limiting vector where x₀ is the initial vector describing the number of inhabitants in the various locations initially. This vector will be an eigenvector for the eigenvalue 1 because

and the sum of its entries will equal the sum of the entries of the initial vector x₀ because this sum is preserved for every multiplication by A since

Here is an example. It is the same example as the one above but here it will involve the long time limit.

Example 12.2.6 Consider the migration matrix for locations 1, 2, and 3. Suppose

initially there are 100 residents in location 1, 200 in location 2 and 400 in location 4. Find the population in the three locations after a long time.

You just need to find the eigenvector which goes with the eigenvalue 1 and then normalize it so the sum of its entries equals the sum of the entries of the initial vector. Thus you need to find a solution to

The augmented matrix is

and its row reduced echelon form is

Therefore, the eigenvectors are

and all that remains is to choose the value of s such that

This yields s = and so the long time limit would equal

You would of course need to round these numbers off. You see that you are not far off after just 10 units of time. Therefore, you might consider this as a useful procedure because it is probably easier to solve a simple system of equations than it is to raise a matrix to a large power.

Example 12.2.7 Suppose a migration matrix is Find the comparison between

the populations in the three locations after a long time.

This amounts to nothing more than finding the eigenvector for λ = 1. Solve

The augmented matrix is

The row echelon form is

and so an eigenvector is

Thus there will be more in location 2 than in location 1. There will be more in location 3 than in location 2.

You see the eigenvalue problem makes these sorts of determinations fairly simple.

There are many other things which can be said about these sorts of migration problems.

They include things like the gambler’s ruin problem which asks for the probability that a compulsive gambler will eventually lose all his money. However those problems are not so easy although they still involve eigenvalues and eigenvectors.

There are many other important applications of eigenvalue problems. We have just given a few such applications here. As pointed out, this is a very hard problem but sometimes you don’t need to find the eigenvalues exactly.

12.3 The Estimation Of Eigenvalues

There are ways to estimate the eigenvalues for matrices from just looking at the matrix. The most famous is known as Gerschgorin’s theorem. This theorem gives a rough idea where the eigenvalues are just from looking at the matrix.

Theorem 12.3.1 Let A be an n×n matrix. Consider the n Gerschgorin discs deﬁned as 

Then every eigenvalue is contained in some Gerschgorin disc.

This theorem says to add up the absolute values of the entries of the ith row which are off the main diagonal and form the disc centered at aii having this radius. The union of these discs contains σ (A) , the spectrum of A.

Proof: Suppose Ax = λx where x 0. Then for A = (aij)

Therefore, picking k such that |xk| ≥ |xj| for all xj, it follows that |xk| 0 since |x|  0 and

Now dividing by |xk|, it follows λ is contained in the kth Gerschgorin disc.

Example 12.3.2 Suppose the matrix is 

Estimate the eigenvalues.

The exact eigenvalues are 35, 56, and 28. The Gerschgorin disks are

and

Gerschgorin’s theorem says these three disks contain the eigenvalues. Now 35 is in D₃, 56 is in D₂ and 28 is in D₁.

More can be said when the Gerschgorin disks are disjoint but this is an advanced topic which requires the theory of functions of a complex variable. If you are interested and have a background in complex variable techniques, this is in [11]

12.4 Exercises

State the eigenvalue problem from an algebraic perspective.

State the eigenvalue problem from a geometric perspective.

If A is the matrix of a linear transformation which rotates all vectors in R² through 30◦, explain why A cannot have any real eigenvalues.

If A is an n × n matrix and c is a nonzero constant, compare the eigenvalues of A and cA.

If A is an invertible n×n matrix, compare the eigenvalues of A and A⁻¹. More generally, for m an arbitrary integer, compare the eigenvalues of A and Am.

Let A, B be invertible n×n matrices which commute. That is, A B = BA. Suppose x is an eigenvector of B. Show that then Ax must also be an eigenvector for B.

Suppose A is an n × n matrix and it satisﬁes Am = A for some m a positive integer larger than 1. Show that if λ is an eigenvalue of A then |λ| equals either 0 or 1.

Show that if Ax = λx and Ay = λy, then whenever a, b are scalars,

A (ax + by) = λ (ax + by) .

Does this imply that ax + by is an eigenvector? Explain.

Find the eigenvalues and eigenvectors of the matrix 

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix 

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

 Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective.

Find the complex eigenvalues and eigenvectors of the matrix

Find the eigenvalues and eigenvectors of the matrix

Determine whether the matrix is defective. 

Find the complex eigenvalues and eigenvectors of the matrix  Determine  whether the matrix is defective.

Find the complex eigenvalues and eigenvectors of the matrixDetermine whether the matrix is defective.

Find the complex eigenvalues and eigenvectors of the matrixDetermine whether the matrix is defective.

Find the complex eigenvalues and eigenvectors of the matrixDetermine whether the matrix is defective.

Let A be a real 3× 3 matrix which has a complex eigenvalue of the form a + ib where b 0. Could A be defective? Explain. Either give a proof or an example.

Let T be the linear transformation which reﬂects vectors about the x axis. Find a matrix for T and then find its eigenvalues and eigenvectors.

Let T be the linear transformation which rotates all vectors in R² counterclockwise through an angle of π/2. Find a matrix of T and then find eigenvalues and eigenvectors.

Let A be the 2 × 2 matrix of the linear transformation which rotates all vectors inR² through an angle of . For which values of does A have a real eigenvalue?

Let T be the linear transformation which reﬂects all vectors in³ through the xy plane. Find a matrix for T and then obtain its eigenvalues and eigenvectors.

Find the Principal direction for stretching for the matrix

The eigenvalues are 2 and 1.

Find the Principal directions for the matrix

Suppose the migration matrix for three locations is

Find a comparison for the populations in the three locations after a long time.

Suppose the migration matrix for three locations is

Find a comparison for the populations in the three locations after a long time.

You own a trailer rental company in a large city and you have four locations, one in the South East, one in the North East, one in the North West, and one in the South West. Denote these locations by SE,NE,NW, and SW respectively. Suppose you observe that in a typical day, .8 of the trailers starting in SE stay in SE, .1 of the trailers in NE go to SE, .1 of the trailers in NW end up in SE, .2 of the trailers in SW end up in SE, .1 of the trailers in SE end up in NE,.7 of the trailers in NE end up in NE,.2 of the trailers in NW end up in NE,.1 of the trailers in SW end up in NE, .1 of the trailers in SE end up in NW, .1 of the trailers in NE end up in NW, .6 of the trailers in NW end up in NW, .2 of the trailers in SW end up in NW, 0 of the trailers in SE end up in SW, .1 of the trailers in NE end up in SW, .1 of the trailers in NW end up in SW, .5 of the trailers in SW end up in SW. You begin with 20 trailers in each location. Approximately how many will you have in each location after a long time? Will any location ever run out of trailers?

Let A be the n × n, n > 1, matrix of the linear transformation which comes from the projection v→projw (v). Show that A cannot be invertible. Also show that A has an eigenvalue equal to 1 and that for λ an eigenvalue, |λ| ≤ 1.

Let v be a unit vector in Rn and let A = I − 2vvT. Show that A has an eigenvalue equal to −1.

Let M be an n × n matrix and suppose x₁,· · ·, xn are n eigenvectors which form a linearly independent set. Form the matrix S by making the columns these vectors. Show that S⁻¹ exists and that S⁻¹MS is a diagonal matrix (one having zeros everywhere except on the main diagonal) having the eigenvalues of M on the main diagonal. When this can be done the matrix is diagonalizable. This is presented in the text. You should write it down in your own words filling in the details without looking at the text.

Show that a matrix M is diagonalizable if and only if it has a basis of eigenvectors. Hint: The first part is done in Problem 39. It only remains to show that if the matrix can be diagonalized by some matrix S giving D = S⁻¹MS for D a diagonal matrix, then it has a basis of eigenvectors. Try using the columns of the matrix S. Like the last problem, you should try to do this yourself without consulting the text. These problems are a nice review of the meaning of matrix multiplication.

Suppose A is an n × n matrix which is diagonally dominant. This means

Show that A⁻¹ must exist.

Is it possible for a nonzero matrix to have only 0 as an eigenvalue?

Let M be an n × n matrix. Then deﬁne the adjoint of M,denoted by M∗ to be the transpose of the conjugate of M. For example,

A matrix M, is self adjoint if M∗ = M. Show the eigenvalues of a self adjoint matrix are all real. If the self adjoint matrix has all real entries, it is called symmetric.

Suppose A is an n × n matrix consisting entirely of real entries but a+ib is a complex eigenvalue having the eigenvector x + iy. Here x and y are real vectors. Show that then a−ib is also an eigenvalue with the eigenvector x−iy. Hint: You should remember that the conjugate of a product of complex numbers equals the product of the conjugates. Here a + ib is a complex number whose conjugate equals a − ib.

Recall an n × n matrix is said to be symmetric if it has all real entries and if A = AT. Show the eigenvectors and eigenvalues of a real symmetric matrix are real.

Recall an n × n matrix is said to be skew symmetric if it has all real entries and if A = −AT. Show that any nonzero eigenvalues must be of the form ib where i² = −1. In words, the eigenvalues are either 0 or pure imaginary. Show also that the eigenvectors corresponding to the pure imaginary eigenvalues are imaginary in the sense that every entry is of the form ix for x ∈ R.

A discrete dynamical system is of the form

x (k + 1) = Ax (k) , x (0) = x₀

where A is an n × n matrix and x (k) is a vector in Rn. Show first that

x (k) = Akx₀

for all k ≥ 1. If A is nondefective so that it has a basis of eigenvectors, {v₁,· · ·, vn} where

Avj = λjvj

you can write the initial condition x₀ in a unique way as a linear combination of these eigenvectors. Thus

Now explain why

which gives a formula for x (k) , the solution of the dynamical system.

Suppose A is an n × n matrix and let v be an eigenvector such that Av = λv. Also suppose the characteristic polynomial of A is

det (λI −A) = λn + an−1λn−1 +· · · + a₁λ + a₀

Explain why

(An + an−1An−1 +· · · + a₁A + a₀I) v = 0

If A is nondefective, give a very easy proof of the Cayley Hamilton theorem based on this. Recall this theorem says A satisﬁes its characteristic equation,

An + an−1An−1 +· · · + a₁A + a₀I = 0.

Suppose an n × n nondefective matrix A has only 1 and −1 as eigenvalues. Find A¹².

Suppose the characteristic polynomial of an n × n. matrix A is 1 − λn Find Amn where m is an integer. Hint: Note first that A is nondefective. Why?

Sometimes sequences come in terms of a recursion formula. An example is the Fibonacci sequence.

x₀ = 1 = x₁, xn+1 = xn + xn−1

Show this can be considered as a discrete dynamical system as follows.

Now use the technique of Problem 47 to find a formula for xn.

Let A be an n × n matrix having characteristic polynomial

det (λI −A) = λn + an−1λn−1 +· · · + a₁λ + a₀

Show that a₀ = (−1)n det (A).

Find  Next find

Find eA where A is the matrixin the above problem.

¹This word has 9 syllables

Matrices And The Inner Product

13.1 Symmetric And Orthogonal Matrices

13.1.1 Orthogonal Matrices

Remember that to find the inverse of a matrix was often a long process. However, it was very easy to take the transpose of a matrix. For some matrices, the transpose equals the inverse and when the matrix has all real entries, and this is true, it is called an orthogonal matrix. Recall the following deﬁnition given earlier.

Deﬁnition 13.1.1 A real n × n matrix U is called an Orthogonal matrix if UUT = UTU = I.

Example 13.1.2 Show the matrix

is orthogonal.

Example 13.1.3 Let U = . Is U orthogonal?

The answer is yes. This is because the columns form an orthonormal set of vectors as well as the rows. As discussed above this is equivalent to UTU = I.

When you say that U is orthogonal, you are saying that

In words, the dot product of the ith row of U with the kth row gives 1 if i = k and 0 if i ≠ k. The same is true of the columns because UTU = I also. Therefore,

which says that the one column dotted with another column gives 1 if the two columns are the same and 0 if the two columns are different.

More succinctly, this states that if u₁,· · ·, un are the columns of U, an orthogonal matrix, then

| [picture] | (13.1)

Deﬁnition 13.1.4 A set of vectors, {u₁,· · ·, un} is said to be an orthonormal set if (13.1).

Theorem 13.1.5 If {u₁,· · ·, um} is an orthonormal set of vectors then it is linearly independent.

Proof: Using the properties of the dot product,

0·u = (0 + 0)·u = 0·u + 0·u

and so, subtracting 0 · u from both sides yields 0 · u = 0. Now suppose ∑jcjuj = 0. Then from the properties of the dot product,

Since k was arbitrary, this shows that each ck = 0 and this has shown that if ∑jcjuj = 0, then each cj = 0. This is what it means for the set of vectors to be linearly independent.

Example 13.1.6 Let U = . Is U an orthogonal matrix?

The answer is yes. This is because the columns (rows) form an orthonormal set of vectors.

The importance of orthogonal matrices is that they change components of vectors relative to different Cartesian coordinate systems. Geometrically, the orthogonal matrices are exactly those which preserve all distances in the sense that if x ∈ Rn and U is orthogonal, then ||Ux|| = ||x|| because

||Ux||² = (Ux)T Ux = xTUTUx = xT Ix = ||x||².

Observation 13.1.7 Suppose U is an orthogonal matrix. Then det (U) = ±1.

This is easy to see from the properties of determinants. Thus

det (U)² = det (UT) det (U) = det (UTU) = det (I) = 1.

Orthogonal matrices are divided into two classes, proper and improper. The proper orthogonal matrices are those whose determinant equals 1 and the improper ones are those whose determinants equal −1. The reason for the distinction is that the improper orthogonal matrices are sometimes considered to have no physical signiﬁcance since they cause a change in orientation which would correspond to material passing through itself in a non physical manner. Thus in considering which coordinate systems must be considered in certain applications, you only need to consider those which are related by a proper orthogonal transformation. Geometrically, the linear transformations determined by the proper orthogonal matrices correspond to the composition of rotations.

13.1.2   Symmetric And Skew Symmetric Matrices

Deﬁnition 13.1.8 A real n × n matrix A, is symmetric if AT = A. If A = −AT, then A is called skew symmetric.

Theorem 13.1.9 The eigenvalues of a real symmetric matrix are real. The eigenvalues of a real skew symmetric matrix are 0 or pure imaginary. 

Proof: The proof of this theorem is in [11]. It is best understood as a special case of more general considerations. However, here is a proof in this special case.

Recall that for a complex number a + ib, the complex conjugate, denoted by is given by the formula The notation, will denote the vector which has every entry replaced by its complex conjugate.

Suppose A is a real symmetric matrix and Ax = λx. Then

Dividing by on both sides yields -λ which says λ is real. (Why?)

Next suppose A = −AT so A is skew symmetric and Ax = λx. Then

and so, dividing by as before, −λ. Letting λ = a + ib, this means a−ib = −a−ib and so a = 0. Thus λ is pure imaginary.

Example 13.1.10 Let A = . This is a skew symmetric matrix. Find its eigenvalues.

Its eigenvalues are obtained by solving the equation det . You see the eigenvalues are ±i, pure imaginary.

Example 13.1.11 Let A = . This is a symmetric matrix. Find its eigenvalues.

Its eigenvalues are obtained by solving the equation, det = −1−4λ + λ² = 0 and the solution is λ = 2 + and λ = 2 − .

Deﬁnition 13.1.12 An n × n matrix A = (aij) is called a diagonal matrix if aij = 0 whenever i ≠ j. For example, a diagonal matrix is of the form indicated below where ∗ denotes a number.

Theorem 13.1.13 Let A be a real symmetric matrix. Then there exists an orthogonal matrix U such that UTAU is a diagonal matrix. Moreover, the diagonal entries are the eigenvalues of A.

Proof: The proof is given later.

Corollary 13.1.14 If A is a real n × n symmetric matrix, then there exists an orthonormal set of eigenvectors, {u₁,· · ·, un}.

Proof: Since A is symmetric, then by Theorem 13.1.13, there exists an orthogonal matrix U such that UTAU = D, a diagonal matrix whose diagonal entries are the eigenvalues of A. Therefore, since A is symmetric and all the matrices are real,

showing D is real because each entry of D equals its complex conjugate.¹

Finally, let

U = ( u₁ u₂ · · · un)

where the ui denote the columns of U and

The equation, UTAU = D implies

where the entries denote the columns of AU and UD respectively. Therefore, Aui = λiui and since the matrix is orthogonal, the ijth entry of UTU equals δij and so

δij = uiT uj = ui ·uj.

This proves the corollary because it shows the vectors {ui} form an orthonormal basis.

Example 13.1.15 Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix

given that the eigenvalues are 3, −1, and 2.

The augmented matrix which needs to be row reduced to find the eigenvectors for λ = 3 is 

and the row reduced echelon form for this is

Therefore, eigenvectors for λ = 3 are

where z ≠ 0.

The augmented matrix, which must be row reduced to find the eigenvectors for λ = −1, is 

and the row reduced echelon form is

Therefore, the eigenvectors for λ = −1 are

The augmented matrix which must be row reduced to find the eigenvectors for λ = 2 is

and its row reduced echelon form is

so the eigenvectors for λ = 2 are

It remains to find an orthonormal basis. You can check that the dot product of any of these vectors with another of them gives zero and so it suﬃces choose z in each case such that the resulting vector has length 1. First consider the vectors for λ = 3. It is required to choose z such that

is a unit vector. In other words, you need

But the above dot product equals and this equals 1 when z = . Therefore, the eigenvector which is desired is

Next find the eigenvector for λ = −1. The same process requires that 1 = which happens when z = Therefore, an eigenvector for λ = −1 which has unit length is

Finally, consider λ = 2. This time you need 1 = which occurs when z = . Therefore, the eigenvector is

Now recall that the vectors form an orthonormal set of vectors if the matrix having them as columns is orthogonal. That matrix is

Is this orthogonal? To find out, multiply by its transpose. Thus

Since the identity was obtained this shows the above matrix is orthogonal and that therefore, the columns form an orthonormal set of vectors. The problem asks for you to find an orthonormal basis. However, you will show in Problem 23 that an orthonormal set of n vectors in Rn is always a basis. Therefore, since there are three of these vectors, they must constitute a basis.

Example 13.1.16 Find an orthonormal set of three eigenvectors for the matrix

given the eigenvalues are 2, and 1.

The eigenvectors which go with λ = 2 are obtained from row reducing the matrix

and its row reduced echelon form is

which shows the eigenvectors for λ = 2 are

and a choice for z which will produce a unit vector is z = . Therefore, the vector we want is

Next consider the eigenvectors for λ = 1. The matrix which must be row reduced is

and its row reduced echelon form is

Therefore, the eigenvectors are of the form

This is a two dimensional eigenspace.

Before going further, we want to point out that no matter how we choose y and z the resulting vector will be orthogonal to the eigenvector for λ = 2. This is a special case of a general result which states that eigenvectors for distinct eigenvalues of a symmetric matrix are orthogonal. This is explained in Problem 15. For this case you need to show the following dot product equals zero.

| [picture] | (13.2)

This is left for you to do.

Continuing with the task of finding an orthonormal basis, Let y = 0 first. This results in eigenvectors of the form

and letting z =  you obtain a unit vector. Thus the second vector will be

It remains to find the third vector in the orthonormal basis. This merely involves choosing y and z in (13.2) in such a way that the resulting vector has dot product with the two given vectors equal to zero. Thus you need

The dot product with the eigenvector for λ = 2 is automatically equal to zero and so all that you need is the above equation. This is satisﬁed when z = Therefore, the vector we want is of the form

and it only remains to choose y in such a way that this vector has unit length. This occurs when y = Therefore, the vector we want is

The three eigenvectors which constitute an orthonormal basis are

To check our work and see if this is really an orthonormal set of vectors, we make them the columns of a matrix and see if the resulting matrix is orthogonal. The matrix is

This matrix times its transpose equals

and so this is indeed an orthonormal basis.

Because of the repeated eigenvalue, there would have been many other orthonormal bases which could have been obtained. It was pretty arbitrary for to take y = 0 in the above argument. We could just as easily have taken z = 0 or even y = z = 1. Any such change would have resulted in a different orthonormal basis. Geometrically, what is happening is the eigenspace for λ = 1 was two dimensional. It can be visualized as a plane in three dimensional space which passes through the origin. There are inﬁnitely many different pairs of perpendicular unit vectors in this plane.

13.1.3 Diagonalizing A Symmetric Matrix

Recall the following deﬁnitions from Chapter 12:

Deﬁnition 13.1.17 An n × n matrix A = (aij) is called a diagonal matrix if aij = 0 whenever i ≠ j. For example, a diagonal matrix is of the form indicated below where ∗ denotes a number.

Deﬁnition 13.1.18 An n × n matrix A is said to be non defective or diagonalizable if there exists an invertible matrix S such that S⁻¹ AS = D where D is a diagonal matrix as described above.

Some matrices are non defective and some are not. As indicated in Theorem 13.1.13 if A is a real symmetric matrix, there exists an orthogonal matrix U such that UTAU = D a diagonal matrix. Therefore, every symmetric matrix is non defective because if U is an orthogonal matrix, its inverse is UT. In the following example, this orthogonal matrix will be found.

Example 13.1.19 Let A = . Find an orthogonal matrix U such that UTAU is a diagonal matrix.

In this case, a tedious computation shows the eigenvalues are 2 and 1. First we will find an eigenvector for the eigenvalue 2. This involves row reducing the following augmented matrix.

The row reduced echelon form is

and so an eigenvector is

However, it is desired that the eigenvectors obtained all be unit vectors and so dividing this vector by its length gives

Next consider the case of the eigenvalue, 1. The matrix which needs to be row reduced in this case is

The row reduced echelon form is

Therefore, the eigenvectors are of the form

Two of these which are orthonormal are

An orthogonal matrix which works in the process is then obtained by letting these vectors be the columns.

It remains to verify this works. UTAU is of the form

the desired diagonal matrix.

13.2 Fundamental Theory And Generalizations

13.2.1 Block Multiplication Of Matrices

Consider the following problem

You know how to do this. You get

Now what if instead of numbers, the entries, A, B, C, D, E, F, G are matrices of a size such that the multiplications and additions needed in the above formula all make sense. Would the formula be true in this case? I will show below that this is true.

Suppose A is a matrix of the form

| [picture] | (13.3)

where Aij is a si × pj matrix where si is constant for j = 1,· · ·, m for each i = 1,· · ·, r. Such a matrix is called a block matrix, also a partitioned matrix. How do you get the block Aij? Here is how for A an m × n matrix:

| [picture] | (13.4)

In the block column matrix on the right, you need to have cj−1 rows of zeros above the small pj × pj identity matrix where the columns of A involved in Aij are cj,· · ·, cj + pj −1 and in the block row matrix on the left, you need to have ri − 1 columns of zeros to the left of the si × si identity matrix where the rows of A involved in Aij are ri,· · ·, ri + si. An important observation to make is that the matrix on the right speciﬁes columns to use in the block and the one on the left speciﬁes the rows used. Thus the block Aij in this case is a matrix of size si × pj. There is no overlap between the blocks of A. Thus the identity n × n identity matrix corresponding to multiplication on the right of A is of the form

these little identity matrices don’t overlap. A similar conclusion follows from consideration of the matrices Isi × si.

Next consider the question of multiplication of two block matrices. Let B be a block matrix of the form

| [picture] | (13.5)

and A is a block matrix of the form

| [picture] | (13.6)

and that for all i, j, it makes sense to multiply BisAsj for all s ∈ {1,· · ·, p}. (That is the two matrices, Bis and Asj are conformable.) and that for fixed ij, it follows BisAsj is the same size for each s so that it makes sense to write ∑s BisAsj.

The following theorem says essentially that when you take the product of two matrices, you can do it two ways. One way is to simply multiply them forming BA. The other way is to partition both matrices, formally multiply the blocks to get another block matrix and this one will be BA partitioned. Before presenting this theorem, here is a simple lemma which is really a special case of the theorem.

Lemma 13.2.1 Consider the following product.

where the first is n × r and the second is r × n. The small identity matrix I is an r × r matrix and there are l zero rows above I and l zero columns to the left of I in the right matrix. Then the product of these matrices is a block matrix of the form

Proof: From the deﬁnition of the way you multiply matrices, the product is

which yields the claimed result. In the formula ej refers to the column vector of length r which has a 1 in the jth position.

Theorem 13.2.2 Let B be a q × p block matrix as in (13.5) and let A be a p × n block matrix as in (13.6) such that Bis is conformable with Asj and each product, BisAsj for s = 1,· · ·, p is of the same size so they can be added. Then BA can be obtained as a block matrix such that the ijth block is of the form

| [picture] | (13.7)

Proof: From (13.4)

where here it is assumed Bis is ri × ps and Asj is ps × qj. The product involves the sth block in the ith row of blocks for B and the sth block in the jth column of A. Thus there are the same number of rows above the Ips×ps as there are columns to the left of Ips×ps in those two inside matrices. Then from Lemma 13.2.1

Since the blocks of small identity matrices do not overlap,

and so

which equals the ijth block of BA. Hence the ijth block of BA equals the formal multiplication according to matrix multiplication,

Example 13.2.3 Let an n × n matrix have the form

where P is n − 1 × n − 1. Multiply it by

where B is also an n × n matrix and Q is n − 1 × n − 1.

You use block multiplication

Note that this all makes sense. For example, b = 1 × n −1 and r = n −1 × 1 so br is a 1 × 1. Similar considerations apply to the other blocks.

Here is an interesting and signiﬁcant application of block multiplication. In this theorem, pM (t) denotes the characteristic polynomial, det (tI − M) . Thus the zeros of this polynomial are the eigenvalues of the matrix M.

Theorem 13.2.4 Let A be an m × n matrix and let B be an n × m matrix for m ≤ n. Then 

pBA (t) =  tn−m ,pAB (t),

so the eigenvalues of BA and AB are the same including multiplicities except that BA has n − m extra zero eigenvalues.

Proof: Use block multiplication to write

Therefore,

Since the two matrices above are similar it follows that and have the same characteristic polynomials. Therefore, noting that BA is an n × n matrix and AB is an m × m matrix,

tm det (tI −BA) = tn det (tI −AB)

and so det (tI −BA) = pBA (t) = tn−m det (tI −AB) = tn−mpAB (t).

13.2.2 Orthonormal Bases, Gram Schmidt Process

Not all bases for Fn are created equal. Recall F equals either C or R and the dot product is given by

The best bases are orthonormal. Much of what follows will be for Fn in the interest of generality.

Deﬁnition 13.2.5 Suppose{v₁,· · ·, vk}. is a set of vectors in F. It is an orthonormal set if 

Every orthonormal set of vectors is automatically linearly independent.

Proposition 13.2.6 Suppose {v₁,· · ·, vk} is an orthonormal set of vectors. Then it is linearly independent.

Proof: Suppose Then taking dot products with vj,

Since j is arbitrary, this shows the set is linearly independent as claimed.

It turns out that if X is any subspace of Fm, then there exists an orthonormal basis for X. This follows from the use of the next lemma applied to a basis for X.

Lemma 13.2.7 Let {x₁,· · ·, xn} be a linearly independent subset of Fp, p ≥ n. Then there exist orthonormal vectors {u₁,· · ·, un} which have the property that for each k ≤ n, span (x₁,· · ·, xk) = span (u₁,· · ·, uk) .

Proof: Let u₁ ≡ x1 / |x₁| . Thus for k = 1, span (u₁) = span (x₁) and {u₁} is an orthonormal set. Now suppose for some k < n, u₁, · · ·, uk have been chosen such that (uj, ul) = δjl and span (x₁,· · ·, xk) = span (u₁,· · ·, uk). Then deﬁne

| [picture] | (13.8)

where the denominator is not equal to zero because the xj form a basis, and so

xk+1 ∉ span (x₁,· · ·, xk) = span (u₁, · · · , uk)

Thus by induction,

uk+1 ∈ span (u₁,· · ·, uk, xk+1) = span (x₁,· · ·, xk, xk+1) . 

Also, xk+1 ∈ span (u₁,· · ·, uk, uk+1) which is seen easily by solving (13.8) for xk+1 and it follows

span (x₁,· · ·, xk, xk+1) = span (u₁,· · ·, uk, uk+1) . 

If l ≤ k,

The vectors, {uj} nj₌₁ , generated in this way are therefore orthonormal because each vector has unit length.

The process by which these vectors were generated is called the Gram Schmidt process. Note that from the construction, each xk is in the span of {u₁,· · ·, uk}. In terms of matrices, this says

(x₁· · ·xn) = (u₁· · ·un) R

where R is an upper triangular matrix. This is closely related to the QR factorization discussed earlier. It is called the thin QR factorization. If the Gram Schmidt process is used to enlarge {u₁· · ·ui} to an orthonormal basis for Fm, {u₁· · ·un, un+1,· · ·, um} then if Q is the matrix which has these vectors as columns and if R is also enlarged to R by adding in rows of zeros, if necessary, to form an m × n matrix, then the above would be of the form

(x₁· · ·xn) = (u₁· · ·um) R′

and you could read off the orthonormal basis for span ( x₁· · ·xn) by simply taking the first n columns of Q = (u₁· · ·um). This is convenient because computer algebra systems are set up to find QR factorizations.

Example 13.2.8 Find an orthonormal basis for span .

This is really easy to do using a computer algebra system.

and so the desired orthonormal basis is

13.2.3 Schur’s Theorem

Every matrix is related to an upper triangular matrix in a particularly signiﬁcant way. This is Schur’s theorem and it is the most important theorem in the spectral theory of matrices. The important result which makes this theorem possible is the Gram Schmidt procedure of Lemma 13.2.7.

Deﬁnition 13.2.9 An n × n matrix U, is unitary if UU∗ = I = U∗U where U∗ is deﬁned to be the transpose of the conjugate of U. Thus . Note that every real orthogonal matrix is unitary. For A any matrix A∗ just deﬁned as the conjugate of the transpose is called the adjoint.

Note that if U =)( v₁ · · · vn ) where the vk are orthonormal vectors in Cn, then U is unitary. This follows because the ijth entry of since the vi are assumed orthonormal.

Lemma 13.2.10 The following holds. (AB)∗ = B∗A∗.

Proof: From the deﬁnition and remembering the properties of complex conjugation,

Theorem 13.2.11 Let A be an n × n matrix. Then there exists a unitary matrix U such that

| U∗AU = T, | (13.9)

where T is an upper triangular matrix having the eigenvalues of A on the main diagonal listed according to multiplicity as roots of the characteristic equation.

Proof: The theorem is clearly true if A is a 1 × 1 matrix. Just let U = 1, the 1×1 matrix which has entry 1. Suppose it is true for (n − 1) × (n − 1) matrices and let A be an n × n matrix. Then let v₁ be a unit eigenvector for A. Then there exists λ₁ such that

Av₁ = λ₁v₁, |v₁| = 1.

Extend {v₁} to a basis and then use the Gram - Schmidt process to obtain {v₁,· · ·, vn}, an orthonormal basis of Cn. Let U₀ be a matrix whose ith column is vi so that U₀ is unitary. Also U∗₀ AU₀ is of the form

where A₁ is an n − 1 × n − 1 matrix. Now by induction, there exists an (n − 1) × (n − 1) unitary matrix such that

an upper triangular matrix. Consider

An application of block multiplication shows that U₁ is a unitary matrix and also that

where T is upper triangular. Then let U = U₀U₁. Since (U₀U₁)∗ = U₁∗U₀∗, it follows that A is similar to T and that U₀U₁ is unitary. Hence A and T have the same characteristic polynomials, and since the eigenvalues of T are the diagonal entries listed with multiplicity, this proves the theorem.

As a simple consequence of the above theorem, here is an interesting lemma.

Lemma 13.2.12 Let A be of the form

where Pk is an mk × mk matrix. Then

Proof: Let Uk be an mk × mk unitary matrix such that 

U∗kPkUk = Tk

where Tk is upper triangular. Then letting

it follows

and

and so

Deﬁnition 13.2.13 An n × n matrix A is called Hermitian if A = A∗. Thus a real symmetric matrix is Hermitian.

Theorem 13.2.14 If A is Hermitian, there exists a unitary matrix U such that

| U∗AU = D | (13.10)

where D is a real diagonal matrix. That is, D has nonzero entries only on the main diagonal and these are real. Furthermore, the columns of U are an orthonormal basis for Fn.

Proof: From Schur’s theorem above, there exists U unitary such that

U∗AU = T

where T is an upper triangular matrix. Then from Lemma 13.2.10

T∗ = (U∗AU)∗ = U∗A∗U = T.

Thus T = T∗ and T is upper triangular. This can only happen if T is really a diagonal matrix having real entries on the main diagonal. (If i ≠ j, one of Tij or Tji equals zero. But Tij = and so they are both zero. Also Tii =

Finally, let

U = ( u₁ u₂ · · · un)

where the ui denote the columns of U and

The equation, U∗AU = D implies

where the entries denote the columns of AU and UD respectively. Therefore, Aui = λiui and since the matrix is unitary, the ijth entry of U∗U equals δij and so

This proves the corollary because it shows the vectors {ui} form an orthonormal basis.

Corollary 13.2.15 If A is a real symmetric (A = AT) matrix, then A is Hermitian and there exists a real unitary matrix U such that UTAU = D where D is a diagonal matrix.

Proof: This follows from Theorem 13.2.14 which says the eigenvalues are all real. Then if Ax = λx, the same is true of . and so in the construction for Schur’s theorem, you can always deal exclusively with real eigenvectors as long as your matrices are real and symmetric. When you construct the matrix which reduces the problem to a smaller one having A₁ in the lower right corner, use the Gram Schmidt process on Rn using the real dot product to construct vectors, v₂,· · ·, vn in Rn such that {v₁,· · ·, vn} is an orthonormal basis for Rn. The matrix A₁ is symmetric also. This is because for j, k ≥ 2

Therefore, continuing this way, the process of the proof delivers only real vectors and real matrices.

13.3 Least Square Approximation

A very important technique is that of the least square approximation.

Lemma 13.3.1 Let A be an m × n matrix and let A (Fn) denote the set of vectors in Fm which are of the form Ax for some x ∈ Fn. Then A (Fn) is a subspace of  Fm.

Proof: Let Ax and Ay be two points of A (Fn). It suffces to verify that if a, b are scalars, then aAx + bAy is also in A (Fn) . But aAx + bAy = A (ax + by) because A is linear.

There is also a useful observation about orthonormal sets of vectors which is stated in the next lemma.

Lemma 13.3.2 Suppose {x₁, x₂,· · ·, xr} is an orthonormal set of vectors. Then if c₁,· · ·, cr are scalars,

Proof: This follows from the deﬁnition. From the properties of the dot product and using the fact that the given set of vectors is orthonormal,

The following theorem gives the equivalence of an orthogonality condition with a minimization condition..

Theorem 13.3.3 Let y ∈ Fm and let A be an m × n matrix. Then there exists x ∈ Fn minimizing the function x → |y−Ax|². Furthermore, x minimizes this function if and only if

(y−Ax)·Aw = 0

for all w. ∈ Fn

Proof: Let {Axk} be an orthonormal basis for AFn. Next note that for a given y,

In particular

for all w ∈ AFn since {Axk} is a basis. Also note that by Lemma 13.3.2,

because the {Axk} are orthonormal. Therefore, letting

It follows that the minimum exists and occurs when yk = (Axk, y) for each k. The above shows that Ax is the closest vector in AFn to y and that (y − Ax, w) = 0 for all w ∈ AFn.

Now suppose x is such that (y − Ax, w) = 0 for all w ∈ AFn. Why is Ax as close as possible to y? Letting z ∈ Fn,

and so the smallest value of |y − Az| is obtained when Ax = Az. Thus Ax is as close as possible to y if the orthogonality condition holds.

Recall the deﬁnition of the adjoint of a matrix.

Deﬁnition 13.3.4 Let A be an m × n matrix. Then

This means you take the transpose of A and then replace each entry by its conjugate. This matrix is called the adjoint. Thus in the case of real matrices having only real entries, the adjoint is just the transpose.

Lemma 13.3.5 Let A be an m × n matrix. Then

Ax · y = x·A∗y

Proof: This follows from the deﬁnition.

The next corollary gives the technique of least squares.

Corollary 13.3.6 A value of x which solves the problem of Theorem 13.3.3 is obtained by solving the equation

A*Ax = A*y

and furthermore, there exists a solution to this system of equations.

Proof: For x the minimizer of Theorem 13.3.3, (y−Ax) ·Aw = 0 for all w ∈ Fn and from Lemma 13.3.5, this is the same as saying

A∗ (y−Ax) · w = 0

for all w ∈ Fn,/i>. This implies

A∗y − A∗Ax = 0.

Therefore, there is a solution to the equation of this corollary, and it solves the minimization problem of Theorem 13.3.3.

Note that x might not be unique but Ax, the closest point of AFn to y is unique. This was shown in the above argument. Sometimes people like to consider the x such that Ax is as close as possible to y and also |x| is as small as possible. It turns out that there exists a unique such x and it is denoted as A⁺y. However, this is as far as I will go with this in this part of the book.

13.3.1 The Least Squares Regression Line

For the situation of the least squares regression line discussed here I will specialize to the case of Rn rather than Fn because it seems this case is by far the most interesting and the extra details are not justiﬁed by an increase in utility. Thus, everywhere you see A∗ it suffices to place AT.

An important application of Corollary 13.3.6 is the problem of finding the least squares regression line in statistics. Suppose you are given points in xy plane

and you would like to find constants m and b such that the line y = mx + b goes through all these points. Of course this will be impossible in general. Therefore, try to find m, b to get as close as possible. The desired system is

which is of the form y = Ax and it is desired to choose m and b to make

as small as possible. According to Theorem 13.3.3 and Corollary 13.3.6, the best values for m and b occur as the solution to

where

Thus, computing ATA,

Solving this system of equations for m and b,

and

One could clearly do a least squares fit for curves of the form y = ax² + bx + c in the same way. In this case you want to solve as well as possible for a, b, and c the system

and one would use the same technique as above. Many other similar problems are important, including many in higher dimensions and they are all solved the same way.

13.3.2 The Fredholm Alternative

The next major result is called the Fredholm alternative. It comes from Theorem 13.3.3 and Lemma 13.3.5.

Theorem 13.3.7 Let A be an m × n matrix. Then there exists x ∈ Fn such that Ax = y if and only if whenever A∗z = 0 it follows that z · y = 0.

Proof: First suppose that for some x ∈ Fn, Ax = y. Then letting A∗z = 0 and using Lemma 13.3.5

y·z = Ax·z = x·A∗z = x·0 = 0.

This proves half the theorem.

To do the other half, suppose that whenever, A∗z = 0 it follows that z·y = 0. It is necessary to show there exists x ∈ Fn such that y = Ax. From Theorem 13.3.3 there exists x minimizing |y−Ax|² which therefore satisﬁes

| (y−Ax)·Aw = 0 | (13.11)

for all w ∈ Fn. Therefore, for all w ∈ Fn,

A∗ (y−Ax)· w = 0

which shows that A∗ (y−Ax) = 0. (Why?) Therefore, by assumption,

(y−Ax) · y = 0.

Now by ( 13.11) with w = x,

(y−Ax) · (y−Ax) = (y−Ax)·y−(y−Ax) · Ax = 0

showing that y = Ax.

The following corollary is also called the Fredholm alternative.

Corollary 13.3.8 Let A be an m × n matrix. Then A is onto if and only if A∗ is one to one.

Proof: Suppose first A is onto. Then by Theorem 13.3.7, it follows that for all y ∈ Fm, y · z = 0 whenever A∗z = 0. Therefore, let y = z where A∗z = 0 and conclude that z · z = 0 whenever A∗z = 0. If A∗x = A∗y, then A∗ (x − y) = 0 and so x − y = 0. Thus A∗ is one to one.

Now let y ∈ Fm be given. y · z = 0 whenever A∗z = 0 because, since A∗ is assumed to be one to one, and 0 is a solution to this equation, it must be the only solution. Therefore, by Theorem 13.3.7 there exists x such that Ax = y therefore, A is onto.

13.4 The Right Polar Factorization∗

The right polar factorization involves writing a matrix as a product of two other matrices, one which preserves distances and the other which stretches and distorts. First here are some lemmas which review and add to many of the topics discussed so far about adjoints and orthonormal sets and such things. This is of fundamental signiﬁcance in geometric measure theory and also in continuum mechanics. Not surprisingly the stress should depend on the part which stretches and distorts. See [6].

Lemma 13.4.1 Let A be a Hermitian matrix such that all its eigenvalues are nonnegative. Then there exists a Hermitian matrix A1/2 such that A1/2 has all nonnegative eigenvalues and (A1/2)² = A.

Proof: Since A is Hermitian, there exists a diagonal matrix D having all real nonnegative entries and a unitary matrix U such that A = U∗DU. Then denote by D1/2 the matrix which is obtained by replacing each diagonal entry of D with its square root. Thus D1/2D1/2 = D. Then deﬁne

A1/2 ≡ U∗D1/2U.

Then

(A1/2)² = U∗D1/2UU∗D1/2U = U∗DU = A.

Since D1/2 is real,

(U∗D1/2U)∗ = U∗(D1/2)∗ (U∗)∗ = U∗D1/2U

so A1/2 is Hermitian.

Next it is helpful to recall the Gram Schmidt algorithm and observe a certain property stated in the next lemma.

Lemma 13.4.2 Suppose {w₁,· · ·, wr, vr+1,· · ·, vp} is a linearly independent set of vectors such that {w₁,· · ·, wr} is an orthonormal set of vectors. Then when the Gram Schmidt process is applied to the vectors in the given order, it will not change any of the w₁,· · ·, wr.

Proof: Let {u₁,· · ·, up} be the orthonormal set delivered by the Gram Schmidt process. Then u₁ = w₁ because by deﬁnition, u₁ ≡ w₁/ | w₁| = w₁. Now suppose uj = wj for all j ≤ k ≤ r. Then if k < r, consider the deﬁnition of uk+1.

By induction, uj = wj and so this reduces to wk+1/ |wk+1| = wk+1.

This lemma immediately implies the following lemma.

Lemma 13.4.3 Let V be a subspace of dimension p and let {w₁,· · ·, wr} be an orthonormal set of vectors in V. Then this orthonormal set of vectors may be extended to an orthonormal basis for V,

{w₁,· · ·,wr , yr + 1, . . . , yp}

Proof: First extend the given linearly independent set {w₁,· · ·, wr} to a basis for V and then apply the Gram Schmidt theorem to the resulting basis. Since {w₁,· · ·, wr} is orthonormal it follows from Lemma 13.4.2 the result is of the desired form, an orthonormal basis extending {w₁,· · ·, wr}.

Here is another lemma about preserving distance.

Lemma 13.4.4 Suppose R is an m × n matrix with m > n and R preserves distances. Then R∗R = I.

Proof: Since R preserves distances, |Rx| = |x| for every x. Therefore from the axioms of the dot product,

and so for all x, y,

(R∗Rx − x, y) + (y, R∗Rx − x) = 0

Hence for all x, y,

Re (R∗Rx − x, y) = 0

Now for a x, y given, choose α ∈ C such that

α (R∗Rx − x, y) = |(R∗Rx−x, y)|

Then

Thus |(R∗Rx−x, y)| = 0 for all x, y because the given x, y were arbitrary. Let y = R∗Rx−x to conclude that for all x,

R∗Rx−x = 0

which says R∗R = I since x is arbitrary.

With this preparation, here is the big theorem about the right polar factorization.

Theorem 13.4.5 Let F be an m × n matrix where m ≥ n. Then there exists a Hermitian n × n matrix U which has all nonnegative eigenvalues and an m × n matrix R which preserves distances and satisﬁes R∗R = I such that

F = RU.

Proof: Consider F∗F. This is a Hermitian matrix because

( F∗F )∗ = F∗ (F∗)∗ = F∗F

Also the eigenvalues of the n × n matrix F∗F are all nonnegative. This is because if x is an eigenvalue,

λ (x, b) = (F∗F x, x) = (Fx,Fx) ≥ 0.

Therefore, by Lemma 13.4.1, there exists an n × n Hermitian matrix U having all nonnegative eigenvalues such that

U² = F∗F.

Consider the subspace U (Fn). Let {Ux₁,· · ·, Uxr} be an orthonormal basis for

U (Fn) ⊆ Fn.

Note that U (Fn) might not be all of Fn. Using Lemma 13.4.3, extend to an orthonormal basis for all of Fn,

{Ux₁,· · ·, Uxr, yr+1,· · ·, yn}.

Next observe that {Fx₁,· · ·, Fxr} is also an orthonormal set of vectors in Fm. This is because

Therefore, from Lemma 13.4.3 again, this orthonormal set of vectors can be extended to an orthonormal basis for Fm,

{Fx₁,· · ·, Fxr, zr+1,· · ·, zm}

Thus there are at least as many zk as there are yj. Now for x ∈ Fn, since

{Ux₁,· · ·, Uxr, yr+1,· · ·, yn}

is an orthonormal basis for Fn, there exist unique scalars,

c₁· · ·, cr, dr+1,· · ·, dn

such that

Define

| [picture] | (13.12)

Then also there exist scalars bk such that

and so from (13.12),

Therefore, and this shows

RUx = Fx.

From (13.12) and Lemma 13.3.2 R preserves distances. Therefore, by Lemma 13.4.4 R∗R = I.

13.5 The Singular Value Decomposition

In this section, A will be an m × n matrix. To begin with, here is a simple lemma.

Lemma 13.5.1 Let A be an m × n matrix. Then A∗A is self adjoint and all its eigenvalues are nonnegative.

Proof: It is obvious that A∗A is self adjoint. Suppose A∗Ax = λx. Then λ|x|² = (λx, x) = (A∗Ax, x) = (Ax,Ax) ≥ 0.

Deﬁnition 13.5.2 Let A be an m × n matrix. The singular values of A are the square roots of the positive eigenvalues of A∗A.

With this deﬁnition and lemma here is the main theorem on the singular value decomposition.

Theorem 13.5.3 Let A be an m × n matrix. Then there exist unitary matrices, U and V of the appropriate size such that

where σ is of the form

for the σi the singular values of A.

Proof: By the above lemma and Theorem 13.2.14 there exists an orthonormal basis, such that where and equals zero if i > k. Thus for i > k, Avi = 0 because.

(Avi,Avi) = (AAvi,vi) = (0,vi) = 0.

For i = 1,· · ·, k, deﬁne ui ∈ Fm by

ui ≡ σi⁻¹Avi.

Thus Avi = σiui. Now

Thus {ui}ki=1 is an orthonormal set of vectors in Fm. Also,

Now extend {ui}ki=1 to an orthonormal basis for all of Fm,{ui}mi=1 and let

U ≡ (u₁· · ·um) 

while V ≡ (v₁· · ·vn) . Thus U is the matrix which has the ui as columns and V is deﬁned as the matrix which has the vi as columns. Then

where σ is given in the statement of the theorem.

The singular value decomposition has as an immediate corollary the following interesting result.

Corollary 13.5.4 Let A be an m × n matrix. Then the rank of A and A∗equals the number of singular values.

Proof: Since V and U are unitary, it follows that

Also since U, V are unitary,

13.6 Approximation In The Frobenius Norm∗

The Frobenius norm is one of many norms for a matrix. It is arguably the most obvious of all norms. Here is its deﬁnition.

Deﬁnition 13.6.1 Let A be a complex m × n matrix. Then

||A||F ≡ (trace (AA∗))1/2

Also this norm comes from the inner product

(A, B)F ≡ trace (AB∗)

Thus ||A||²F is easily seen to equal Σij |aij|2 so essentially, it treats the matrix as a vector in Fm × n.

Lemma 13.6.2 Let A be an m × n complex matrix with singular matrix

with σ as deﬁned above. Then

| [picture] | (13.13)

and the following hold for the Frobenius norm. If U, V are unitary and of the right size,

| ||UA||F = ||A||F , ||UAV ||F = ||A||F . | (13.14)

Proof: From the deﬁnition and letting U, V be unitary and of the right size,

Also,

It follows

Now consider (13.13). From what was just shown,

Of course, this shows that

the sum of the squares of the singular values of A.

Why is the singular value decomposition important? It implies

where σ is the diagonal matrix having the singular values down the diagonal. Now sometimes A is a huge matrix, 1000 × 2000 or something like that. This happens in applications to situations where the entries of A describe a picture. What also happens is that most of the singular values are very small. What if you deleted those which were very small, say for all i ≥ l and got a new matrix,

Then the entries of A 'would end up being close to the entries of A but there is much less information to keep track of. This turns out to be very useful. More precisely, letting

Thus A is approximated by A′ where A′ has rank l < r. In fact, it is also true that out of all matrices of rank l, this A′ is the one which is closest to A in the Frobenius norm. Here is why.

Let B be a matrix which has rank l. Then from Lemma 13.6.2

and since the singular values of A decrease from the upper left to the lower right, it follows that for B to be closest as possible to A in the Frobenius norm,

which implies B = A′ above. This is really obvious if you look at a simple example. Say

for example. Then what rank 1 matrix would be closest to this one in the Frobenius norm? Obviously

13.7 Moore Penrose Inverse∗

The singular value decomposition also has a very interesting connection to the problem of least squares solutions. Recall that it was desired to find x such that |Ax − y| is as small as possible. Lemma 13.3.3 shows that there is a solution to this problem which can be found by solving the system A∗Ax = A∗y. Each x which solves this system, solves the minimization problem as was shown in the lemma just mentioned. Now consider this equation for the solutions of the minimization problem in terms of the singular value decomposition.

Therefore, this yields the following upon using block multiplication and multiplying on the left by V

| [picture] | (13.15)

One solution to this equation which is very easy to spot is

| [picture] | (13.16)

This special x is denoted by A⁺y. The matrix is denoted by A⁺. Thus x just deﬁned is a solution to the least squares problem of finding the x such that Ax is as close as possible to y. Suppose now that z is some other solution to this least squares problem. Thus from the above,

and so, multiplying both sides by

To make V ∗z as small as possible, you would have only the first r entries of V ∗z be nonzero since the later ones will be zeroed out anyway so they are unnecessary. Hence

and consequently

However, minimizing |V ∗z| is the same as minimizing |z| because V is unitary. Hence A⁺y is the solution to the least squares problem which has smallest norm.

13.8 Exercises

Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal. If the matrix is orthogonal, determine whether it is proper or improper.

Show that every real matrix may be written as the sum of a skew symmetric and a symmetric matrix. Hint: If A is an n × n matrix, show that is skew symmetric.

Let x be a vector in Rn and consider the matrix Show this matrix is both symmetric and orthogonal.

For U an orthogonal matrix, explain why ||Ux|| = ||x|| for any vector x. Next explain why if U is an n × n matrix with the property that ||Ux|| = ||x|| for all vectors, x, then U must be orthogonal. Thus the orthogonal matrices are exactly those which preserve distance.

A quadratic form in three variables is an expression of the form a₁x² + a₂y² + a₃z² + a₄xy + a₅xz + a₆yz. Show that every such quadratic form may be written as

where A is a symmetric matrix.

Given a quadratic form in three variables, x, y, and z, show there exists an orthogonal matrix U and variables x′, y′, z′ such that

with the property that in terms of the new variables, the quadratic form is

λ₁ (x′)² + λ₂ (y′)² + λ₃ (z′)²

where the numbers, λ₁, λ₂, and λ₃ are the eigenvalues of the matrix A in Problem 5.

If A is a symmetric invertible matrix, is it always the case that A⁻¹ must be symmetric also? How about Ak for k a positive integer? Explain.

If A, B are symmetric matrices, does it follow that AB is also symmetric?

Suppose A, B are symmetric and AB = BA. Does it follow that AB is symmetric?

Here are some matrices. What can you say about the eigenvalues of these matrices just by looking at them?

Find the eigenvalues and eigenvectors of the matrix Here b, c are real numbers.

Find the eigenvalues and eigenvectors of the matrix Here a, b, c are real numbers.

Find the eigenvalues and an orthonormal basis of eigenvectors for A.

Hint: Two eigenvalues are 12 and 18.

Find the eigenvalues and an orthonormal basis of eigenvectors for A.

Hint: One eigenvalue is 3.

Show that if A is a real symmetric matrix and λ and µ are two different eigenvalues, then if x is an eigenvector for λ and y is an eigenvector for µ, then x·y = 0. Also all eigenvalues are real. Supply reasons for each step in the following argument. First

and so . This shows that all eigenvalues are real. It follows all the eigenvectors are real. Why? Now let x, y, µ and λ be given as above.

λ (x·y) = λx·y = Ax·b = x·Ay = x·µy = µ (x·y) = µ (x·y)

and so

(λ−µ) x·y = 0.

Since λ ̸= µ, it follows x · y = 0.

Suppose U is an orthogonal n × n matrix. Explain why rank (U) = n.

Show that if A is an Hermitian matrix and λ and µ are two different eigenvalues, then if x is an eigenvector for λ and y is an eigenvector for µ, then x·y = 0. Also all eigenvalues are real. Supply reasons for each step in the following argument. First

and so This shows that all eigenvalues are real. Now let x, y, µ and λ be given as above.

and so

(λ−µ) x·y = 0.

Since λ ≠ µ, it follows x·y = 0.

Show that the eigenvalues and eigenvectors of a real matrix occur in conjugate pairs.

If a real matrix A has all real eigenvalues, does it follow that A must be symmetric. If so, explain why and if not, give an example to the contrary.

Suppose A is a 3× 3 symmetric matrix and you have found two eigenvectors which form an orthonormal set. Explain why their cross product is also an eigenvector.

Study the deﬁnition of an orthonormal set of vectors. Write it from memory.

Determine which of the following sets of vectors are orthonormal sets. Justify your answer.

Show that if {u₁,· · ·, un} is an orthonormal set of vectors in Fn, then it is a basis. Hint: It was shown earlier that this is a linearly independent set. If you wish, replace Fn with Rn. Do this version if you do not know the dot product for vectors in Cn.

Fill in the missing entries to make the matrix orthogonal.

Fill in the missing entries to make the matrix orthogonal.

Fill in the missing entries to make the matrix orthogonal.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: One eigenvalue is -2.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: Two eigenvalues are 18 and 24.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: Two eigenvalues are 12 and 18.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: The eigenvalues are −3,−2, 1.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: The eigenvalues are 0, 2, 2 where 2 is listed twice because it is a root of multiplicity 2. 

Find the eigenvalues and an orthonormal basis of eigenvectors for A. Diagonalize A by finding an orthogonal matrix U and a diagonal matrix D such that UTAU = D.

Hint: The eigenvalues are 2, 1, 0.

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix

Hint: The eigenvalues are 1, 2,−2. 

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix

Hint: The eigenvalues are −1, 2,−1 where −1 is listed twice because it has multiplicity 2 as a zero of the characteristic equation.

Explain why a matrix A is symmetric if and only if there exists an orthogonal matrix U such that A = UTDU for D a diagonal matrix.

The proof of Theorem 13.3.3 concluded with the following observation. If −ta + t²b ≥ 0 for all t ∈ R and b ≥ 0, then a = 0. Why is this so?

Using Schur’s theorem, show that whenever A is an n × n matrix, det (A) equals the product of the eigenvalues of A.

In the proof of Theorem 13.3.7 the following argument was used. If x · w = 0 for all w ∈ Rn, then x = 0. Why is this so?

Using Corollary 13.3.8 show that a real m × n matrix is onto if and only if its transpose is one to one.

Suppose A is a 3 × 2 matrix. Is it possible that AT is one to one? What does this say about A being onto? Prove your answer.

Find the least squares solution to the following system.

  x + 2y = 1

2x + 3y = 2

3x + 5y = 4

You are doing experiments and have obtained the ordered pairs, 

(0, 1) , (1, 2) , (2, 3.5) , (3, 4)

Find m and b such that y = mx + b approximates these four points as well as possible. Now do the same thing for y = ax² + bx + c, finding a, b, and c to give the best approximation.

Suppose you have several ordered triples, (xi, yi, zi) . Describe how to find a polynomial,

z = a + bx + cy + dxy + ex² + fy²

for example giving the best fit to the given ordered triples. Is there any reason you have to use a polynomial? Would similar approaches work for other combinations of functions just as well?

Find an orthonormal basis for the spans of the following sets of vectors.

(a) (3 ,−4, 0) , (7,−1, 0) , (1, 7, 1).
(b) (3, 0,−4) , (11, 0, 2) , (1, 1, 7)
(c) (3, 0,−4) , (5, 0, 10) , (−7, 1, 1)

Using the Gram Schmidt process or the QR factorization, find an orthonormal basis for the span of the vectors, (1, 2, 1) , (2,−1, 3) , and (1, 0, 0) .

Using the Gram Schmidt process or the QR factorization, find an orthonormal basis for the span of the vectors, (1, 2, 1, 0) , (2,−1, 3, 1) , and (1, 0, 0, 1) .

The set, V ≡ {(x, y, z) : 2x + 3y −z = 0} is a subspace of R³. Find an orthonormal basis for this subspace.

The two level surfaces, 2x + 3y −z + w = 0 and 3x−y + z + 2w = 0 intersect in a subspace of R⁴, find a basis for this subspace. Next find an orthonormal basis for this subspace.

Let A, B be a m × n matrices. Deﬁne an inner product on the set of m × n matrices by

(A, B)F ≡ trace (AB∗) .

Show this is an inner product satisfying all the inner product axioms. Recall for M an n × n matrix, trace The resulting norm, ||·||F is called the Frobenius norm and it can be used to measure the distance between two matrices.

Let A be an m × n matrix. Show

where the σj are the singular values of A.

The trace of an n × n matrix M is deﬁned as ∑i Mii. In other words it is the sum of the entries on the main diagonal. If A, B are n × n matrices, show trace (AB) = trace (BA). Now explain why if A = S⁻¹BS it follows trace (A) = trace (B). Hint: For the first part, write these in terms of components of the matrices and it just falls out.

Using Problem 53 and Schur’s theorem, show that the trace of an n × n matrix equals the sum of the eigenvalues.

If A is a general n × n matrix having possibly repeated eigenvalues, show there is a sequence {Ak} of n × n matrices having distinct eigenvalues which has the property that the ijth entry of Ak converges to the ijth entry of A for all ij. Hint: Use Schur’s theorem.

Prove the Cayley Hamilton theorem as follows. First suppose A has a basis of eigenvectors {vk}nk=1 , Avk = λkvk. Let p (λ) be the characteristic polynomial. Show p (A) vk = p (λk) vk = 0. Then since {vk} is a basis, it follows p (A) x = 0 for all x and so p (A) = 0. Next in the general case, use Problem 55 to obtain a sequence {Ak} of matrices whose entries converge to the entries of A such that Ak has n distinct eigenvalues and therefore by Theorem 12.1.13 Ak has a basis of eigenvectors. Therefore, from the first part and for pk (λ) the characteristic polynomial for Ak, it follows pk (Ak) = 0. Now explain why and the sense in which

lim pk (Ak) = p (A) .

k→∞

Show that the Moore Penrose inverse A⁺ satisﬁes the following conditions.

AA⁺A = A, A⁺AA⁺ = A⁺, A⁺A, AA⁺ are Hermitian.

Next show that if A₀ satisﬁes the above conditions, then it must be the Moore Penrose inverse and that if A is an n × n invertible matrix, then A⁻¹ satisﬁes the above conditions. Thus the Moore Penrose inverse generalizes the usual notion of inverse but does not contradict it. Hint: Let

and suppose

where P is the same size as σ. Now use the conditions to identify P = σ, Q = 0 etc.

Find the least squares solution to

Next suppose ε is so small that all ε² terms are ignored by the computer but the terms of order ε are not ignored. Show the least squares equations in this case reduce to

Find the solution to this and compare the y values of the two solutions. Show that one of these is −2 times the other. This illustrates a problem with the technique for finding least squares solutions presented as the solutions to A∗Ax = A∗y. One way of dealing with this problem is to use the QR factorization. This is illustrated in the next problem. It turns out that this helps alleviate some of the round off difficulties of the above.

Show that the equations A∗Ax = A∗y can be written as R∗Rx = R∗Q∗y where R is upper triangular and R∗ is lower triangular. Explain how to solve this system efficiently. Hint: You first find Rx and then you find x which will not be hard because R is upper triangular.

Show that A⁺ = (A∗A)⁺A∗.  Hint: You might use the description of A⁺ in terms of the singular value decomposition.

¹Recall that for a complex number, x + iy, the complex conjugate, denoted by is deﬁned as x - iy.

Numerical Methods For Solving Linear Systems

14.1 Iterative Methods For Linear Systems

Consider the problem of solving the equation

| Ax = b | (14.1)

where A is an n × n matrix. In many applications, the matrix A is huge and composed mainly of zeros. For such matrices, the method of Gauss elimination (row operations) is not a good way to solve the system because the row operations can destroy the zeros and storing all those zeros takes a lot of room in a computer. These systems are called sparse. To solve them it is common to use an iterative technique. The idea is to obtain a sequence of approximate solutions which get close to the true solution after a sufficient number of iterations.

Deﬁnition 14.1.1 Let be a sequence of vectors in Fn. Say

Then this sequence is said to converge to the vector x = (x₁,· · ·, xn) ∈ Fn, written as

if for each j = 1, 2,· · ·, n,

In words, the sequence converges if the entries of the vectors in the sequence converge to the corresponding entries of x.

Example 14.1.2 Consider xk = . Find

From the above deﬁnition, this limit is the vector (0, 1, 0) because

A more complete mathematical explanation is given in Linear Algebra. Linear Algebra

14.1.1 The Jacobi Method

The first technique to be discussed here is the Jacobi method which is described in the following deﬁnition. In this technique, you have a sequence of vectors, {xk} which converge to the solution to the linear system of equations and to get the ith component of the xk+1, you use all the components of xk except for the ith. The precise description follows.

Deﬁnition 14.1.3 The Jacobi iterative technique, also called the method of simultaneous corrections, is deﬁned as follows. Let x¹ be an initial vector, say the zero vector or some other vector. The method generates a succession of vectors, x², x³, x⁴,· · · and hopefully this sequence of vectors will converge to the solution to (14.1). The vectors in this list are called iterates and they are obtained according to the following procedure. Letting A = (aij) ,

| [picture] | (14.2)

In terms of matrices, letting

The iterates are deﬁned as

| [picture] | (14.3)

The matrix on the left in (14.3) is obtained by retaining the main diagonal of A and setting every other entry equal to zero. The matrix on the right in (14.3) is obtained from A by setting every diagonal entry equal to zero and retaining all the other entries unchanged.

Example 14.1.4 Use the Jacobi method to solve the system

In terms of the matrices, the Jacobi iteration is of the form

Now iterate this starting with

Solving this system yields

Then you use x² to find

The solution is

Now use this as the new data to find

Thus you find

Then another iteration for x⁵ gives

and so

The solution to the system of equations obtained by row operations is

so already after only five iterations the iterates are pretty close to the true solution. How well does it work?

A few more iterates will yield a better solution.

14.1.2 The Gauss Seidel Method

The Gauss Seidel method differs from the Jacobi method in using for all j < i in going from xk to xk+1. This is why it is called the method of successive corrections. The precise description of this method is in the following deﬁnition.

Deﬁnition 14.1.5 The Gauss Seidel method, also called the method of successive corrections is given as follows. For A = (aij) , the iterates for the problem Ax = b are obtained according to the formula

| [picture] | (14.4)

In terms of matrices, letting

The iterates are deﬁned as

| [picture] | (14.5)

In words, you set every entry in the original matrix which is strictly above the main diagonal equal to zero to obtain the matrix on the left. To get the matrix on the right, you set every entry of A which is on or below the main diagonal equal to zero. Using the iteration procedure of (14.4) directly, the Gauss Seidel method makes use of the very latest information which is available at that stage of the computation.

The following example is the same as the example used to illustrate the Jacobi method.

Example 14.1.6 Use the Gauss Seidel method to solve the system

In terms of matrices, this procedure is

As before, let x¹ be the zero vector. Thus the first iteration is to solve

Hence

Thus is given by

And so

Another iteration for x⁴ involves solving

and so

Recall the answer is

so the iterates are already pretty close to the answer. You could continue doing these iterates and it appears they converge to the solution. Now consider the following example.

Example 14.1.7 Use the Gauss Seidel method to solve the system

The exact solution is given by doing row operations on the augmented matrix. When this is done the row echelon form is

and so the solution is

The Gauss Seidel iterations are of the form

and so, multiplying by the inverse of the matrix on the left, the iteration reduces to the following in terms of matrix multiplication.

This time, we will pick an initial vector close to the answer. Let

This is very close to the answer. Now lets see what the Gauss Seidel iteration does to it.

You can’t expect to be real close after only one iteration. Lets do another.

The iterates seem to be getting farther from the actual solution. Why is the process which worked so well in the other examples not working here? A better question might be: Why does either process ever work at all? A complete answer to this question is given in more advanced linear algebra books. You can also see it in Linear Algebra.

Both iterative procedures for solving

| Ax = b | (14.6)

are of the form

Bxr+1 = −Cxr + b

where A = B + C. In the Jacobi procedure, the matrix C was obtained by setting the diagonal of A equal to zero and leaving all other entries the same while the matrix B was obtained by making every entry of A equal to zero other than the diagonal entries which are left unchanged. In the Gauss Seidel procedure, the matrix B was obtained from A by making every entry strictly above the main diagonal equal to zero and leaving the others unchanged, and C was obtained from A by making every entry on or below the main diagonal equal to zero and leaving the others unchanged. Thus in the Jacobi procedure, B is a diagonal matrix while in the Gauss Seidel procedure, B is lower triangular. Using matrices to explicitly solve for the iterates, yields

| xr+1 = −B⁻¹Cxr + B⁻¹b. | (14.7)

This is what you would never have the computer do, but this is what will allow the statement of a theorem which gives the condition for convergence of these and all other similar methods.

Theorem 14.1.8 Let A = B + C and suppose all eigenvalues of B⁻¹C have absolute value less than 1 where A = B + C. Then the iterates in (14.7) converge to the unique solution of (14.6).

A complete explanation of this important result is found in more advanced linear algebra books. You can also see it in Linear Algebra. It depends on a theorem of Gelfand which is completely proved in this reference. Theorem 14.1.8 is very remarkable because it gives an algebraic condition for convergence, which is essentially an analytical question.

14.2 The Operator Norm∗

Recall that for x ∈ Cn,

Also recall Theorem 3.2.17 which says that

| |z| ≥ 0 and |z| = 0 if and only if z = 0 | (14.8)

| If α is a scalar, |αz| = |α| |z| | (14.9)

| |z + w| ≤ |z| +|w|. | (14.10)

If you have the above axioms holding for ||·|| replacing |·|, then ||·|| is called a norm. For example, you can easily verify that

||x|| ≡ max { |xi|, i = 1,· · ·, n : x = (x₁,· · ·, xn) }

is a norm. However, there are many other norms.

One important observation is that x ||x|| is a continuous function. This follows from the observation that from the triangle inequality,

||x − y|| + ||y|| ≥ ||x||

 ||x − y|| + ||x|| = ||y − x|| + ||x|| ≥ ||y||

Hence

||x|| − ||y|| ≤ ||x − y||
||y|| − ||x|| ≤ ||x − y||

and so

| ||x|| − ||y|| | ≤ ||x − y||

This section will involve some analysis. If you want to talk about norms, this is inevitable. It will need some of the theorems of calculus which are usually neglected. In particular, it needs the following result which is a case of the Heine Borel theorem. To see this proved, see any good calculus book.

Theorem 14.2.1 Let S denote the points x ∈ Fn such that |x| = 1. Then if is any sequence of points of S, there exists a subsequence which converges to a point of S.

Deﬁnition 14.2.2 Let A be an m × n matrix. Let ||·||k denote a norm on Ck. Then the operator norm is deﬁned as follows.

||A|| ≡ max { ||Ax||m : ||x||n ≤ 1 }

Lemma 14.2.3 The operator norm is well deﬁned and is in fact a norm on the vector space of m × n matrices.

Proof: It has already been observed that the m × n matrices form a vector space starting on Page 78. Why is ||A|| < ∞?

claim: There exists c > 0 such that whenever ||x|| ≤ 1, it follows that |x| ≤ c.

Proof of the claim: If not, then there exists {xk} such that ||xk|| ≤ 1 but |xk| > k for k = 1, 2,· · ·. Then |xk/|xk|| = 1 and so by the Heine Borel theorem from calculus, there exists a further subsequence, still denoted by k such that

Letting

It follows that ak → a in Fn. Hence

which converges to 0. However,

and so, by continuity of ||·|| mentioned above,

Therefore, y = 0 but also |y| = 1, a contradiction. This proves the claim.

Now consider why ||A|| < ∞. Let c be as just described in the claim.

sup { ||Ax||m : ||x||n ≤ 1} ≤ sup { ||Ax||m : |x| ≤ c}

Consider for x, y with |x| , |y| ≤ c

for some constant C. So x  Ax is continuous. Since the norm ||·||m is continuous also, it follows from the extreme value theorem of calculus that ||Ax||m achieves its maximum on the compact set {x : |x| ≤ c}. Thus ||A|| is well deﬁned. The only other issue of signiﬁcance is the triangle inequality. However,

≤ max { ||Ax||m : ||x||n ≤ 1} + max { ||Bx||m : ||x||n ≤ 1}
= ||A|| + ||B||

Obviously ||A|| = 0 if and only if A = 0. The rule for scalars is also immediate.

The operator norm is one way to describe the magnitude of a matrix. Earlier the Frobenius norm was discussed. The Frobenius norm is actually not used as much as the operator norm. Recall that the Frobenius norm involved considering the m × n matrix as a vector in Fmn and using the usual Euclidean norm. It can be shown that it really doesn’t matter which norm you use in terms of estimates because they are all equivalent. This is discussed in Problem 25 below for those who have had a legitimate calculus course, not just the usual undergraduate offering.

14.3 The Condition Number∗

Let A be an m × n matrix and consider the problem Ax = b where it is assumed there is a unique solution to this problem. How does the solution change if A is changed a little bit and if b is changed a little bit? This is clearly an interesting question because you often do not know A and b exactly. If a small change in these quantities results in a large change in the solution x, then it seems clear this would be undesirable. In what follows ||·|| when applied to a matrix will always refer to the operator norm.

Lemma 14.3.1 Let A, B be m × n matrices. Then for ||·|| denoting the operator norm,

||AB|| ≤ ||A|| ||B||.

Proof: This follows from the deﬁnition. Letting ||x|| ≤ 1, it follows from the deﬁnition of the operator norm that

||ABx|| ≤ ||A|| ||Bx|| ≤ ||A|| ||B|| ||x|| ≤ ||A|| ||B||

and so

Lemma 14.3.2 Let A, B be m × n matrices such that A⁻¹ exists, and suppose ||B||< 1/ A⁻¹ . Then (A + B)⁻¹ exists and

The above formula makes sense because ||A⁻¹B|| < 1.

Proof: By Lemma 14.3.1,

Suppose (A + B) x = 0. Then 0 = A (I + A⁻¹B) x and so since A is one to one, (I + A⁻¹B) x = 0. Therefore,

a contradiction. This also shows (I + A⁻¹B) is one to one. Therefore, both (A + B)⁻¹ and (I + A⁻¹B)−1 exist. Hence

(A + B)⁻¹ = (A (I + A⁻¹B))⁻¹ = (I + A⁻¹B)⁻¹A⁻¹

Now if

x = (I + A⁻¹B)⁻¹y

for ||y|| ≤ 1, then

(I + A⁻¹B) x = y

and so

and so

Since ||y|| ≤ 1 is arbitrary, this shows

Therefore,

Proposition 14.3.3 Suppose A is invertible, b ≠ 0, Ax = b, and A₁x₁ = b₁ where || A − A₁ || < 1/ || A⁻¹. Then

| [picture] | (14.11)

Proof: It follows from the assumptions that

Ax − A₁x + A₁x −A₁x₁ = b−b₁.

Hence

A₁ (x−x₁) = (A₁ −A) x + b−b₁.

Now A₁ = (A + (A₁ −A)) and so by the above lemma, A₁⁻¹ exists and so

(x − x₁) = A⁻¹(A₁ −A) x + A⁻¹(b−b₁)
= (A + (A₁ −A))⁻¹ (A₁ −A) x + (A + (A₁ − A))⁻¹ (b−b₁) .

By the estimate in Lemma 14.3.2,

Dividing by ||x||,

| [picture] | (14.12)

Now b = Ax = A (A⁻¹ b) and so ||b|| ||A|| ≤ ||A⁻¹b|| and so

||x|| = ||A-1 b|| ≥ ||b|| / ||A||.

Therefore, from (14.12),

This shows that the number, A⁻¹ ||A||, controls how sensitive the relative change in the solution of Ax = b is to small changes in A and b. This number is called the condition number. It is bad when it is large because a small relative change in b, for example, could yield a large relative change in x.

14.4 Exercises

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

Solve the system

using the Gauss Seidel method and the Jacobi method. Check your answer by also solving it using row operations.

If you are considering a system of the form Ax = b and A⁻¹ does not exist, will either the Gauss Seidel or Jacobi methods work? Explain. What does this indicate about using either of these methods for finding eigenvectors for a given eigenvalue?

Verify that

||x||∞ ≡ max { |xi|, i = 1,· · ·, n : x = (x₁,· · ·, xn)}

is a norm. Next verify that

is also a norm on Fn.

Let A be an n × n matrix. Denote by ||A||₂ the operator norm taken with respect to the usual norm on Fn. Show that

||A||₂ = σ₁

where σ₁ is the largest singular value. Next explain why ||A⁻¹||₂ = 1/σn where σn is the smallest singular value of A. Explain why the condition number reduces to σ₁/σn if the operator norm is deﬁned in terms of the usual norm,

Let p, q > 1 and 1/p + 1/q = 1. Consider the following picture.

Using elementary calculus, verify that for a, b > 0,

↑For p > 1, the p norm on Fn is deﬁned by

In fact, this is a norm and this will be shown in this and the next problem. Using the above problem in the context stated there where p, q > 1 and 1/p+1/q = 1, verify Holder’s inequality

Hint: You ought to consider the following.

Now use the result of the above problem.

↑ Now for p > 1, verify that ||x + y||p ≤ ||x||p+||y||p. Then verify the other axioms of a norm. This will give an inﬁnite collection of norms for Fn. Hint: You might do the following.

Now explain why p − 1 = p/q and use the Holder inequality.

This problem will reveal the best kept secret in undergraduate mathematics, the deﬁnition of the derivative of a function of n variables. Let ||·|| be a norm on Fn and also denote by ||·|| a norm on Fm. If you like, just use the standard norm on both Fn and Fm. It can be shown that this doesn’t matter at all (See Problem 25 on 368.) but to avoid possible confusion, you can be speciﬁc about the norm. A set U ⊆ Fn is said to be open if for every x ∈U, there exists some rx > 0 such that B (x, rx) ⊆U where

B (x, r) ≡ {y ∈ Fn : ||y−x|| < r}

This just says that if U contains a point x then it contains all the other points suﬃciently near to x. Let f : U Fm be a function deﬁned on U having values in Fm. Then f is differentiable at x ∈U means that there exists an m × n matrix A such that for every ε > 0, there exists a δ > 0 such that whenever 0 < |v| < δ, it follows that

Stated more simply,

Show that A is unique and verify that the ith column of A is

so in particular, all partial derivatives exist. This unique m × n matrix is called the derivative of f. It is written as Df (x) = A.

Numerical Methods For Solving
The Eigenvalue Problem

15.1 The Power Method For Eigenvalues

This chapter presents some simple ways to find eigenvalues and eigenvectors. It is only an introduction to this important subject. However, I hope to convey some of the ideas which are used. As indicated earlier, the eigenvalue eigenvector problem is extremely difficult. Consider for example what happens if you find an eigenvalue approximately. Then you can’t find an approximate eigenvector by the straight forward approach because A−λI is invertible whenever λ is not exactly equal to an eigenvalue.

The power method allows you to approximate the largest eigenvalue and also the eigenvector which goes with it. By considering the inverse of the matrix, you can also find the smallest eigenvalue. The method works in the situation of a nondefective matrix A which has a real eigenvalue of algebraic multiplicity 1, λn which has the property that |λk|< |λn| for all k ≠ n. Such an eigenvalue is called a dominant eigenvalue.

Let {x₁,· · ·, xn} be a basis of eigenvectors for Fn such that Axn = λnxn. Now let u₁ be some nonzero vector. Since {x₁,· · ·, xn} is a basis, there exists unique scalars, ci such that

Assume you have not been so unlucky as to pick u₁ in such a way that cn = 0. Then let Auk = uk+1 so that

| [picture] | (15.1)

For large m the last term, , determines quite well the direction of the vector on the right. This is because |λn| is larger than |λk| for k < n and so for a large m, the sum, , on the right is fairly insigniﬁcant. Therefore, for large m, um is essentially a multiple of the eigenvector xn, the one which goes with λn. The only problem is that there is no control of the size of the vectors um. You can fix this by scaling. Let S₂ denote the entry of Au₁ which is largest in absolute value. We call this a scaling factor. Then u₂ will not be just Au₁ but Au₁/S₂. Next let S₃ denote the entry of Au₂ which has largest absolute value and deﬁne u₃ ≡ Au₂/S₃. Continue this way. The scaling just described does not destroy the relative insigniﬁcance of the term involving a sum in (15.1). Indeed it amounts to nothing more than changing the units of length. Also note that from this scaling procedure, the absolute value of the largest element of uk is always equal to 1. Therefore, for large m,

Therefore, the entry of Aum which has the largest absolute value is essentially equal to the entry having largest absolute value of

and so for large m, it must be the case that λn ≈ Sm+1. This suggests the following procedure.

 Finding the largest eigenvalue with its eigenvector.

Start with a vector u₁ which you hope has a component in the direction of xn. The vector (1,· · ·, 1)T is usually a pretty good choice.

If uk is known,

where Sk+1 is the entry of Auk which has largest absolute value.

When the scaling factors, Sk are not changing much, Sk+1 will be close to the eigenvalue and uk+1 will be close to an eigenvector.

Check your answer to see if it worked well.

In finding an initial vector, it is clear that if you start with a vector which isn’t too far from an eigenvector, the process will work faster. Also, the computer is able to raise the matrix to a power quite easily. You might start with Apx for large p. As explained above, this will point in roughly the right direction. Then normalize it by dividing by the largest entry and use the resulting vector as your initial approximation. This ought to be close to an eigenvector and so the process would be expected to converge rapidly for this as an initial choice.

Example 15.1.1 Find the largest eigenvalue of

I will use the above suggestion.

Now divide by the largest entry to get the initial approximation for an eigenvector

The power method will now be applied to find the largest eigenvalue for the above matrix beginning with this vector.

Scaling this vector by dividing by the largest entry gives

Now lets do it again.

The new scaling factor is very close to the one just encountered. Therefore, it seems this is a good place to stop. The eigenvalue is approximately 11. 999 and the eigenvector is close to the one obtained above. How well does it work? With the above equation, consider

These are clearly very close so this is a good approximation. In fact, the exact eigenvalue is 12 and an eigenvector is

15.2 The Shifted Inverse Power Method

This method can find various eigenvalues and eigenvectors. It is a signiﬁcant generalization of the above simple procedure and yields very good results. The situation is this: You have a number α which is close to λ, some eigenvalue of an n × n matrix A. You don’t know λ but you know that α is closer to λ than to any other eigenvalue. Your problem is to find both λ and an eigenvector which goes with λ. Another way to look at this is to start with α and seek the eigenvalue λ, which is closest to α along with an eigenvector associated with λ. If α is an eigenvalue of A, then you have what you want. Therefore, we will always assume α is not an eigenvalue of A and so (A−αI)⁻¹ exists. When using this method it is nice to choose α fairly close to an eigenvalue. Otherwise, the method will converge slowly. In order to get some idea where to start, you could use Gerschgorin’s theorem to get a rough idea where to look. The method is based on the following lemma.

Lemma 15.2.1 Let be the eigenvalues of A, αnot an eigenvalue. Then xk is an eigenvector of A for the eigenvalue λk, if and only if xk is an eigenvector for (A − αI)⁻¹ corresponding to the eigenvalue

Proof: Let λk and xk be as described in the statement of the lemma. Then

if and only if

In explaining why the method works, we will assume A is nondefective. This is not necessary! One can use Gelfand’s theorem on the spectral radius which is presented in [11] and invariance of (A − αI)⁻¹ on generalized eigenspaces to prove more general results. It suffices to assume that the eigenspace for λk has dimension equal to the multiplicity of the eigenvalue λk but even this is not necessary to obtain convergence of the method. This method is better than might be supposed from the following explanation.

Pick u₁, an initial vector and let Axk = λkxk, where {x₁,· · ·, xn} is a basis of eigenvectors which exists from the assumption that A is nondefective. Assume α is closer to λn than to any other eigenvalue. Since A is nondefective, there exist constants, ak such that

Possibly λn is a repeated eigenvalue. Then combining the terms in the sum which involve eigenvectors for λn, a simpler description of u₁ is

where y is an eigenvector for λn which is assumed not equal to 0. (If you are unlucky in your choice for u₁, this might not happen and things won’t work.) Now the iteration procedure is deﬁned as

where Sk+1 is the element of (A−αI)⁻¹uk which has largest absolute value. From Lemma 15.2.1,

Now it is being assumed that λn is the eigenvalue which is closest to α and so for large k, the term,

is very small, while for every k ≥ 1, uk is a moderate sized vector because every entry has absolute value less than or equal to 1. Thus

where Ek → 0, y is some eigenvector for λn, and Ck is of moderate size, remaining bounded as k → ∞. Therefore, for large k,

uk+1 − Cky = CkEk≈ 0

and multiplying by (A − αI)⁻¹ yields

Therefore, for large k, uk is approximately equal to an eigenvector of (A − αI)⁻¹. Therefore,

and so you could take the dot product of both sides with uk and approximate λn by solving the following for λn.

How else can you find the eigenvalue from this? Suppose uk = (w₁,· · ·, wn)T and from the construction |wi| ≤ 1 and wk = 1 for some k. Then

Hence the entry of (A − αI)⁻¹uk which has largest absolute value is approximately and so it is likely that you can estimate λn using the formula

Of course this would fail if (A − αI)⁻¹uk had consistently more than one entry having equal absolute value, but this is unlikely.

Here is how you use the shifted inverse power method to find the eigenvalue and eigenvector closest to α.

Find (A − αI)⁻¹.

Pick u₁. It is important that where y is an eigenvector which goes with the eigenvalue closest to α and the sum is in an “invariant subspace corresponding to the other eigenvalues”. Of course you have no way of knowing whether this is so but it typically is so. If things don’t work out, just start with a different u₁. You were phenomenally unlucky in your choice.

If uk has been obtained,

where Sk+1 is the element of (A - αI)⁻¹uk which has largest absolute value.

When the scaling factors, Sk₊₁ are not changing much and the uk are not changing much, find the approximation to the eigenvalue by solving

for λ. The eigenvector is approximated by uk+1.

Check your work by multiplying by the original matrix to see how well what you have found works.

Also note that this is just the power method applied to (A − λI)⁻¹. The eigenvalue you want is the one which makes as large as possible for all λ ∈ σ (A). This is because making λ − α small is the same as making (λ − α)⁻¹ large.

Example 15.2.2 Find the eigenvalue of which is

closest to −7. Also find an eigenvector which goes with this eigenvalue.

In this case the eigenvalues are −6, 0, and 12 so the correct answer is −6 for the eigenvalue.

To get a good initial vector, follow the shortcut described earlier which works for the power method.

That middle term looks a lot like 0 so for an initial guess I will simply take it equal to 0. Also the other term is very close to 1. Therefore, I will take it to equal 1. Then

It looks like we have found an eigenvector for (A + 7I)⁻¹. Then to find the eigenvalue desired, solve

This yields λ = −6 which is actually the exact eigenvalue closest to −7.

Example 15.2.3 Consider the symmetric matrix Find

the middle eigenvalue and an eigenvector which goes with it.

Since A is symmetric, it follows it has three real eigenvalues which are solutions to

If you use your graphing calculator to graph this polynomial, you find there is an eigenvalue somewhere between −.9 and −.8 and that this is the middle eigenvalue. Of course you could zoom in and find it very accurately without much trouble but what about the eigenvector which goes with it? If you try to solve

there will be only the zero solution because the matrix on the left will be invertible and the same will be true if you replace −.8 with a better approximation like −.86 or −.855. This is because all these are only approximations to the eigenvalue and so the matrix in the above is nonsingular for all of these. Therefore, you will only get the zero solution and

However, there exists such an eigenvector and you can find it using the shifted inverse power method. Pick α = −.855. Then

Next use the power method for this matrix.

Now divide by the largest entry to get a good first approximation to the eigenvector.

Now divide by the largest entry to get the next approximation

Clearly this vector has not changed much and so the next scaling factor will be very close to the one just considered. Hence the eigenvalue will be determined by solving

It follows the approximate eigenvalue is about

λ = −0.856 95

The approximate eigenvector is

How well does it work?

This is pretty close. If you wanted to get closer, you could simply do more iterations. For practical purposes, the eigenvalue and eigenvector have been obtained.

Example 15.2.4 Find the eigenvalues and eigenvectors of the matrix

This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just get the characteristic equation, graph it using a calculator and zoom in to find the eigenvalues. If you do this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5. (The characteristic equation is 2 + 8λ + 4λ² −λ³ = 0.) Of course we have no idea what the eigenvectors are.

Lets first try to find the eigenvector and a better approximation for the eigenvalue near −1.2. In this case, let α = −1.2. Then

Then

The initial approximation for an eigenvector will then be the above divided by its largest entry.

Next

Normalizing this, the next approximation is

which is essentially the same vector. Hence this vector is the approximate eigenvector and to find the eigenvalue you solve

Thus the eigenvalue is λ = −1. 218 5. How well does it work?

while

For practical purposes, this has found the eigenvalue near −1.2 as well as an eigenvector associated with it.

Next we shall find the eigenvector and a more precise value for the eigenvalue near −.4. In this case,

The first approximation to an eigenvector can be obtained as before.

The first choice for an approximate eigenvector is

Now

Hence the next approximate eigenvector is

This is pretty close to the same vector and so the next scaling factor will be essentially the same as the one just used. Thus the above vector is a good approximate eigenvector and an approximate eigenvalue is obtained by solving

Hence λ = −0.297 57. How well does it work?

It works pretty well. For practical purposes, the eigenvalue and eigenvector have now been found.
If you want better accuracy, you could just continue iterating.

Next we will find the eigenvalue and eigenvector for the eigenvalue near 5.5. In this case,

As before, I have no idea what the eigenvector is but to avoid giving the impression that you always need to start with the vector (1, 1, 1)T, let u₁ = (1, 2, 3)T. What follows is the iteration without all the comments between steps.

S₁ = 86. 4.

S₂ = 95. 379 629.

S₃ = 62. 321 522.

S₄ = 62. 321 329. Looks like it is time to stop because this scaling factor is not changing much from S3.

Then the approximation of the eigenvalue is gotten by solving

which gives λ = 5. 516 045 9. Lets see how well it works.

15.2.1 Complex Eigenvalues

What about complex eigenvalues? If your matrix is real, you won’t see these by graphing the characteristic equation on your calculator. Will the shifted inverse power method find these eigenvalues and their associated eigenvectors? The answer is yes. However, for a real matrix, you must pick α to be complex. This is because the eigenvalues occur in conjugate pairs, so if you don’t pick it complex, it will be the same distance between any conjugate pair of complex numbers and so nothing in the above argument for convergence implies you will get convergence to a complex number. Also, the process of iteration will yield only real vectors and scalars.

Example 15.2.5 Find the complex eigenvalues and corresponding eigenvectors for the matrix

Here the characteristic equation is λ³ − 5λ² + 8λ− 6 = 0. One solution is λ = 3. The other two are 1 + i and 1−i. Apply the process to α = i so we will find the eigenvalue closest to i.

I shall use the trick of raising to a high power to begin with.

Now normalize by dividing by the largest entry to get a good initial vector.

Then

It looks like this has found the eigenvector exactly. Then to get the eigenvalue, you need to solve

Thus λ = 1 + i. How well does the above vector work?

It appears that this has found the exact answer.

This illustrates an interesting topic which leads to many related topics. If you have a polynomial, x⁴ + ax³ + bx² + cx + d, you can consider it as the characteristic polynomial of a certain matrix, called a companion matrix. In this case,

The above example was just a companion matrix for λ³ − 5λ² + 8λ − 6. You can see the pattern which will enable you to obtain a companion matrix for any polynomial of the form λn + a₁λn−1 + · · · + an−1λ + an. This illustrates that one way to find the complex zeros of a polynomial is to use the shifted inverse power method on a companion matrix for the polynomial. Doubtless there are better ways but this does illustrate how impressive this procedure is. Do you have a better way?

15.3 The Rayleigh Quotient

There are many specialized results concerning the eigenvalues and eigenvectors for Hermitian matrices. A matrix A is Hermitian if A = A∗ where A∗ means to take the transpose of the conjugate of A. In the case of a real matrix, Hermitian reduces to symmetric. Recall also that for x ∈ Fn,

The following corollary gives the theoretical foundation for the spectral theory of Hermitian matrices. This is a corollary of a theorem which is proved Corollary 13.2.14 and Theorem 13.2.14 on Page 264.

Corollary 15.3.1 If A is Hermitian, then all the eigenvalues of A are real and there exists an orthonormal basis of eigenvectors.

Thus for this orthonormal basis,

For x ∈ Fn, x ≠ 0, the Rayleigh quotient is defined by

Now let the eigenvalues of A be λ₁ ≤ λ₂ ≤ · · · ≤ λn and Axk = λkxk where is the above orthonormal basis of eigenvectors mentioned in the corollary. Then if x is an arbitrary vector, there exist constants, ai such that

Also,

Therefore,

In other words, the Rayleigh quotient is always between the largest and the smallest eigenvalues of A. When x = xn, the Rayleigh quotient equals the largest eigenvalue and when x = x₁ the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate a Rayleigh quotient. How close is it to some eigenvalue?

Theorem 15.3.2 Let x ≠ 0 and form the Rayleigh quotient,

Then there exists an eigenvalue of A, denoted here by λq such that

| [picture] | (15.2)

Proof: Let x = where is the orthonormal basis of eigenvectors.

Now pick the eigenvalue, λq which is closest to q. Then

which implies (15.2).

Example 15.3.3 Consider the symmetric matrix Let

x = (1, 1, 1)T.

How close is the Rayleigh quotient to some eigenvalue of A? Find the eigenvector and eigenvalue to several decimal places.

Everything is real and so there is no need to worry about taking conjugates. Therefore, the Rayleigh quotient is

According to the above theorem, there is some eigenvalue of this matrix, λq such that

Could you find this eigenvalue and associated eigenvector? Of course you could. This is what the inverse shifted power method is all about.

Solve

In other words solve

and divide by the entry which is largest, 3. 870 7, to get

Now solve

and divide by the entry with largest absolute value, 2. 997 9 to get

Now solve

and divide by the entry with largest absolute value, 3. 045 4, to get

Solve

and divide by the largest entry, 3. 042 1 to get

You can see these scaling factors are not changing much. The predicted eigenvalue is obtained by solving

to obtain λ = 6. 6621. How close is this?

while

You see that for practical purposes, this has found the eigenvalue and an eigenvector.

15.4 The QR Algorithm

15.4.1 Basic Considerations

The QR algorithm is one of the most remarkable techniques for finding eigenvalues. In this section, I will discuss this method. To see more on this algorithm, consult Golub and Van Loan [5]. For an explanation of why the algorithm works see Wilkinson [16]. There is also more discussion in Linear Algebra. This will only discuss real matrices for the sake of simplicity. Also, there is a lot more to this algorithm than will be presented here. First here is an introductory lemma.

Lemma 15.4.1 Suppose A is a block upper triangular matrix,

This means that the Bi are ri × ri matrices whose diagonals are subsets of the main diagonal of A. Then 

Proof: Say  where Ti is upper triangular. Such unitary matrices exist by Schur’sitheorem. Then consider the similarity transformation,

By block multiplication this equals

Now this is a real upper triangular matrix and the eigenvalues of A consist of the union of the eigenvalues of the Ti which is the same as the union of the eigenvalues of the Bi.

Here is the description of the great and glorious QR algorithm.

The QR Algorithm

Let A be an n x n real matrix. Let A₀ = A. Suppose that Ak₋₁ has been found. To find Ak let

Ak−1 = QkRk, Ak = RkQk,

where QkRk is a QR factorization of Ak−1. Thus R is upper triangular with nonnegative entries on the main diagonal and Q is real and unitary (orthogonal).

The main signiﬁcance of this algorithm is in the following easy but important theorem.

Theorem 15.4.2 Let A be any n × n complex matrix and let {Ak} be the sequence of matrices described above by the QR algorithm. Then each of these matrices is unitarily similar to A.

Proof: Clearly A₀ is orthogonally similar to A because they are the same thing. Suppose then that

Ak−1 = Q∗ AQ

Then from the algorithm,

Therefore, from the algorithm,

and so Ak is unitarily similar to A also.

Although the sequence {Ak} may fail to converge, it is nevertheless often the case that for large k, Ak is of the form

where the Bi are blocks which run down the diagonal of the matrix, and all of the entries below this block diagonal are very small. Then letting TB denote the matrix obtained by setting all of these small entries equal to zero, one can argue, using methods of analysis, that the eigenvalues of Ak are close to the eigenvalues of TB. From Lemma 15.4.1 the eigenvalues of TB are the eigenvalues of the blocks Bi. Thus, the eigenvalues of A are the same as those of Ak and these are close to the eigenvalues of TB.

In proving things about this algorithm and also for the sake of convenience, here is a technical result.

Corollary 15.4.3 For Qk, Rk, Ak given in the QR algorithm,

| [picture] | (15.3)

For Q(k) ≡ Q₁· · ·Qk and R(k) ≡ Rk· · ·R₁, it follows that

Ak = Q(k)R(k)

Here Ak is the usual thing, A raised to the kth power.

Proof: From the algorithm,

Hence

Suppose the formula (15.3) holds for k. Then from the algorithm,

Hence and so

This shows the first part.

The second part is clearly true from the algorithm if k = 1. Then from the first part and the algorithm,

It follows that

Hence

Now suppose that A⁻¹ exists. How do two QR factorizations compare? Since A⁻¹ exists, it would require that if A = QR, then R⁻¹ must exist. Now an upper triangular matrix has inverse which is also upper triangular. This follows right away from the algorithm presented early in the book for finding the inverse. If A = Q₁R₁ = Q₂R₂, then and so R₁R₂−1 is an upper triangular matrix which is also unitary and in addition has all positive entries down the diagonal. For simplicity, call it R. Thus R is upper triangular and RR∗ = R∗R = I. It follows easily that R must equal I and so R₁ = R₂ which requires Q₁ = Q₂.

Now in the above corollary, you know that

Also, from this corollary, you know that

Ak = Q(k)R(k)

You could also simply take the QR factorization of Ak to obtain Ak = QR. Then from what was just pointed out, if A⁻¹ exists,

Q(k) = Q

Thus from the above corollary,

Therefore, in using the QR algorithm in the case where A has an inverse, it suﬃces to take

Ak = QR

and then consider the matrix

Q∗AQ = Ak.

This is so theoretically. In practice it might not work out all that well because of round off errors.

There is also an interesting relation to the power method. Let

Then from the way we multiply matrices,

and for large k, Akai would be expected to point roughly in the direction of the eigenvector corresponding to the largest eigenvalue. Then if you form the QR factorization,

Ak+1 = QR

the columns of Q are an orthonormal basis obtained essentially from the Gram Schmidt procedure. Thus the first column of Q has roughly the direction of an eigenvector associated with the largest eigenvalue of A. It follows that the first column of Q∗AQ is approximately equal to and so the top entry will be close to and the entries below it are close to 0. Thus the eigenvalues₁ of the matrix should be close to this top entry of the first column along with the eigenvalues of the (n − 1) × (n − 1) matrix in the lower right corner. If this is a 2 × 2 you can find the eigenvalues using the quadratic formula. If it is larger, you could just use the same procedure for finding its eigenvalues but now you are dealing with a smaller matrix.

Example 15.4.4 Find the eigenvalues of the matrix

First use the computer to raise this matrix to a large power.

Now find the QR factorization of this last matrix. The Q equals

Next examine

You see now that this is essentially equal to the block upper triangular matrix

and the eigenvalues of this matrix can be easily identiﬁed. They are 2 and the eigenvalues of the block

But these can be found easily using the quadratic formula. This yields i,−i. In fact the exact eigenvalues for this matrix are 2, i,−i.

15.4.2    The Upper Hessenberg Form

Actually, when using the QR algorithm, contrary to what I have done above, you should always deal with a matrix which is similar to the given matrix which is in upper Hessenberg form. This means all the entries below the sub diagonal equal 0. Here is an easy lemma.

Lemma 15.4.5 Let A be an n × n matrix. Then it is unitarily similar to a matrix in upper Hessenberg form and this similarity can be computed.

Proof: Let A be an n × n matrix. Suppose n > 2. There is nothing to show otherwise.

where A₁ is n − 1 × n − 1. Consider the n − 1 × 1 matrix d. Then let Q be a Householder reﬂection such that

Then

By similar reasoning, there exists an n − 1 × n − 1 matrix

such that

Thus

will have all zeros below the first two entries on the sub diagonal. Continuing this way shows the result.

The reason you should use a matrix which is upper Hessenberg and similar to A in the QR algorithm is that the algorithm keeps returning a matrix in upper Hessenberg form and if you are looking for block upper triangular matrices, this will force the size of the blocks to be no larger than 2 × 2 which are easy to handle using the quadratic formula. This is in the following lemma.

Lemma 15.4.6 Let {Ak} be the sequence of iterates from the QR algorithm, A⁻¹ exists. Then if Ak is upper Hessenberg, so is Ak+1.

Proof: The matrix is upper Hessenberg means that Aij = 0 whenever i − j ≥ 2.

Ak+₁ = RkQk

where Ak = QkRk. Therefore AkR⁻¹ = Qk and so

Ak₊₁ = RkQk = RkAkRk ⁻¹

Let the ijth entry of Ak be . Then if i − j ≥ 2

It is given that whenever p − q ≥ 2. However, from the above sum,

p − q ≥ i −j ≥ 2,

and so the sum equals 0.

Example 15.4.7 Find the solutions to the equation using the QR algorithm.

This is the characteristic equation of the matrix

Since the constant term in the equation is not 0, it follows that the matrix has an inverse. It is already in upper Hessenberg form. Lets apply the algorithm.

Then when you take the QR factorization of this, you find Q =

Then you look at

which yields

Of course the entries in the bottom left should all be 0. They aren’t because of round off error. The only other entry of importance is 1. 043 4× 10⁻³ which is small. Hence the eigenvalues are close to the eigenvalues of the two blocks

This yields

0.982 47 + 0.982 09i,    0.982 47 − 0.982 09i

1. 017 5 + 1. 017 2i,   1. 017 5 − 1. 017 2i

The real solutions are 1 + i, 1 + i, 1−i, and 1−i. You could of course use the shifted inverse power method to get closer and to also obtain eigenvectors for the matrix.

Example 15.4.8 Find the eigenvalues for the symmetric matrix

Also find an eigenvector.

I should work with a matrix which is upper Hessenberg which is similar to the given matrix. However, I don’t feel like it. It is easier to just raise to a power and see if things work. This is what I will do.

Now take the QR factorization of this. When you do, Q =

Thus you look at QTAQ, a matrix which is similar to A and which equals

It follows that the eigenvalues are approximately 6. 642 9 and the eigenvalues of the 3 × 3 matrix in the lower right corner. I shall use the same technique to find its eigenvalues.

Then take the QR factorization of this to get Q =

Then you look at QTBQ which equals

Thus the eigenvalues are approximately 6 . 642 9,−4. 101 8, and the eigenvalues of the matrix in the lower right corner in the above. These are 2. 458 9,−1. 480 0×10⁻⁶. In fact, 0 is an eigenvalue of the original matrix and it is being approximated by −1. 480 0×10⁻⁶. To summarize, the eigenvalues are approximately

6. 642 9, −4. 101 8, 2. 458 9, 0

Of course you could now use the shifted inverse power method to find these more exactly if desired and also to find the eigenvectors.

If you wanted to find an eigenvector, you could start with one of these approximate eigenvalues and use the shifted inverse power method to get an eigenvector. For example, pick 6. 642 9.

So try for the next approximation

Next one is

This isn’t changing by much from the above vector and so the scaling factor will be about 15993. Now solve

The solution is 6 . 643 0. The approximate eigenvector is what I just got. Lets check it.

This is clearly very close. This has essentially found the eigenvalues and an eigenvector for the largest eigenvalue.

15.5 Exercises

Using the power method, find the eigenvalue correct to one decimal place having largest absolute value for the matrix along with an eigenvector associated with this

eigenvalue.

Using the power method, find the eigenvalue correct to one decimal place having largest absolute value for the matrix along with an eigenvector associated with this

eigenvalue.

Using the power method, find the eigenvalue correct to one decimal place having largest absolute value for the matrix along with an eigenvector associated with this

eigenvalue.

Using the power method, find the eigenvalue correct to one decimal place having largest absolute value for the matrix along with an eigenvector associated with

this eigenvalue.

In Example 15.3.3 an eigenvalue was found correct to several decimal places along with an eigenvector. Find the other eigenvalues along with their eigenvectors.

Find the eigenvalues and eigenvectors of the matrix numerically. In this case the exact eigenvalues are

±√3, 6. Compare with the exact answers.

Find the eigenvalues and eigenvectors of the matrix numerically. The exact eigenvalues are 2, 4 +√15,

4−√15. Compare your numerical results with the exact values. Is it much fun to compute the exact eigenvectors?

Find the eigenvalues and eigenvectors of the matrix numerically. We don’t know the exact

eigenvalues in this case. Check your answers by multiplying your numerically computed eigenvectors by the matrix.

Find the eigenvalues and eigenvectors of the matrix numerically. We don’t know the exact

eigenvalues in this case. Check your answers by multiplying your numerically computed eigenvectors by the matrix.

Consider the matrix and the vector (1, 1, 1)T.

Estimate the distance between the Rayleigh quotient determined by this vector and some eigenvalue of A.

Consider the matrix and the vector (1, 1, 1)T.

Estimate the distance between the Rayleigh quotient determined by this vector and some eigenvalue of A.

Using Gerschgorin’s theorem, find upper and lower bounds for the eigenvalues of

The QR algorithm works very well on general matrices. Try the QR algorithm on the following matrix which happens to have some complex eigenvalues.

Use the QR algorithm to get approximate eigenvalues and then use the shifted inverse power method on one of these to get an approximate eigenvector for one of the complex eigenvalues.

Use the QR algorithm to approximate the eigenvalues of the symmetric matrix

Try to find the eigenvalues of the matrix using the

QR algorithm. It has eigenvalues 1, i,−i. You will see the algorithm won’t work well.

Let q (λ) = a₀ + a₁λ + · · · + an−1λn−1 + λn. Now consider the companion matrix,

Show that q (λ) is the characteristic equation for C. Thus the roots of q (λ) are the eigenvalues of C. You can prove something similar for

Hint: The characteristic equation is

Expand along the first column. Thus

Now use induction on the first term and for the second, note that you can expand along the top row to get

(−1)n−2a₀(−1)n = a₀.

Suppose A isAreal symmetric, invertible, matrix, or more generally one which has real eigenvalues. Then as described above, it is typically the case that

Ap = Q₁R

and

where e is very small. Then you can do the same thing with A₁ to obtain another smaller orthogonal matrix Q₂ such that

Explain why

where the ei are very small. Explain why one can construct an orthogonal matrix Q such that

QTAQ = (T + E)

where T is an upper triangular matrix and E is very small. In case A is symmetric, explain why T is actuallyAdiagonal matrix. Next explain why, in the case of A symmetric, that the columns of Q are an orthonormal basis of vectors, each of which is close to an eigenvector. Thus this will compute, not just the eigenvalues but also the eigenvectors.

Explain how one could use the QR algorithm or the above procedure to compute the singular value decomposition of an arbitrary real m × n matrix. In fact, there are other algorithms which will work better.

Vector Spaces

16.1 Algebraic Considerations

16.1.1 The Definition

It is time to consider the idea of an abstract Vector space.

Definition 16.1.1 A vector space is an Abelian group of “vectors” denoted here by bold face letters, satisfying the axioms of an Abelian group,

v + w = w + v,

the commutative law of addition,

(v + w) + z = v + (w + z) ,

the associative law for addition,

v + 0 = v,

the existence of an additive identity,

v + (−v) = 0,

the existence of an additive inverse, along with a field of “scalars” F which are allowed to multiply the vectors according to the following rules. (The Greek letters denote scalars.)

α (v + w) = αv + αv,

(α + β) v = αv + βv,

α (βv) = αβ (v) ,

1v = v.

(16.1)

(16.2)

(16.3)

(16.4)

The field of scalars is usually R or C and the vector space will be called real or complex depending on whether the field is R or C. However, other fields are also possible. For example, one could use the field of rational numbers or even the field of the integers mod p for p a prime. A vector space is also called a linear space.

For example, Rn with the usual conventions is an example of a real vector space and Cn is an example of a complex vector space. Up to now, the discussion has been for Rn or Cn and all that is taking place is an increase in generality and abstraction. We no longer know what the vectors are. We also have no idea what field is being considered, at least in the next section.

If you are interested in considering other fields, you should have some examples other than C,R,Q. Some of these are discussed in the following exercises. If you are happy with only considering R and C, skip these exercises.

16.2 Exercises

Prove the Euclidean algorithm: If m, n are positive integers, then there exist integers q, r ≥ 0 such that r < m and

n = qm + r

Hint: You might try considering

S ≡ {n − km : k ∈ N and n − km < 0}

and picking the smallest integer in S or something like this.

↑The greatest common divisor of two positive integers m, n, denoted as q is a positive number which divides both m and n and if p is any other positive number which divides both m, n, then p divides q. Recall what it means for p to divide q. It means that q = pk for some integer k. Show that the greatest common divisor of m, n is the smallest positive integer in the set S

S ≡ {xm + yn : x, y ∈ Z and xm + yn > 0}

Two positive integers are called relatively prime if their greatest common divisor is 1.

↑A positive integer larger than 1 is called a prime number if the only positive numbers which divide it are 1 and itself. Thus 2,3,5,7, etc. are prime numbers. If m is a positive integer and p does not divide m where p is a prime number, show that p and m are relatively prime.

↑There are lots of fields. This will give an example of a finite field. Let Z denote the set of integers. Thus Z = { · · ·,−3,−2,−1, 0, 1, 2, 3,· · · }. Also let p be a prime number. We will say that two integers, a, b are equivalent and write a ~ b if a − b is divisible by p. Thus they are equivalent if a b = px for some integer x. First show that a ~ a. Next show that if a ~ b then b ~ a. Finally show that if a ~ b and b ~ c then a ~ c. For a an integer, denote by [a] the set of all integers which is equivalent to a, the equivalence class of a. Show first that is suffices to consider only [a] for a = 0, 1, 2, · · · , p − 1 and that for 0 ≤a < b ≤p− 1, [a] ̸= [b]. That is, [a] = [r] where r ∈ {0, 1, 2, · · · , p− 1}. Thus there are exactly p of these equivalence classes. Hint: Recall the Euclidean algorithm. For a > 0, a = mp + r where r < p. Next define the following operations.

[a] + [b] ≡ [a + b]
[a] [b] ≡ [ab]

Show these operations are well defined. That is, if [a] = [a′] and [b] = [b′] , then [a] + [b] = [a′] + [b′] with a similar conclusion holding for multiplication. Thus for addition you need to verify [a + b] = [a′ + b′] and for multiplication you need to verify [ab] = [a′b′]. For example, if p = 5 you have [3] = [8] and [2] = [7] . Is [2× 3] = [8× 7]? Is [2 + 3] = [8 + 7]? Clearly so in this example because when you subtract, the result is divisible by 5. So why is this so in general? Now verify that {[0] , [1] , · · · , [p− 1]} with these operations is a Field. This is called the integers modulo a prime and is written Zp. Since there are infinitely many primes p, it follows there are infinitely many of these finite fields. Hint: Most of the axioms are easy once you have shown the operations are well defined. The only two which are tricky are the ones which give the existence of the additive inverse and the multiplicative inverse. Of these, the first is not hard. −[x] = [−x]. Since p is prime, there exist integers x, y such that 1 = px + ky and so 1−ky = px which says 1 ~ ky and so [1] = [ky] . Now you finish the argument. What is the multiplicative identity in this collection of equivalence classes?

16.2.1 Linear Independence And Bases

Just as in the case of Fn one has a concept of subspace, linear independence, and bases.

Definition 16.2.1 If {v₁, · · · , vn} ⊆V, a vector space, then

A subset, W ⊆ V is said to be a subspace if it is also a vector space with the same field of scalars. Thus W ⊆ V is a subspace if ax + by ∈ W whenever a, b ∈ F and x, y ∈ W. The span of a set of vectors as just described is an example of a subspace.

Definition 16.2.2 If {v₁, · · · , vn} ⊆V, the set of vectors is linearly independent if

implies

α₁ = · · · = αn = 0

and {v₁, · · · , vn} is called a basis for V if

span (v₁, · · · , vn) = V

and {v₁, · · · , vn} is linearly independent. The set of vectors is linearly dependent if it is not linearly independent.

The next theorem is called the exchange theorem. It is very important that you understand this theorem. There are two kinds of people who go further in linear algebra, those who understand this theorem and its corollary presented later and those who don’t. Those who do understand these theorems are able to proceed and learn more linear algebra while those who don’t are doomed to wander in the wilderness of confusion and sink into the swamp of despair. Therefore, I am giving multiple proofs. Try to understand at least one of them. Several amount to the same thing, just worded differently. Before giving the proof, here is some important notation.

Notation 16.2.3 Let wij ∈ V, a vector space and let 1 ≤ i ≤ r while 1 ≤ j ≤ s. Thus these vectors can be listed in a rectangular array.

Then means to sum the vectors in each column and then to add the s sums which result while means to sum the vectors in each row and then to add the r sums which result. Either way you simply get the sum of all the vectors in the above array. This is because you can add vectors in any order and you get the same answer.

Theorem 16.2.4 Let { x₁, · · · , xr } be a linearly independent set of vectors such that each xi is in the span{ y₁, · · · , ys }. Then r ≤ s.

Proof 1: Let

If r > s, then the matrix A = (ajk) has more columns than rows. By Corollary 8.2.8 one of these columns is a linear combination of the others. This implies there exist scalars c₁, · · · , cr such that

Then

which contradicts the assumption that { x₁, · · · , xr } is linearly independent. Hence r ≤ s.

Proof 2: Define span(y₁, · · · , ys) ≡ V, it follows there exist scalars, c₁, · · · , cs such that

| [picture] | (16.5)

Not all of these scalars can equal zero because if this were the case, it would follow that x₁ = 0 and so { x₁, · · · , xr } would not be linearly independent. Indeed, if x₁ = 0, 1x1 + 0xi = x₁ = 0 and so there would exist a nontrivial linear combination of the vectors {x₁, · · · ,xr which equals zero.

Say ck0. Then solve ((16.5)) for yk and obtain

Define {z₁, · · · , zs−1} by

Therefore, span (x₁, z₁, · · · , zs−1) = V because if v ∈V, there exist constants c₁, · · · , cs such that

Replace the yk in the above with a linear combination of the vectors,

to obtain v ∈ span (x₁, z₁, · · · , zs−1) . The vector yk, in the list {y₁, · · · , ys}, has now been replaced with the vector x₁ and the resulting modified list of vectors has the same span as the original list of vectors, {y₁, · · · , ys}.

Now suppose that r > s and that span (x₁, · · · , xl, z₁, · · · , zp) = V, where the vectors, z₁, · · · , zp are each taken from the set, {y₁, · · · , ys} and l + p = s. This has now been done for l = 1 above. Then since r > s, it follows that l ≤ s < r and so l + 1 ≤ r. Therefore, xl+1 is a vector not in the list, {x₁, · · · , xl} and since span (x₁, · · · , xl, z₁, · · · , zp) = V there exist scalars, ci and dj such that

| [picture] | (16.6)

Not all the dj can equal zero because if this were so, it would follow that {x₁, · · · , xr} would be a linearly dependent set because one of the vectors would equal a linear combination of the others. Therefore, ((16.6)) can be solved for one of the zi, say zk, in terms of xl+1 and the other zi and just as in the above argument, replace that zi with xl+1 to obtain

Continue this way, eventually obtaining

span (x₁, · · · ,xs) = V

But then xr ∈ span (x₁, · · · , xs) contrary to the assumption that {x₁, · · · , xr} is linearly independent. Therefore, r ≤ s as claimed.  

Proof 3: Let V ≡ span (y₁, · · · , ys) and suppose r > s. Let Al ≡ {x₁, · · · , xl}, A₀ = ∅, and let Bs−l denote a subset of the vectors, {y₁, · · · , ys} which contains s − l vectors and has the property that span (Al, Bs−l) = V. Note that the assumption of the theorem says span (A₀, Bs) = V.

Now an exchange operation is given for span (Al, Bs−l) = V . Since r > s, it follows l < r. Letting

it follows there exist constants, ci and di such that

and not all the di can equal zero. (If they were all equal to zero, it would follow that the set, {x₁, · · · , xr} would be dependent since one of the vectors in it would be a linear combination of the others.)

Let dk ̸0. Then zk can be solved for as follows.

This implies V = span (Al+1, Bs−l−1), where Bs−l−1 ≡ Bs−l \ {zk}, a set obtained by deleting zk from Bk−l. You see, the process exchanged a vector in Bs−l with one from {x₁, · · · , xr} and kept the span the same. Starting with V = span (A₀, Bs) , do the exchange operation until V = span (As−1, z) where z ∈ {y₁, · · · , ys}. Then one more application of the exchange operation yields V = span (As) . But this implies xr ∈ span (As) = span (x₁, · · · , xs) , contradicting the linear independence of {x₁, · · · , xr}. It follows that r ≤s as claimed.

Proof 4: Suppose r > s. Let zk denote a vector of {y₁, · · · , ys}. Thus there exists j as small as possible such that

span (y₁, · · · , ys) = span (x₁, · · · , xm, z₁, · · · , zj)

where m + j = s. It is given that m = 0, corresponding to no vectors of {x₁, · · · , xm} and j = s, corresponding to all the yk results in the above equation holding. If j > 0 then m < s and so

Not all the bi can equal 0 and so you can solve for one of them in terms of

xm+1, xm, · · · , x₁,

and the other zk. Therefore, there exists

{z₁, · · · , zj₋₁} ⊆ {y₁, · · · , ys}

such that

span (y₁, · · · , ys) = span (x₁, · · · , xm+1, z₁, · · · , zj−1)

contradicting the choice of j. Hence j = 0 and

span (y₁, · · · , ys) = span (x₁, · · · , xs)

It follows that

xs+1 ∈ span (x₁, · · · , xs)

contrary to the assumption the xk are linearly independent. Therefore, r ≤ s as claimed.

Corollary 16.2.5 If {u₁, · · · , um} and {v₁, · · · , vn} are two bases for V, then m = n.

Proof: By Theorem 16.2.4, m ≤ n and n ≤ m.

This corollary is very important so here is another proof of it given independent of the exchange theorem above.

Theorem 16.2.6 Let V be a vector space and suppose {u₁, · · · , uk} and {v₁, · · · , vm} are two bases for V . Then k = m.

Proof: Suppose k > m. Then since the vectors, {u₁, · · · , uk} span V, there exist scalars, cij such that

Therefore,

if and only

Now since{v₁, · · · , vn} is independent, this happens if and only if

However, this is a system of m equations in k variables, d₁, · · · , dk and m < k. Therefore, there exists a solution to this system of equations in which not all the dj are equal to zero. Recall why this is so. The augmented matrix for the system is of the form ( C  0 ) where C is a matrix which has more columns than rows. Therefore, there are free variables and hence nonzero solutions to the system of equations. However, this contradicts the linear independence of {u₁, · · · , uk} because, as explained above, Similarly it cannot happen that m > k.

Definition 16.2.7 A vector space V is of dimension n if it has a basis consisting of n vectors. This is well defined thanks to Corollary 16.2.5. It is always assumed here that n < ∞ in this case, such a vector space is said to be finite dimensional.

Theorem 16.2.8 If V = span (u₁, · · · , un) then some subset of {u₁, · · · , un} is a basis for V. Also, if {u₁, · · · , uk} ⊆V is linearly independent and the vector space is finite dimensional, then the set {u₁, · · · , uk}, can be enlarged to obtain a basis of V.

Proof: Let

S = {E ⊆ {u₁, · · · , un} such that span (E) = V}.

For E ∈S, let |E| denote the number of elements of E. Let

m ≡ min{|E| such that E ∈S}.

Thus there exist vectors

{v₁, · · · , vm} ⊆ {u₁, · · · , un}

such that

span (v₁, · · · , vm) = V

and m is as small as possible for this to happen. If this set is linearly independent, it follows it is a basis for V and the theorem is proved. On the other hand, if the set is not linearly independent, then there exist scalars,

c₁, · · · , cm

such that

and not all the ci are equal to zero. Suppose ck 0. Then the vector vk may be solved for in terms of the other vectors. Consequently,

V = span (v₁, · · · , vk−1, vk+1, · · · , vm)

contradicting the definition of m. This proves the first part of the theorem.

To obtain the second part, begin with {u₁, · · · , uk} and suppose a basis for V is

{v₁, · · · , vn}.

If

span (u₁, · · · , uk) = V,

then k = n. If not, there exists a vector

Then {u₁, · · · , uk, uk+1} is also linearly independent. Continue adding vectors in this way until n linearly independent vectors have been obtained. Then

span (u₁, · · · , un) = V

because if it did not do so, there would exist un+1 as just described and {u₁, · · · , un+1} would be a linearly independent set of vectors having n + 1 elements even though {v₁, · · · , vn} is a basis. This would contradict Theorem 16.2.4. Therefore, this list is a basis.

It is useful to emphasize some of the ideas used in the above proof.

Lemma 16.2.9 Suppose v ∉ span (u₁, · · · , uk) and {u₁, · · · , uk} is linearly independent. Then {u₁, · · · , uk, v} is also linearly independent.

Proof: Suppose It is required to verify that each ci = 0 and that d = 0. But if d0, then you can solve for v as a linear combination of the vectors, {u₁, · · · , uk},

contrary to assumption. Therefore, d = 0. But then and the linear independence ofi=1 {u₁, · · · , uk} implies each ci = 0 also.

Theorem 16.2.10 Let V be a nonzero subspace of a finite dimensional vector space, W of dimension, n. Then V has a basis with no more than n vectors.

Proof: Let v₁ ∈V where v₁0. If span (v₁) = V, stop. {v₁} is a basis for V . Otherwise, there exists v₂ ∈V which is not in span (v₁) . By Lemma 16.2.9 {v₁, v₂} is a linearly independent set of vectors. If span (v₁, v₂) = V stop, {v₁, v₂} is a basis for V. If span (v₁, v₂) ̸V, then there exists v₃ ∉ span (v₁, v₂) and {v₁, v₂, v₃} is a larger linearly independent set of vectors. Continuing this way, the process must stop before n + 1 steps because if not, it would be possible to obtain n + 1 linearly independent vectors contrary to the exchange theorem, Theorems 16.2.4.

16.3 Vector Spaces And Fields∗

16.3.1 Irreducible Polynomials

There exist very interesting examples of vector spaces which are not Fn and for which the field of scalars is not R or C. This section gives a convincing application of the value of the above abstract theory and provides many further examples of fields. It is an optional section which you might read if you find it interesting.

Here I will give some basic algebra relating to polynomials. This is interesting for its own sake but also provides the basis for constructing many different kinds of fields. The first is the Euclidean algorithm for polynomials.

Definition 16.3.1 A polynomial is an expression of the form p where as usual is defined to equal 1. Two polynomials are said to be equal if their corresponding coefficients are the same. Thus, in particular, p (λ) = 0 means each of the ak = 0. An element of the field λ is said to be a root of the polynomial if p (λ) = 0 in the sense that when you plug in λ into the formula and do the indicated operations, you get 0. The degree of a nonzero polynomial is the highest exponent appearing on λ. The degree of the zero polynomial p (λ) = 0 is not defined. You add and multiply polynomials using the standard conventions learned in junior high school.

Example 16.3.2 Consider the polynomial p (λ) = λ₂ + λ where the coefficients are in Z₂. Is this polynomial equal to 0? Not according to the above definition, because its coefficients are not all equal to 0. However, p (1) = p (0) = 0 so it sends every element of Z₂ to 0. Note the distinction between saying it sends everything in the field to 0 with having the polynomial be the zero polynomial.

Lemma 16.3.3 Let f (λ) and g (λ) ̸0 be polynomials. Then there exists a polynomial q (λ) such that

where the degree of r (λ) is less than the degree of g (λ) or r (λ) = 0.

Proof: Consider the polynomials of the form f (λ)−g (λ) l (λ) and out of all these polynomials, pick one which has the smallest degree. This can be done because of the well ordering of the natural numbers. Let this take place when l (λ) = q₁ (λ) and let

It is required to show degree of r (λ) < degree of g (λ) or else r (λ) = 0.

Suppose f (λ)−g (λ) l (λ) is never equal to zero for any l (λ). Then r (λ) 0. It is required to show that the degree of r (λ) is smaller than the degree of g (λ) . If this doesn’t happen, then the degree of r ≥ the degree of g. Let

where m ≥ n and bm and an are nonzero. Then let r₁ (λ) be given by

which has smaller degree than m, the degree of r (λ). But

and this is not zero by the assumption that f (λ)−g (λ) l (λ) is never equal to zero for any l (λ) yet has smaller degree than r (λ) which is a contradiction to the choice of r (λ).

Now with this lemma, here is another one which is very fundamental. First here is a definition. A polynomial is monic means it is of the form

That is, the leading coefficient is 1. In what follows, the coefficients of polynomials are in F, a field of scalars which is completely arbitrary. Think R if you need an example.

Definition 16.3.4 A polynomial f is said to divide a polynomial g if g (λ) = f (λ) r (λ) for some polynomial r (λ). Let {i (λ)} be a finite set of polynomials. The greatest common divisor will be the monic polynomial q such that q (λ) divides each i (λ) and if p (λ) divides each i (λ) , then p (λ) divides q (λ) . The finite set of polynomials {i} is said to be relatively prime if their greatest common divisor is 1. A polynomial f (λ) is irreducible if there is no polynomial with coefficients in F which divides it except nonzero scalar multiples of f (λ) and constants.

Proposition 16.3.5 The greatest common divisor is unique.

Proof: Suppose both q (λ) and q′ (λ) work. Then q (λ) divides q′ (λ) and the other way around and so

Therefore, the two must have the same degree. Hence l′ (λ) , l (λ) are both constants. However, this constant must be 1 because both q (λ) and q′ (λ) are monic.

Theorem 16.3.6 Let ψ (λ) be the greatest common divisor of {i (λ)}, not all of which are zero polynomials. Then there exist polynomials ri (λ) such that

Furthermore, ψ (λ) is the monic polynomial of smallest degree which can be written in the above form.

Proof: Let S denote the set of monic polynomials which are of the form

where ri (λ) is a polynomial. Then S because some i (λ) 0. Then let the ri be chosen such that the degree of the expressionis as small as possible. Letting ψ (λ) equal thisi=1 sum, it remains to verify it is the greatest common divisor. First, does it divide each i (λ)? Suppose it fails to divide ₁ (λ) . Then by Lemma 16.3.3

where degree of r (λ) is less than that of ψ (λ). Then dividing r (λ) by the leading coefficient if necessary and denoting the result by ψ₁ (λ) , it follows the degree of ψ₁ (λ) is less than the degree of ψ (λ) and ψ₁ (λ) equals

for a suitable a ∈ F. This is one of the polynomials in S. Therefore, ψ (λ) does not have the smallest degree after all because the degree of ψ₁ (λ) is smaller. This is a contradiction. Therefore, ψ (λ) divides i (λ) . Similarly it divides all the other i (λ).

If p (λ) divides all the i (λ) , then it divides ψ (λ) because of the formula for ψ (λ) which equals

Lemma 16.3.7 Suppose (λ) and ψ (λ) are monic polynomials which are irreducible and not equal. Then they are relatively prime.

Proof 1: Suppose η (λ) is a nonconstant polynomial. If η (λ) divides (λ) , then since (λ) is irreducible, η (λ) equals a (λ) for some a ∈ F. If η (λ) divides ψ (λ) then it must be of the form bψ (λ) for some b ∈ F and so it follows

but both ψ (λ) and (λ) are monic polynomials which implies a = b and so ψ (λ) = (λ). This is assumed not to happen. It follows the only polynomials which divide both ψ (λ) and (λ) are constants and so the two polynomials are relatively prime. Thus a polynomial which divides them both must be a constant, and if it is monic, then it must be 1. Thus 1 is the greatest common divisor.

Lemma 16.3.8 Let ψ (λ) be an irreducible monic polynomial not equal to 1 which divides

where each i (λ) is an irreducible monic polynomial. Then ψ (λ) equals some i (λ)

Proof : Suppose ψ (λ) i (λ) for all i. Then by Lemma 16.3.7, there exist polynomials mi (λ) , ni (λ) such that

Hence

Then, letting and applying the binomial theorem, there exists a polynomial h (λ) such thati=1

Thus, using the fact that ψ (λ) divides for a suitable polynomial g (λ),

which is impossible if ψ (λ) is non constant, as assumed.

Now here is a simple lemma about canceling monic polynomials.

Lemma 16.3.9 Suppose p (λ) is a monic polynomial and q (λ) is a polynomial such that

p (λ) q = 0.

Then q (λ) = 0. Also if

p (λ) q₁ = p (λ) q₂ (λ)

then q₂ (λ) = q₂ (λ).

Proof: Let

Then the product equals

Then look at those terms involving and is given to be 0. Since pk = 1, it follows qn = 0. Thus

Then consider the term involving λn−1+k and conclude that since pk = 1, it follows qn−1 = 0. Continuing this way, each qi = 0. This proves the first part. The second follows from

As a simple application, one can prove uniqueness of q (λ) and r (λ) in Lemma 16.3.3. Suppose q also work in the conclusion of this lemma. Then

and so

If (λ)r (λ) , then the degree of the right is less than the degree of the left which is impossible. Thus (λ) = r (λ). Hence

Therefore,

The following is the analog of the fundamental theorem of arithmetic for polynomials.

Theorem 16.3.10 Let f (λ) be a nonconstant polynomial with coefficients in F. Then there is some a ∈ F such that f where i (λ) is an irreducible nonconstant monic polynomial and repeats are allowed. Furthermore, this factorization is unique in the sense that any two of these factorizations have the same nonconstant factors in the product, possibly in different order and the same constant a.

Proof: That such a factorization exists is obvious. If f (λ) is irreducible, you are done. Factor out the leading coefficient. If not, then f (λ) = a₁ (λ) ₂ (λ) where these are monic polynomials. Continue doing this with the i and eventually arrive at a factorization of the desired form.

It remains to argue the factorization is unique except for order of the factors. Suppose

where the ₁ (λ) and the ψi (λ) are all irreducible monic nonconstant polynomials and a, b ∈ F. If n > m, then by Lemma 16.3.8, each ψi (λ) equals one of the j (λ) . By the above cancellation lemma, Lemma 16.3.9, you can cancel all these ψi (λ) with appropriate j (λ) and obtain a contradiction because the resulting polynomials on either side would have different degrees. Similarly, it cannot happen that n < m. It follows n = m and the two products consist of the same polynomials. Then it follows a = b.

The following corollary will be well used. This corollary seems rather believable but does require a proof.

Corollary 16.3.11 Let q where the ki are positive integers and the i (λ) are irreducible monic polynomials. Suppose also that p (λ) is a monic polynomial which divides q (λ) . Then

where ri is a nonnegative integer no larger than ki.

Proof: Using Theorem 16.3.10, let p where the ψi (λ) are each irreducible and monic and b ∈ F. Since p (λ) is monic, b = 1. Then there exists a polynomial g (λ) such that

Hence g (λ) must be monic. Therefore,

for ηj monic and irreducible. By uniqueness, each ψi equals one of the j (λ) and the same holding true of the ηi (λ). Therefore, p (λ) is of the desired form.

Is there a way to compute the greatest common divisor of two polynomials? Let m (λ) , n (λ) be two polynomials, not equal since otherwise there is nothing to show. Then by Lemma 16.3.3, there exist unique q (λ) , m₁ (λ) where m₁ (λ) either equals 0 or has degree less than the degree of m (λ) such that

(q (λ) will refer to a generic polynomial in what follows.) Then if l (λ) divides both n (λ) and m (λ) , then l (λ) must also divide m₁ (λ) . Hence in determining the greatest common divisor, one can consider the two polynomials m₁ (λ) , n (λ) instead. Then write

where the degree of n₁ (λ) is less than the degree of n (λ) or else equal to 0. Continuing this way, one obtains a sequence of polynomials (mi (λ) , ni (λ)) which have the same greatest common divisor as the original n (λ) and m (λ) but such that the sum of the degrees of mi (λ) and ni (λ) is a strictly decreasing sequence. From the construction just described, it must be the case that eventually some mi (λ) or ni (λ) is a constant. Say this happens when

where nk is a constant. If it is 0 then the greatest common divisor is just nk−1 (λ) normalized to make the leading coefficient equal to 1. If it is not zero, then the greatest common divisor is 1.

Example 16.3.12 Find the greatest common divisor of x² + 2x + 1 and x³ + 4x² + 5x + 2.

By the Euclidean algorithm for polynomials

x³ + 4x² + 5x + 2 = (x² + 2x + 1)(x + 2) + 0

Hence the greatest common divisor is x² + 2x + 1.

Example 16.3.13 Find the greatest common divisor of x₂ + 3x + 2 and x³ + 3x² −x− 3.

By the Euclidean algorithm for polynomials,

x³ + 3x² - x - 3 = (x² + 3x + 2) x + (-3x - 3)

Thus you can consider the two polynomials (x³ + 4x² + 5x + 2,−3x− 3). Now divide again.

and so it follows that the greatest common divisor is x + 1.

Example 16.3.14 Find the greatest common divisor of x² + 3x + 2 and x² + 2x− 3.

By Euclidean algorithm for polynomials,

x² + 3x + 2 = x² + 2x− 3 + (x− 1)

Now you can consider x² + 3x + 2 and x − 1. Do another division.

x² + 3x + 2 = (x + 4) (x− 1) + 6

Then it follows that the greatest common divisor is 1. Thus the two original polynomials are relatively prime.

16.3.2 Polynomials And Fields

When you have a polynomial like x² − 3 which has no rational roots, it turns out you can enlarge the field of rational numbers to obtain a larger field such that this polynomial does have roots in this larger field. I am going to discuss a systematic way to do this. It will turn out that for any polynomial with coefficients in any field, there always exists a possibly larger field such that the polynomial has roots in this larger field. This book has mainly featured the field of real or complex numbers but this procedure will show how to obtain many other fields which could be used in most of what was presented earlier in the book. Here is an important idea concerning equivalence relations.

Definition 16.3.15 Let S be a set. The symbol, ∼ is called an equivalence relation on S if it satisfies the following axioms.

1. x ∼ x for all x ∈S. (Reflexive)

2. If x ∼ y then y ∼ x. (Symmetric)

3. If x ∼ y and y ∼ z, then x ∼ z. (Transitive)

Definition 16.3.16 [x] denotes the set of all elements of S which are equivalent to x and [x] is called the equivalence class determined by x or just the equivalence class of x.

Also recall the notion of equivalence classes.

Theorem 16.3.17 Let ∼ be an equivalence class defined on a set, S and let H denote the set of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either x ∼ y and [x] = [y] or it is not true that x ∼ y and [x] ∩ [y] = ∅.

Definition 16.3.18 Let F be a field, for example the rational numbers, and denote by F [x] the polynomials having coefficients in F. Suppose p (x) is a polynomial. Let a (x) ∼ b (x) (a (x) is similar to b (x)) when

a (x) − b (x) = k (x) p (x)

for some polynomial k (x) .

Proposition 16.3.19 In the above definition, ∼ is an equivalence relation.

Proof: First of all, note that a (x) ∼ a (x) because their difference equals 0p (x) . If a (x) ∼ b (x) , then a (x)−b (x) = k (x) p (x) for some k (x) . But then b (x)−a (x) = −k (x) p (x) and so b (x) ∼a (x). Next suppose a (x) ∼b (x) and b (x) ∼c (x) . Then a (x)−b (x) = k (x) p (x) for some polynomial k (x) and also b (x)−c (x) = l (x) p (x) for some polynomial l (x). Then

a (x) − c (x) = a (x) −b (x) + b (x) − c (x)

= k (x) p (x) + l (x) p (x) = (l (x) + k (x)) p (x)

and so a (x) ∼ c (x) and this shows the transitive law.

With this proposition, here is another definition which essentially describes the elements of the new field. It will eventually be necessary to assume the polynomial p (x) in the above definition is irreducible so I will begin assuming this.

Definition 16.3.20 Let F be a field and let p (x) ∈ F[x] be irreducible. This means there is no polynomial which divides p (x) except for itself and constants. For the similarity relation defined in Definition 16.3.18, define the following operations on the equivalence classes. [a (x)] is an equivalence class means that it is the set of all polynomials which are similar to a (x).

[a (x)] + [b (x)] ≡ [a (x) + b (x)]

[a (x)] [b (x)] ≡ [a (x) b (x)]

This collection of equivalence classes is sometimes denoted by F[x] / (p (x)).

Proposition 16.3.21 In the situation of Definition 16.3.20, p (x) and q (x) are relatively prime for any q (x) ∈ F[x] which is not a multiple of p (x). Also the definitions of addition and multiplication are well defined. In addition, if a, b ∈ F and [a] = [b] , then a = b.

Proof: First consider the claim about p (x) , q (x) being relatively prime. If ψ (x) is the greatest common divisor, it follows ψ (x) is either equal to p (x) or 1. If it is p (x) , then q (x) is a multiple of p (x) . If it is 1, then by definition, the two polynomials are relatively prime.

To show the operations are well defined, suppose

[a (x)] = [a′ (x)] , [b (x)] = [b′ (x)]

It is necessary to show

[a (x) + b (x)] = [a′ (x) + b′ (x)]
[a (x) b (x)] = [a′ (x) b′ (x)]

Consider the second of the two.

Now by assumption ( a′ (x)−a (x)) is a multiple of p (x) as is (b′ (x)−b (x)) , so the above is a multiple of p (x) and by definition this shows [a (x) b (x)] = [a′ (x) b′ (x)]. The case for addition is similar.

Now suppose [ a] = [b] . This means a − b = k (x) p (x) for some polynomial k (x) . Then k (x) must equal 0 since otherwise the two polynomials a−b and k (x) p (x) could not be equal because they would have different degree.

Note that from this proposition and math induction, if each ai ∈ F,

With the above preparation, here is a definition of a field in which the irreducible polynomial p (x) has a root.

Definition 16.3.22 Let p (x) ∈ F[x] be irreducible and let a (x) ∼b (x) when a (x) −b (x) is a multiple of p (x) . Let G denote the set of equivalence classes as described above with the operations also described in Definition 16.3.20.

Also here is another useful definition and a simple proposition which comes from it.

Definition 16.3.23 Let F ⊆ K be two fields. Then clearly K is also a vector space over F. Then also, K is called a finite field extension of F if the dimension of this vector space, denoted by [K : F] is finite.

There are some easy things to observe about this.

Proposition 16.3.24 Let F ⊆ K ⊆ L be fields. Then [L : F] = [L : K] [K : F].

Proof: Let be a basis for L over K and let be a basis of K over F . Then if {li} l ∈L, there exist unique scalars xi in K such that

Now xi ∈ K so there exist fji such that

Then it follows that

It follows that {kjli} is a spanning set. If

Then, since the liare independent, it follows that

and since { kj } is independent, each fji = 0 for each j for a given arbitrary i. Therefore, { kjli } is a basis.

Theorem 16.3.25 The set of all equivalence classes G ≡ F/ (p (x)) described above with the multiplicative identity given by [1] and the additive identity given by [0] along with the operations of Definition 16.3.20, is a field and p ([x]) = [0] . (Thus p has a root in this new field.) In addition to this, [G :F] = n, the degree of p (x) .

Proof: Everything is obvious except for the existence of the multiplicative inverse and the assertion that p ([x]) = 0. Suppose then that [a (x)][0] . That is, a (x) is not a multiple of p (x). Why does [a (x)]⁻¹ exist? By Theorem 16.3.6, a (x) , p (x) are relatively prime and so there exist polynomials ψ (x) , (x) such that

1 = ψ (x) p (x) + a (x) (x)

and so

1−a (x) (x) = ψ (x) p (x)

which, by definition implies

[1 − a (x) (x)] = [1] − [a (x) (x)] = [1]− [a (x)] [ (x)] = [0]

and so [ (x)] = [a (x)]⁻¹. This shows G is a field.

Now if p (x) = anxn + an−1xn−1 + · · · + a₁x + a₀, p ([x]) = 0 by (16.7) and the definition which says [p (x)] = [0].

Consider the claim about the dimension. It was just shown that [1] , [x] ,[x²], · · · , [xn] is linearly dependent. Also [1] , [x] ,[x²], · · · ,[xn−1] is independent because if not, there would exist a polynomial q (x) of degree n − 1 which is a multiple of p (x) which is impossible. Now for [q (x)] ∈ G, you can write

q (x) = p (x) l (x) + r (x)

where the degree of r (x) is less than n or else it equals 0. Either way, [q (x)] = [r (x)] which is a linear combination of [1] , [x] ,[x²], · · · ,[xn−1]. Thus [G :F] = n as claimed.

Note that if p (x) were not irreducible, then you could find a field extension G such that [G :F] ≤ n. You could do this by working with an irreducible factor of p (x).

Usually, people simply write b rather than [b] if b ∈ F. Then with this convention,

[b (x)] = [b] [ (x)] = b [ (x)] .

This shows how to enlarge a field to get a new one in which the polynomial has a root. By using a succession of such enlargements, called field extensions, there will exist a field in which the given polynomial can be factored into a product of polynomials having degree one. The field you obtain in this process of enlarging in which the given polynomial factors in terms of linear factors is called a splitting field.

Theorem 16.3.26 Let p (x) = xn + an−1xn−1 + · · · + a₁x + a₀ be a polynomial with coefficients in a field of scalars F. There exists a larger field G such that there exist {z₁, · · · , zn} listed according to multiplicity such that.

This larger field is called the splitting field. Furthermore,

[G : F] ≤ n!

Proof: From Theorem 16.3.25, there exists a field F₁ such that p (x) has a root, z₁ (= [x] if p is irreducible.) Then by the Euclidean algorithm

p (x) = (x−z₁) q₁ (x) + r

where r ∈ F₁. Since p (z₁) = 0, this requires r = 0. Now do the same for q₁ (x) that was done for p (x) , enlarging the field to F₂ if necessary, such that in this new field

q₁ (x) = (x − z₂) q₂ (x) .

and so

p (x) = (x − z₁) (x − z₂) q₂ (x)

After n such extensions, you will have obtained the necessary field G.

Finally consider the claim about dimension. Then, by Theorem 16.3.25, there is a larger field G₁ such that p (x) has a root a₁ in G₁ and [G :F] ≤ n. Then

p (x) = (x − a₁) q (x)

Continue this way until the polynomial equals the product of linear factors. Then by Proposition 16.3.24 applied multiple times, [G : F] ≤ n!.

Example 16.3.27 The polynomial x²+1 is irreducible in R (x) , polynomials having real coefficients. To see this is the case, suppose ψ (x) 0 x² + 1. Then

x² + 1 = ψ (x) q (x)

If the degree of ψ (x) is less than 2, then it must be either a constant or of the form ax + b. In the latter case, −b/a must be a zero of the right side, hence of the left but x² + 1 has no real zeros. Therefore, the degree of ψ (x) must be two and q (x) must be a constant. Thus the only polynomial which divides x²+1 are constants and multiples of x²+1. Therefore, this shows x²+1 is irreducible. Find the inverse of [x² + x + 1] in the space of equivalence classes, R/ (x² + 1).

You can solve this with partial fractions

and so

1 = ( -x)(x² + x + 1) + (x + 1) (x² + 1)

1 ~ (-x) (x² + x + 1)

and so the inverse is [−x] .

The following proposition is interesting. It was essentially proved above but to emphasize it, here it is again.

Proposition 16.3.28 Suppose p (x) ∈ F [x] is irreducible and has degree n. Then every element of G =F[x] / (p (x)) is of the form [0] or [r (x)] 0 (x) is less than n.

Proof: This follows right away from the Euclidean algorithm for polynomials. If k (x) has degree larger than n − 1, then

k (x) = q (x) p (x) + r (x)

where r (x) is either equal to 0 or has degree less than n. Hence

[k (x)] = [r (x)].

Example 16.3.29 In the situation of the above example, find [ax + b]⁻¹ assuming a²+b²0. Note this includes all cases of interest thanks to the above proposition.

You can do it with partial fractions as above.

and so

Thus

and so

You might find it interesting to recall that (ai + b)⁻¹ =

16.3.3 The Algebraic Numbers

Each polynomial having coefficients in a field F has a splitting field. Consider the case of all polynomials p (x) having coefficients in a field F ⊆ G and will look at all roots which are also in G. The theory of vector spaces is very useful in the study of these algebraic numbers.

Definition 16.3.30 The algebraic numbers A are those numbers which are in G and also roots of some polynomial p (x) having coefficients in F. Here is a definition.

Theorem 16.3.31 Let a ∈ A. Then there exists a unique monic irreducible polynomial p (x) having coefficients in F such that p (a) = 0. This is called the minimal polynomial for a.

Proof: By definition, there exists a polynomial q (x) having coefficients in F such that q (a) = 0. If q (x) is irreducible, divide by the leading coefficient and this proves the existence. If q (x) is not irreducible, then there exist nonconstant polynomials r (x) and k (x) such that q (x) = r (x) k (x). Then one of r (a) , k (a) equals 0. Pick the one which equals zero and let it play the role of q (x). Continuing this way, in finitely many steps one obtains an irreducible polynomial p (x) such that p (a) = 0. Now divide by the leading coefficient and this proves existence. Suppose pi, i = 1, 2 both work and they are not equal. Then by Lemma 16.3.7 they must be relatively prime because they are both assumed to be irreducible and so there exist polynomials l (x) , k (x) such that

1 = l (x) p₁ (x) + k (x) p₂ (x)

But now when a is substituted for x, this yields 0 = 1, a contradiction. The polynomials are equal after all.

Definition 16.3.32 For a an algebraic number, let deg (a) denote the degree of the minimal polynomial of a.

Also, here is another definition.

Definition 16.3.33 Let a₁, · · · , am be in A. A polynomial in {a₁, · · · , am} will be an expression of the form

where the ak1···kn are in F, each kj is a nonnegative integer, and all but finitely many of the ak1···kn equal zero. The collection of such polynomials will be denoted by

F[a₁, · · · , am].

Now notice that for a an algebraic number, F[a] is a vector space with field of scalars F. Similarly, for {a₁, · · · , am} algebraic numbers, F[a₁, · · · , am] is a vector space with field of scalars F. The following fundamental proposition is important.

Proposition 16.3.34 Let {a₁, · · · , am} be algebraic numbers. Then

and for an algebraic number a,

dim F[a] = deg (a)

Every element of F [a₁, · · · , am] is in A and F[a₁, · · · , am] is a field.

Proof: First consider the second assertion. Let the minimal polynomial of a be

p (x) = xn + an−1xn−1 + · · · + a₁x + a₀.

Since p (a) = 0, it follows { 1, a, a², · · · , an} is linearly dependent. However, if the degree of q (x) is less than the degree of p (x) , then if q (x) is not a constant, the two must be relatively prime because p (x) is irreducible and so there exist polynomials k (x) , l (x) such that

1 = l (x) q (x) + k (x) p (x)

and this is a contradiction if q (a) = 0 because it would imply upon replacing x with a that 1 = 0. Therefore, no polynomial having degree less than n can have a as a root. It follows

{ 1, a, a², · · · , an−1}

is linearly independent. Thus dim F [a] = deg (a) = n. Here is why this is. If q (a) is any element of F [a] ,

q (x) = p (x) k (x) + r (x)

where deg r (x) < deg p (x) and so q (a) = r (a) and r (a)(1, a, a₂, · · · , an−1).

Now consider the first claim. By definition, F [a₁, · · · , am] is obtained from all linear combinations  where the ki are nonnegative integers. From the first part, it suffices to consider only kj ≤ deg (aj). Therefore, there exists a spanning set for F [a₁, · · · , am] which has

entries. By Theorem 16.2.4 this proves the first claim.

Finally consider the last claim. Let g (a₁, · · · , am) be a polynomial in {a₁, · · · , am} in F [a₁, · · · , am].Since

it follows

1, g (a₁, · · · , am) , g (a₁, · · · , am)², · · · , g (a₁, · · · , am)p

are dependent. It follows g (a₁, · · · , am) is the root of some polynomial having coefficients in F. Thus everything in F [a₁, · · · , am] is algebraic. Why is F [a₁, · · · , am] a field? Let g (a₁, · · · , am) be as just mentioned. Then it has a minimal polynomial,

p (x) = xp + ap−1xp−1 + · · · + a₁x + a₀

where the ai ∈ F. Then a₀ 0 or else the polynomial would not be minimal. Therefore,

and so the multiplicative inverse for g (a₁, · · · , am) is

The other axioms of a field are obvious.

Now from this proposition, it is easy to obtain the following interesting result about the algebraic numbers.

Theorem 16.3.35 The algebraic numbers A, those roots of polynomials in F [x] which are in G, are a field.

Proof: Let a be an algebraic number and let p (x) be its minimal polynomial. Then 0 (x) is of the form

xn + an−1xn−1 + · · · + a₁x + a₀

where a₀0. Then plugging in a yields

and so By the proposition, every element of F [a] is in A and this shows that for every element of A, its inverse is also in A. What about products and sums of things in A? Are they still in A? Yes. If a, b ∈ A, then both a + b and ab ∈ F [a, b] and from the proposition, each element of F [a, b] is in A.

A typical example of what is of interest here is when the field F of scalars is Q, the rational numbers and the field G is R or C. However, you can certainly conceive of many other examples by considering the integers mod a prime, for example (See Problems 1 - Problem 4 on Page 326 for example.) or any of the fields which occur as field extensions in the above.

16.3.4 The Lindemann Weierstrass Theorem And Vector Spaces

As another application of the abstract concept of vector spaces, there is an amazing theorem due to Weierstrass and Lindemann. There is a proof of this theorem in [9]. It is also in an appendix of Linear Algebra.

Theorem 16.3.36 Suppose a₁, · · · , an are algebraic numbers and suppose α₁, · · · , αn are distinct algebraic numbers. Then

In other words, the {eα1, · · · , eαn} are independent as vectors with field of scalars equal to the algebraic numbers.

A number is transcendental if it is not a root of a polynomial which has integer coefficients. Most numbers are this way but it is hard to verify that specific numbers are transcendental. That π is transcendental follows from

e⁰ + eiπ = 0.

By the above theorem, this could not happen if π were algebraic because then iπ would also be algebraic. Recall these algebraic numbers form a field and i is clearly algebraic, being a root of x² + 1. This fact about π was first proved by Lindemann in 1882 and then the general theorem above was proved by Weierstrass in 1885. This fact that π is transcendental solved an old problem called squaring the circle which was to construct a square with the same area as a circle using a straight edge and compass. It can be shown that the fact π is transcendental implies this problem is impossible.¹

16.4 Exercises

Let M ={u = (u₁, u₂, u₃, u₄) ∈ R⁴ : |u₁| ≤ 4}. Is M a subspace? Explain.

Let M ={u = (u₁, u₂, u₃, u₄) ∈ R₄ : sin (u₁) = 1}. Is M a subspace? Explain.

If you have 5 vectors inF⁵ and the vectors are linearly independent, can it always be concluded they span F⁵? Here F is an arbitrary field. Explain.

If you have 6 vectors in F⁵, is it possible they are linearly independent? Here F is an arbitrary field. Explain.

Show in any vector space, 0 is unique.

↑In any vector space, show that if x + y = 0, then y = −x.

↑Show that in any vector space, 0x = 0. That is, the scalar 0 times the vector x gives the vector 0.

↑Show that in any vector space, (−1) x = −x.

Let X be a vector space and suppose {x₁, · · · , xk} is a set of vectors from X. Show that 0 is in span (x₁, · · · , xk) .

Let X consist of the real valued functions which are defined on an interval [a, b] . For i, g ∈ X, f + g is the name of the function which satisfies (f + g) (x) = f (x) + g (x) and for α a real number, (αf) (x) ≡α (f (x)). Show this is a vector space with field of scalars equal to R. Also explain why it cannot possibly be finite dimensional.

Let S be a nonempty set and let V denote the set of all functions which are defined on S and have values in W a vector space having field of scalars F. Also define vector addition according to the usual rule, (f + g) (s) ≡ f (s) + g (s) and scalar multiplication by (αf) (s) ≡αf (s). Show that V is a vector space with field of scalars F.

Verify that any field F is a vector space with field of scalars F. However, show that R is a vector space with field of scalars Q.

Let F be a field and consider functions defined on {1, 2, · · · , n} having values in F. Explain how, if V is the set of all such functions, V can be considered as Fn.

Let V be the set of all functions defined on N ≡ {1, 2,· · · } having values in a field F such that vector addition and scalar multiplication are defined by (f + g) (s) ≡ f (s) + g (s) and (αf) (s) ≡ αf (s) respectively, for f, g ∈V and α ∈ F. Explain how this is a vector space and show that for ei given by

the vectors are linearly independent.

Suppose, in the context of Problem 10 you have smooth functions {y₁, y₂, · · · , yn} (all derivatives exist) defined on an interval [a, b] . Then the Wronskian of these functions is the determinant 

Show that if W (y₁, · · · , yn) (x)  0 for some x, then the functions are linearly independent. 

Give an example of two functions, y₁, y₂ defined on [−1, 1] such that

W (y₁, y₂) (x) = 0

for all x ∈ [−1, 1] and yet {y₁, y₂} is linearly independent.

Let the vectors be polynomials of degree no more than 3. Show that with the usual definitions of scalar multiplication and addition wherein, for p (x) a polynomial, (αp) (x) = αp (x) and for p, q polynomials (p + q) (x) ≡ p (x) + q (x) , this is a vector space.

In the previous problem show that a basis for the vector space is {1, x, x², x³}.

Let V be the polynomials of degree no more than 3. Determine which of the following are bases for this vector space.

(a) {x + 1, x³ + x² + 2x, x² + x, x³ + x² + x}

(b) {x³ + 1, x² + x, 2x³ + x², 2x³ −x² − 3x + 1}

In the context of the above problem, consider polynomials

  {aix³ + bix² + cix + di, i = 1, 2, 3, 4}

Show that this collection of polynomials is linearly independent on an interval [a, b] if and only if

is an invertible matrix.

Suppose V is a finite dimensional vector space. Based on the exchange theorem above, it was shown that any two bases have the same number of vectors in them. Give a different proof of this fact using the earlier material in the book. Hint: Suppose {x₁, · · · , xn} and {y₁, · · · , ym} are two bases with m < n. Then define

by

Consider the linear transformation,  Argue it is a one to one and onto mapping from Fn to Fm. Now consider a matrix of this linear transformation and its row reduced echelon form.

This and the following problems will present most of a differential equations course. To begin with, consider the scalar initial value problem

When a is real, show the unique solution to this problem is y = y₀ea(t−t₀). Next suppose

| [picture] | (16.8)

where y (t) = u (t) + iv (t) . Show there exists a unique solution and it is

| [picture] | (16.9)

Next show that for a real or complex there exists a unique solution to the initial value problem

and it is given by

Hint: For the first part write as y′ − ay = 0 and multiply both sides by e−at. Then explain why you get

Now you finish the argument. To show uniqueness in the second part, suppose

and verify this requires y (t) = 0. To do this, note

and that

Thus from the first part |y (t)|² = 0e-2at = 0. Finally observe by a simple computation that (16.8) is solved by (16.9). For the last part, write the equation as

y′ − ay = f

and multiply both sides by e−at and then integrate from t₀ to t using the initial condition.

↑Now consider A an n × n matrix. By Schur’s theorem there exists unitary Q such that

Q⁻¹AQ = T

where T is upper triangular. Now consider the first order initial value problem

x′ = Ax, x (t₀) = x₀.

Show there exists a unique solution to this first order system. Hint: Let y = Q⁻¹x and so the system becomes

| y′ = Ty, y (t₀) = Q⁻¹x₀ | (16.10)

Now letting y = (y₁, · · · , yn)T, the bottom equation becomes

Then use the solution you get in this to get the solution to the initial value problem which occurs one level up, namely

Continue doing this to obtain a unique solution to (16.10). 

↑Now suppose Φ (t) is an n × n matrix of the form

| Φ (t) =( x₁ (t) · · · xn (t))  | (16.11)

where

Explain why

Φ′ (t) = AΦ (t)

if and only if Φ (t) is given in the form of (16.11). Also explain why if c ∈ Fn,

y (t) ≡ Φ (t) c

solves the equation 

y′ (t) = Ay (t) .

↑In the above problem, consider the question whether all solutions to

| x′ = Ax | (16.12)

are obtained in the form Φ (t) c for some choice of c ∈ Fn. In other words, is the general solution to this equation Φ (t) c for c ∈ Fn? Prove the following theorem using linear algebra.

Theorem 16.4.1 Suppose Φ (t) is an n × n matrix which satisfies 

Φ′ (t) = AΦ(t) .

Then the general solution to (16.12) is Φ (t) c if and only if Φ (t)⁻¹ exists for some t. Furthermore, if Φ′ (t) = AΦ (t) , then either Φ (t)⁻¹ exists for all t or Φ (t)⁻¹ never exists for any t.

(det (Φ (t)) is called the Wronskian and this theorem is sometimes called the Wronskian alternative.)

Hint: Suppose first the general solution is of the form Φ (t) c where c is an arbitrary constant vector in Fn. You need to verify Φ (t)⁻¹ exists for some t. In fact, show Φ (t)⁻¹ exists for every t. Suppose then that Φ (t₀)⁻¹ does not exist. Explain why there exists c ∈ Fn such that there is no solution x to

c = Φ (t₀) x

By the existence part of Problem 24 there exists a solution to

x′ = Ax, x (t₀) = c

but this cannot be in the form Φ (t) c. Thus for every t, Φ (t)⁻¹ exists. Next suppose for some t₀, Φ (t₀)⁻¹ exists. Let z′ = Az and choose c such that

z (t₀) = Φ (t₀) c

Then both z (t) , Φ (t) c solve

x′ = Ax, x (t₀) = z (t₀)

Apply uniqueness to conclude z = Φ (t) c. Finally, consider that Φ (t) c for c ∈ Fn either is the general solution or it is not the general solution. If it is, then Φ (t)⁻¹ exists for all t. If it is not, then Φ (t)⁻¹ cannot exist for any t from what was just shown.

↑Let Φ′ (t) = AΦ (t) . Then Φ (t) is called a fundamental matrix if Φ (t)⁻¹ exists for all t. Show there exists a unique solution to the equation

| x′ = Ax + f, x (t₀) = x₀ | (16.13)

and it is given by the formula

Now these few problems have done virtually everything of significance in an entire undergraduate differential equations course, illustrating the superiority of linear algebra. The above formula is called the variation of constants formula.

Hint: Uniqueness is easy. If x₁, x₂ are two solutions then let u (t) = x₁ (t) − x₂ (t) and argue u′ = Au, u (t₀) = 0. Then use Problem 24. To verify there exists a solution, you could just differentiate the above formula using the fundamental theorem of calculus and verify it works. Another way is to assume the solution in the form

x (t) = Φ (t) c (t)

and find c (t) to make it all work out. This is called the method of variation of parameters.

↑Show there exists a special Φ such that Φ′ (t) = AΦ (t) , Φ (0) = I, and Φ (t)⁻¹ exists for all t. Show using uniqueness that

Φ (−t) = Φ (t)⁻¹

and that for all t, s ∈ R 

Φ (t + s) = Φ (t) Φ (s)

Explain why with this special Φ, the solution to (16.13) can be written as

Hint: Let Φ (t) be such that the jth column is xj (t) where

Use uniqueness as required.

¹ Gilbert, the librettist of the Savoy operas, may have heard about this great achievement. In Princess Ida which opened in 1884 he has the following lines. “As for fashion they forswear it, so the say - so they say; and the circle they will square it some fine day some fine day.” Of course it had been proved impossible to do this a couple of years before.

16.5 Inner Product Spaces

16.5.1 Basic Deﬁnitions And Examples

An inner product space V is a vector space which also has an inner product. It is usually assumed, when considering inner product spaces that the field of scalars is either F = R or C. This terminology has already been considered in the context of Fn. In this section, it will be assumed that the field of scalars is C, the complex numbers, unless speciﬁed to be something else. An inner product is a mapping <·,·> : V × V → C which satisﬁes the following axioms.

Axioms For Inner Product

(u, v) ∈ C, (u, v) = (v, u).

If a, b are numbers and u, v, z are vectors then ((au + bv) , z) = a(u, z) + b(v, z).

(u, u) ≥ 0 and it equals 0 if and only if u = 0.

Note this implies (x,αy) =α(x, y) because

Example 16.5.1 Let V be the continuous complex valued functions deﬁned on a finite closed interval I. Deﬁne an inner product as follows.

where p (x) some function which is strictly positive on the closed interval I. It is understood in writing this that

Then with this convention, the usual calculus theorems hold about evaluating integrals using the fundamental theorem of calculus and so forth. You simply apply these theorems to the real and imaginary parts of a complex valued function.

Example 16.5.2 Let V be the polynomials of degree at most n which are deﬁned on a closed interval I and let {x₀, x₁, · · · , xn} be n + 1 distinct points in I. Then deﬁne

This last example clearly satisﬁes all the axioms for an inner product except for the one which says that <u, u> = 0 if and only if u = 0. Suppose then that <f, f> = 0. Then f must vanish at n + 1 distinct points but f is a polynomial of degree n. Therefore, it has at most n zeros unless it is identically equal to 0. Hence the second case holds and so f equals 0.

Example 16.5.3 Let V be any complex vector space and let {v₁, · · · , vn} be a basis. Decree that

Then deﬁne

This makes the complex vector space into an inner product space.

Example 16.5.4 Let V consist of sequences a = {ak}∞ , ak ∈ C, with the property that

and the inner product is then deﬁned as

All of the axioms of the inner product are obvious for this example except the most basic one which says that the inner product has values in C. Why does the above sum even converge? It converges from a comparison test.

and by assumption,

and therefore, the given sum which deﬁnes the inner product is absolutely convergent. Therefore, thanks to completeness of C this sum also converges. This fact should be familiar to anyone who has had a calculus class in the context that the sequences are real valued. The case where they are complex valued follows right away from a consideration of real and imaginary parts.

By far the most important example of an inner product space is L² (Ω), the space of Lebesgue measurable square integrable functions deﬁned on Ω. However, this is a book on algebra, not analysis, so this example will be ignored.

16.5.2 The Cauchy Schwarz Inequality And Norms

The most fundamental theorem relative to inner products is the Cauchy Schwarz inequality.

Theorem 16.5.5 (Cauchy Schwarz)The following inequality holds for x and y ∈ V, an inner product space.

Equality holds in this inequality if and only if one vector is a multiple of the other.

Proof: Let θ ∈ C such that |θ| = 1 and

Consider p (t) ≡ (x +θty, x + tθy) where t ∈ R. Then from the above list of properties of the dot product,

and this must hold for all t ∈ R. Therefore, if (y, y) = 0 it must be the case that |(x, y)| = 0 also since otherwise the above inequality would be violated for large negative t. Therefore, in this case,

In the other case, if (y, y) ̸= 0, then p(t) ≥ 0 for all t means the graph of y = p(t) is a parabola which opens up and it either has exactly one real zero in the case its vertex touches the t axis or it has no real zeros.

From the quadratic formula this happens exactly when

which is equivalent to (16.14).

It is clear from a computation that if one vector is a scalar multiple of the other that equality holds in (16.14). Conversely, suppose equality does hold. Then this is equivalent to saying 4|(x, y)|² − 4(x, x) (y, y) = 0 and so from the quadratic formula, there exists one real zero to p (t) = 0. Call it t₀. Then

and so x = −θt₀y.

So what does the Cauchy Schwarz inequality say in the above examples? In Example 16.5.1 it says that

With the Cauchy Schwarz inequality, it is possible to obtain the triangle inequality. This is the inequality in the next theorem. First it is necessary to deﬁne the norm or length of a vector. This is what is in the next deﬁnition.

Deﬁnition 16.5.6 Let V be an inner product space and let z ∈ V. Then |z| ≡ (z, z)1/2. |z| is called the norm of z and also the length of z.

With the deﬁnition of length of a vector, here are the main properties of length.

Theorem 16.5.7 For length deﬁned in Deﬁnition 16.5.6, the following hold.

Proof: The first two claims are left as exercises. To establish the third,

The properties (16.16) - (16.18) are the axioms for a norm. A vector space which has a norm is called a normed linear space or a normed vector space.

16.5.3 The Gram Schmidt Process

The Gram Schmidt process is also valid in an inner product space. If you have a linearly independent set of vectors, there is an orthonormal set of vectors which has the same span. Recall the deﬁnition of an orthonormal set. It is the same as before.

Deﬁnition 16.5.8 Let V be an inner product space and let {ui} be a collection of vectors. It is an orthonormal set if

As before, every orthonormal set of vectors is linearly independent. If

where {uk}nk=1 is an orthonormal set of vectors, why is each ck = 0?

This is true because you can take the inner product of both sides with uj. Then

The right side equals 0 because

Subtracting (0, u) from both sides shows that (0, u) = 0. Therefore, from the properties of the inner product,

Since cj was arbitrary, this veriﬁes that an orthonormal set of vectors is linearly independent.

Now consider the Gram Schmidt process.

Lemma 16.5.9 Let {x₁, · · · , xn} be a linearly independent subset of an inner product space V. Then there exists an orthonormal set of vectors {u₁, · · · , un} which has the property that for each k ≤ n, span (x₁, · · · , xk) = span (u₁, · · · , uk) .

Proof: Let u₁ ≡ x₁/|x₁|. Thus for k = 1, span (u₁) = span (x₁) and {u₁} is an orthonormal set. Now suppose for some k < n, u₁, · · ·, uk have been chosen such that (uj, ul) = δjl and span (x₁, · · · , xk) = span (u₁, · · · , uk). Then deﬁne

where the denominator is not equal to zero because the xj form a basis, and so

Thus by induction,

Also, xk+1 ∈ span (u₁, · · · , uk, uk+1) which is seen easily by solving (16.19) for xk+1 and it follows

If l ≤ k,

The vectors, {uj}nj=1, generated in this way are therefore orthonormal because each vector has unit length.

As in the case of Fn, if you have a finite dimensional subspace of an inner product space, you can begin with a basis and then apply the Gram Schmidt process above to obtain an orthonormal basis.

There is nothing wrong with the above algorithm, but when you use it, it tends to get pretty intricate and it is easy to get lost in the details. There is a way to simplify it to produce fewer steps using matrices. I will illustrate in the case of three vectors. Say {u₁, u₂, u₃} is linearly independent and you wish to find an orthonormal set {v₁, v₂, v₃} which has the same span such that span (u₁, · · · , uk) = span (v₁, · · · , vk) for each k = 1, 2, 3. Then you would have

where R is an upper triangular matrix. Then

Let G be the matrix whose rs entry is (ur, us). This is called the Grammian matrix. Then the above reduces to the following matrix equation.

Taking inverses of both sides yields

Then it follows that

Example 16.5.10 Let the real inner product space V consist of the continuous functions deﬁned on [0, 1] with the inner product given by

and consider the functions (vectors) {1, x, x², x³}. Show this is a linearly independent set of vectors and obtain an orthonormal set of vectors having the same span.

First, why is this a linearly independent set of vectors? This follows easily from Problem 15 on Page 346. You consider the Wronskian of these functions.

Therefore, the vectors are linearly independent. Now following the above procedure with matrices, let

Also it is necessary to compute the Grammian.

You also need the inverse of this Grammian. However, a computer algebra system can provide this right away

Now it just remains to find the ai.

Thus you can take (There is more than one solution.)

Thus the desired orthonormal basis is given by

which yields

16.5.4 Approximation And Least Squares

Let V be an inner product space and let U be a finite dimensional subspace. Given y ∈V, how can you find the vector of U which is closest to y out of all such vectors in U? Does there even exist such a closest vector? The following picture is suggestive of the conclusion of the following lemma. It turns out that pictures like this do not mislead when you are dealing with inner product spaces in any number of dimensions.

Note that in the picture, z is a point in U and also w is a point of U. The following lemma states that for z to be closest to y out of all vectors in U, the vector from z to y should be perpendicular to any vector w ∈ U. Since U is a subspace, this is the same as saying that the vector → zy is perpendicular to the vector → zw which is the situation illustrated by the above picture.

Lemma 16.5.11 Suppose y ∈ V, an inner product space and U is a subspace of V. Then

for all w ∈ U if and only if, for all w ∈ U, 

Furthermore, there is at most one z which minimizes |y−w| for w ∈U.

Proof: First suppose condition (16.21). Letting w ∈U, and using the properties of the inner product and the deﬁnition of the norm,

It follows then that |y−w| is minimized when w = z.Next suppose z is a minimizer. Then pick w ∈U and let t ∈ R. Let θ ∈ C be such that |θ| = 1 and θ(y−z, w) = |(y−z, w)|. Then

has a minimum when t = 0. But from the axioms of the inner product and deﬁnition of the norm, this function of t equals

Hence its derivative when t = 0 which is 2|(y − z, w)| equals 0.

Suppose now that zi, i = 1, 2 both are minimizers. Then, as above,

and this is a contradiction unless z₁ = z₂ because |y−z₁|² = |y−z₂|².

This z described above is called the orthogonal projection of y onto U. The picture suggests why it is called this. The vector y − z is perpendicular to the vectors in U.

With the above lemma, here is a theorem about existence, uniqueness and properties of a minimizer. The following theorem shows that the orthogonal projection is obtained by the linear transformation given by the formula

Note that the formula as well as the geometric interpretation suggested in the above picture shows that T² = T .

Theorem 16.5.12 Let V be an inner product space and let U be an n dimensional subspace of V . Then if y ∈ V is given, there exists a unique x ∈ U such that

for all w ∈ U and in addition, there is a formula for x in terms of any orthonormal basis for U,{u₁, · · · , un},

Proof: By Lemma 16.5.11 there is at most one minimizer and it is characterized by the condition

for all w ∈U Let {uk}nk=1 be an orthonormal basis for U. By the Gram Schmidt process, Lemma . 16.5.9, there exists such an orthonormal basis. Now it only remains to verify that

for all w. Since {uk}nk=1is a basis, it suﬃces to verify that

However, from the properties of the inner product,

Note it follows that for any orthonormal basis {uk}nk=1, the same unique vector x is obtained as

and it is the unique minimizer of w→|y−w|. This is stated in the following corollary for the sake of emphasis.

Corollary 16.5.13 Let V be an inner product space and let U be an n dimensional subspace of V. Then for y given in V, and {uk}n ,{vk}n two orthonormal bases for U,

The scalars (y, uk) are called the Fourier coeﬃcients.

Example 16.5.14 Let V denote the real inner product space consisting of continuous functions deﬁned on [0, 1] with the inner product

Let U = span (1, x, x²). It is desired to find the vector (function) in U which is closest to sin in the norm determined by this inner product. Thus it is desired to minimize

out of all functions p contained in U.

By Example 16.5.10, an orthonormal basis for U is

Then by Theorem 16.5.12, the closest vector (function) in U to sin can be computed as follows. First determine the Fourier coeﬃcients.

Next, from Theorem 16.5.12, the closest point to sin is

Simplifying and approximating things like sin 1, this yields the following for the approximation to sin x.

– 0.23546x² + 1.0913x – 7.4649 x 10⁻³

If this is graphed along with sin x for x ∈ [0, 1] the result is as follows.

One of the functions is represented by the solid line and the other by the dashed line. Now by contrast, consider the Taylor series for sin x up to degree 2. This is just f (x) = x + 0x².

You see the difference. The approximation using the inner product norm, called mean square approximation, attempts to approximate the given function on the whole interval while the Taylor series approximation is only good for small x.

16.5.5 Fourier Series

One of the most important applications of these ideas about approximation is to Fourier series. Much more can be said about these than will be presented here. However, Theorem 16.5.12 is a very useful framework for discussing these series.

For x ∈ R, deﬁne eix by the following formula

The reason for deﬁning it this way is that ei0 = 1, and (eix)= ieix if you use this deﬁnition. Also it follows from the trigonometry identities that

This is because

In addition,

because

It follows that the functions  for

form an orthonormal set in V, the inner product space of continuous functions deﬁned on [−π, π] with the inner product given by

I will verify this now. Let k, l be two integers in In, kl

Also

Example 16.5.15 Let V be the inner product space of piecewise continuous functions deﬁned on [−π, π] and let U denote the span of the functions (vectors){eikx}nk=–n. Let

Find the vector of U which is closest to f in the mean square sense (In the norm deﬁned by this inner product).

First of all, you need to find the Fourier coeﬃcients. Since x → cos (x) is even and x → sin (x) is odd

Therefore, the best approximation is

The term for k can be combined with the term for −k to yield

The terms when k is even are all 0. Therefore, the above reduces to

In the case where n = 4, the graph of the function f being approximated along with the above function which is approximating it are as shown in the following picture.

This sum which delivers the closest point in U will be denoted by Snf.

Note how the approximate function, closest in the mean square norm, is not equal to the given function at very many points but is trying to be close to it across the entire interval [−π, π], except for a small interval centered at 0. You might try doing similar graphs on a calculator or computer in which you take larger and larger values of n. What will happen is that there will be a little bump near the point of discontinuity which won’t go away, but this little bump will get thinner and thinner. The reason this must happen is roughly because the functions in the sum are continuous and the function being approximated is not. Therefore, convergence cannot take place uniformly. This is all I will say about these considerations because this is not an analysis book. See Problem 20 below for a discussion of where the Fourier series does converge at jumps.

16.5.6 The Discrete Fourier Transform

Everything done above for the Fourier series on [−π, π] could have been done just as easily on [0, 2π] because all the functions are periodic of period 2π. Thus, for f a function deﬁned on [0, 2fπ] , you could consider the partial sums of the Fourier series on [0, 2π]

where

and all the results of the last section continue to apply except now it is on the new interval. This is done to make the presentation of what follows easier to write.

The idea is that maybe you don’t know what the function f is at all points, only at certain points

Then instead of the integral given above, you could write a Riemann sum which approximates it. I will simply write the left Riemann sum. This yields the approximation bk for ak. Assuming f is continuous, the approximation would improve as N → ∞.

where yj is deﬁned to be the value of the function at xj. This is called the discrete Fourier transform. In terms of matrix multiplication, let . Then

Thus you can find these approximate values by matrix multiplication.

Example 16.5.16 Suppose you have the following table of values for the function f.

Note that the above only uses the first four values.

In this case, N = 4 and so ωN = e−i(π/2) = −i. Then the approximate Fourier coeﬃcients are given by

It follows that the approximate Fourier series for the given function is

This simpliﬁes to

If you graph this, it will not do all that well in approximating some functions which have the given values at the given points. This is not surprising since only four points were considered. This is why in practice, people like to use a large number of points and when you do, the computations become suﬃciently long that a special algorithm was developed for doing them. It is called the fast Fourier transform. So when you see this mentioned, this is what it is about, eﬃciently computing the discrete Fourier transform which can be thought of as a way to approximate the Fourier coeﬃcients based on incomplete information for a given function.

16.6 Exercises

Verify that Examples 16.5.1 - 16.5.4 are each inner product spaces.

In each of the examples 16.5.1 - 16.5.4 write the Cauchy Schwarz inequality.

Verify (16.16) and (16.17).

Consider the Cauchy Schwarz inequality. Show that it still holds under the assumptions (u, v) = (v, u),((au + bv) , z) = a(u, z) + b(v, z), and (u, u) ≥ 0. Thus it is not necessary to say that (u, u) = 0 only if u = 0. It is enough to simply state that (u, u) ≥ 0.

Consider the integers modulo a prime, Zp. This is a field of scalars. Now let the vector space be (Zp)n where n ≥ p. Deﬁne now

Does this satisfy the axioms of an inner product? Does the Cauchy Schwarz inequality hold for this <>? Does the Cauchy Schwarz inequality even make any sense?

If you only know that (u, u) ≥ 0 along with the other axioms of the inner product and if you deﬁne |z| the same way, how do the conclusions of Theorem 16.5.7 change?

In an inner product space, an open ball is the set

B (x, r) ≡ {y : |y−x|< r}.

If z ∈ B (x, r) , show there exists δ > 0 such that B(z, δ) ⊆ B(x, r). In words, this says that an open ball is open. Hint: This depends on the triangle inequality.

Let V be the real inner product space consisting of continuous functions deﬁned on [−1, 1] with the inner product given by

Show that {1, x, x²} are linearly independent and find an orthonormal basis for the span of these vectors.

A regular Sturm Liouville problem involves the differential equation for an unknown function of x which is denoted here by y,

and it is assumed that p(t) , q(t) > 0 for any t along with boundary conditions,

where

There is an immense theory connected to these important problems. The constant λ is called an eigenvalue. Show that if y is a solution to the above problem corresponding to λ = λ₁ and if z is a solution corresponding to λ = λ₂ ̸= λ₁, then

| [picture] | (16.22)

Hint: Do something like this:

Now subtract and either use integration by parts or show

and then integrate. Use the boundary conditions to show that y′(a) z(a)−z′(a)y(a) = 0 and y′ (b) z (b)−z′ (b) y (b) = 0.

Using the above problem or standard techniques of calculus, show that

are orthonormal with respect to the inner product

Hint: If you want to use the above problem, show that sin (nx) is a solution to the boundary value problem

y′′ + n²y = 0, y (0) = y (π) = 0

Find S₅f (x) where f (x) = x on [−π, π] . Then graph both S₅f (x) and f (x) if you have access to a system which will do a good job of it.

Find S₅f (x) where f (x) = |x| on [−π, π] . Then graph both S₅f (x) and f (x) if you have access to a system which will do a good job of it.

Find S₅f (x) where f (x) = x² on [−π, π] . Then graph both S₅f (x) and f (x) if you have access to a system which will do a good job of it.

Let V be the set of real polynomials deﬁned on [0, 1] which have degree at most 2. Make this into a real inner product space by deﬁning

(f, g) ≡ f (0) g (0) + f (1/2) g (1/2) + f (1) g (1)

Find an orthonormal basis and explain why this is an inner product.

Consider Rn with the following deﬁnition.

Does this deﬁne an inner product? If so, explain why and state the Cauchy Schwarz inequality in terms of sums.

From the above, for f a piecewise continuous function,

Show this can be written in the form

where

This is called the Dirichlet kernel. Show that

For V the vector space of piecewise continuous functions, deﬁne Sn : V → V by

Show that Sn is a linear transformation. (In fact, Snf is not just piecewise continuous but inﬁnitely differentiable. Why?) Explain why ∫ππ Dn (t) dt = 1. Hint: To obtain the formula, do the following.

Change the variable of summation in the bottom sum and then subtract and solve for Dn (t).

↑Let V be an inner product space and let U be a finite dimensional subspace with an orthonormal basis {ui}ni=1 . If y ∈V, show

Now suppose that {uk}∞k=1 is an orthonormal set of vectors of V . Explain why

When applied to functions, this is a special case of the Riemann Lebesgue lemma.

↑Let f be any piecewise continuous function which is bounded on [−π, π] . Show, using the above problem, that

↑∗Let f be a function which is deﬁned on (−π, π]. The 2π periodic extension is given by the formula f (x + 2π) = f (x) . In the rest of this problem, f will refer to this 2π periodic extension. Assume that f is piecewise continuous, bounded, and also that the following limits

Here it is assumed that

both exist at every point. The above conditions rule out functions where the slope taken from either side becomes inﬁnite. Justify the following assertions and eventually conclude that under these very reasonable conditions

the mid point of the jump. In words, the Fourier series converges to the midpoint of the jump of the function.

Now apply some trig. identities and use the result of Problem 18 to conclude that both of these terms must converge to 0.

↑Using the Fourier series obtained in Problem 11 and the result of Problem 19 above, find an interesting formula by examining where the Fourier series converges when x = π/2. Of course you can get many other interesting formulas in the same way. Hint: You should get

Let V be an inner product space and let K be a convex subset of V . This means that if x, z ∈K, then the line segment x + t (z−x) = (1−t) x + tz is contained in K for all t ∈ [0, 1] . Note that every subspace is a convex set. Let y ∈V and let x ∈K. Show that x is the closest point to y out of all points in K if and only if for all w ∈K,

Re(y − x, w − x) ≤ 0.

In  Rn, a picture of the above situation where x is the closest point to y is as follows.

The condition of the above variational inequality is that the angle θ shown in the picture is larger than 90 degrees. Recall the geometric description of the dot product presented earlier. See Page 41.

Show that in any inner product space the parallelogram identity holds.

Next show that in a real inner product space, the polarization identity holds.

∗This problem is for those who know about Cauchy sequences and completeness of Fp and about closed sets. Suppose K is a closed nonempty convex subset of a finite dimensional subspace U of an inner product space V . Let y ∈V. Then show there exists a unique point x ∈K which is closest to y. Hint: Let

Let {xn} be a minimizing sequence,

Use the parallelogram identity in the above problem to show that {xn} is a Cauchy sequence. Now let {uk}pk=1 be an orthonormal basis for U. Say

Verify that for cn ≡ (cn₁ −cmp)Fp.

Now use completeness of Fp and the assumption that K is closed to get the existence of x ∈K such that |x−y| = λ.

∗Let K be a closed nonempty convex subset of a finite dimensional subspace U of a real inner product space V . (It is true for complex ones also.) For x ∈V, denote by Px the unique closest point to x in K. Verify that P is Lipschitz continuous with Lipschitz constant 1,

Hint: Use Problem 21.

∗This problem is for people who know about compactness. It is an analysis problem. If you have only had the usual undergraduate calculus course, don’t waste your time with this problem. Suppose V is a finite dimensional normed linear space. Recall this means that there exists a norm ∥·∥ deﬁned on V as described above,

Let |·| denote the norm which comes from Example 16.5.3, the inner product by decree. Show |·| and ∥·∥ are equivalent. That is, there exist constants δ, ∆ > 0 such that for all x ∈V,

In explain why every two norms on a finite dimensional vector space must be equivalent in the above sense.

Linear Transformations

17.1 Matrix Multiplication As A Linear Transformation

Deﬁnition 17.1.1 Let V and W be two finite dimensional vector spaces. A function, L which maps V to W is called a linear transformation and L ∈ L(V, W ) if for all scalars α and β, and vectors v, w,

These linear transformations are also called homomorphisms. If one of them is one to one, it is called injective and if it is onto, it is called surjective. When a linear transformation is both one to one and onto, it is called bijective. ,

An example of a linear transformation is familiar matrix multiplication. Let A = (aij) be an m × n matrix. Then an example of a linear transformation L :Fn → Fm is given by

Here

17.2 L(V, W) As A Vector Space

In what follows I will denote vectors in bold face. However, this does not mean they are in Fn.

Deﬁnition 17.2.1 Given L, M ∈ L(V, W) deﬁne a new element of L(V, W) , denoted by L + M according to the rule

For α a scalar and L ∈ L(V,W) , deﬁne αL ∈ L(V,W) by

You should verify that all the axioms of a vector space hold for L(V,W) with the above deﬁnitions of vector addition and scalar multiplication. What about the dimension of L(V,W)?

Theorem 17.2.2 Let V and W be finite dimensional linear spaces of dimension n and m respectively Then dim (L(V,W)) = mn.

Proof: Let the two sets of bases be

for X and Y respectively. Let Eik ∈ L(V,W) be the linear transformation deﬁned on the basis, {v₁, • • • , vn}, by

where δik = 1 if i = k and 0 if i ≠ k. Thus

Then let L ∈ L(V,W). Since {w₁, • • • , wm} is a basis, there exist constants djk such that

Also

It follows that

because the two linear transformations agree on a basis. Since L is arbitrary, this shows

spans L(V,W).

If

then

and so, since {w₁, • • • , wm} is a basis, dil = 0 for each i = 1, • • • , m. Since l is arbitrary, this shows dil = 0 for all i and l. Thus these linear transformations form a basis and this shows the dimension of L(V,W) is mn as claimed.

17.3 Eigenvalues And Eigenvectors Of Linear Transformations

Here is a very useful theorem due to Sylvester.

Theorem 17.3.1 Let A ∈ L(V,W) and B ∈ L(W,U) where V, W, U are all vector spaces over a field F. Suppose also that ker (A) and A (ker (BA)) are finite dimensional subspaces. Then

Proof: If x ∈ ker (BA) , then Ax ∈ ker (B) and so A(ker(BA)) ⊆ ker (B) . The following picture may help.

Now let {x₁, • • • , xn} be a basis of ker (A) and let {Ay₁, • • • , Aym} be a basis for A (ker (BA)) . Take any z ∈ ker (BA) . Then Az = ∑mi=1 aiAyi and so

which means z – ∑ mi=1aiyi ∈ ker (A) and so there are scalars bi such that

It follows span (x₁, • • • , xn, y₁, • • • , ym) ⊇ ker (BA) and so by the first part, (See the picture.)

Of course this result holds for any finite product of linear transformations by induction. One way this is quite useful is in the case where you have a finite product of linear transformations ∏l Lii=1 all in L(V,V) . Then

and so if you can find a linearly independent set of vectors in ker (∏li = 1 Li) of size

then it must be a basis for ker.

Deﬁnition 17.3.2 Let {Vi}ri₌₁ be subspaces of V. Then

denotes all sums of the form ∑ri=1 vi where vi ∈ Vi. If whenever

it follows that vi = 0 for each i, then a special notation is used to denote ∑ri=1 Vi. This notation is

and it is called a direct sum of subspaces.

Lemma 17.3.3 If V = V₁ ⊕ • • • ⊕Vr and if βi = {vi₁, • • • , vimi } is a basis for Vi, then a basis for V is {β₁, • • • , βr}.

Proof: Suppose ∑ri=1 ∑mij=1 cijvij = 0. then since it is a direct sum, it follows for each i,

and now since {vii=1, • • • , vimi } is a basis, each cij = 0.

Here is a useful lemma.

Lemma 17.3.4 Let Li be in L(V,V) and suppose for i ≠ j, LiLj = LjLi and also Li is one to one on ker (Lj) whenever i ≠ j. Then

Here ∏pi=1 Li is the product of all the linear transformations. A symbol like ∏j≠iLj is the product of all of them but Li.

Proof: Note that since the operators commute, Lj : ker (Li) → ker (Li). Here is why. If Liy = 0 so that y ∈ ker (Li) , then

LiLjy = LjLiy = Lj0 = 0

and so Lj : ker (Li) → ker (Li). Suppose

but some vi ≠ 0. Then do ∏j≠i Lj to both sides. Since the linear transformations commute, this results in

which contradicts the assumption that these Lj are one to one and the observation that they map ker (Li) to ker (Li). Thus if

then each vi = 0.

Let βi  = {vi₁, • • • , vimi } be a basis for ker (Li). Then from what was just shown and Lemma 17.3.3, { β₁, • • • , βp } must be linearly independent and a basis for

It is also clear that since these operators commute,

Therefore, by Sylvester’s theorem and the above,

Now in general, if W is a subspace of V, a finite dimensional vector space and the two have the same dimension, then W = V . This is because W has a basis and if v is not in the span of this basis, then v adjoined to the basis of W would be a linearly independent set so the dimension of V would then be strictly larger than the dimension of W.

It follows

Here is a situation in which the above holds. ker (A – λiI)r is sometimes called a generalized eigenspace. The following is an important result on generalized eigenspaces.

Theorem 17.3.5 Let V be a vector space of dimension n and A a linear transformation and suppose {λ₁, • • • , λk} are distinct scalars. Deﬁne for ri ∈ N

Then

Proof: It is obvious the linear transformations (A−λiI)ri commute. Now here is a claim.

Claim : Let µ≠ λi, Then (A − µI)m : Vi →Vi and is one to one and onto for every m ∈ N. 

Proof: It is clear (A – µI)m maps Vi to Vi because if v ∈ Vi then (A−λiI)ri v = 0. Consequently,

which shows that (A − µI)mv ∈ Vi.

It remains to verify (A – µI)m is one to one. This will be done by showing that (A – µI) is one to one. Let w ∈ Vi and suppose (A – µI) w = 0 so that Aw = µw. Then for m ≡ ri, (A−λiI)m w = 0 and so by the binomial theorem,

Therefore, since µ ≠ λi, it follows w = 0 and this veriﬁes (A – µI) is one to one. Thus (A – µI)m is also one to one on Vi. Letting {ui₁, • • • , ui rk} be a basis for Vi, it follows.

is also a basis and so (A – µI)m is also onto. The desired result now follows from Lemma 17.3.4.

Let V be a finite dimensional vector space with field of scalars C. For example, it could be a subspace of Cn. Also suppose A ∈ L(V,V) . Does A have eigenvalues and eigenvectors just like the case where A is a n x n matrix?

Theorem 17.3.6 Let V be a nonzero finite dimensional vector space of dimension n. Suppose also the field of scalars equals C.¹ Suppose A ∈ L(V,V) . Then there exists v ≠ 0 and λ ∈ C such that

  Av = λv.

Proof: Consider the linear transformations, I, A, A², • • • , An2. There are n² + 1 of these transformations and so by Theorem 17.2.2 the set is linearly dependent. Thus there exist constants, ci ∈ C such that

This implies there exists a polynomial, q (λ) which has the property that q (A) = 0. In fact, q (λ) ≡ c₀ + ∑n2k=1 ckλk. Dividing by the leading term, it can be assumed this polynomial is of the form λm + cm-1λm−1 + • • • + c₁λ + c₀, a monic polynomial. Now consider all such monic polynomials, q such that q (A) = 0 and pick one which has the smallest degree. This is called the minimal polynomial and will be denoted here by p(λ) . By the fundamental theorem of algebra, p(λ) is of the form

where some of the λk might be repeated. Thus, since p has minimal degree,

Therefore, there exists u ≠ 0 such that

But then

As a corollary, it is good to mention that the minimal polynomial just discussed is unique.

Corollary 17.3.7 Let A ∈ L(V,V) where V is an n dimensional vector space, the field of scalars being F. Then there exists a polynomial q (λ) having coeﬃcients in F such that q (A) = 0. Letting p (λ) be the monic polynomial having smallest degree such that p (A) = 0, it follows that p (λ) is unique.

Proof: The existence of p(λ) follows from the above theorem. Suppose then that p₁ (λ) is another one. That is, it has minimal degree of all polynomials q(λ) satisfying q (A) = 0 and is monic. Then by Lemma 16.3.3 there exists r (λ) which is either equal to 0 or has degree smaller than that of p(λ) and a polynomial l(λ) such that

p₁ (λ) = p (λ) l (λ) + r (λ)

By assumption, r (A) = 0. Therefore, r (λ) = 0. Also by assumption, p₁ (λ) and p (λ) have the same degree and so l (λ) is a scalar. Since p₁ (λ) and p (λ) are both monic, it follows this scalar must equal 1. This shows uniqueness.

Corollary 17.3.8 In the above theorem, each of the scalars λk has the property that there exists a nonzero v such that (A – λiI) v = 0. Furthermore the λi are the only scalars with this property.

Proof: For the first claim, just factor out (A – λiI) instead of (A – λmI) . Next suppose

(A – µI) v = 0

for some µ and v ≠ 0. Then

continuing this way yields

a contradiction unless µ = for some λk.

Therefore, these are eigenvectors and eigenvalues with the usual meaning. This leads to the following deﬁnition.

Deﬁnition 17.3.9 For A ∈ L(V,V) where dim (V) = n, the scalars, λk in the minimal polynomial,

are called the eigenvalues of A. In the last expression, λk is a repeated root which occurs rk times. The collection of eigenvalues of A is denoted by σ (A). The generalized eigenspaces are

ker (A−λkI)rk ≡Vk.

Theorem 17.3.10 In the situation of the above deﬁnition,

V = V₁ ⊕ • • • ⊕Vp

That is, the vector space equals the direct sum of its generalized eigenspaces.

Proof: Since V = ker (∏pk=1 (A−λkI)rk) , the conclusion follows from Theorem 17.3.5.

17.4 Block Diagonal Matrices

In this section the vector space will be Cn and the linear transformations will be those which result by multiplication by n × n matrices.

Deﬁnition 17.4.1 Let A and B be two n × n matrices. Then A is similar to B, written as A ∼B when there exists an invertible matrix S such that A = S⁻¹BS.

Theorem 17.4.2 Let A be an n ×n matrix. Letting λ₁, λ₂, • • • , λr be the distinct eigenvalues of A,arranged in some order, there exist square matrices P₁, • • • , Pr such that A is similar to the block diagonal matrix

in which Pk has the single eigenvalue λk. Denoting by rk the size of Pk it follows that rk equals the dimension of the generalized eigenspace for λk. Furthermore, if S is the matrix satisfying

S⁻¹AS = P,

then S is of the form

( B₁ • • • Br )

where Bk = ( uk₁ • • • ukrk ) in which the columns,{uk, • • • , ukrk } = Dk constitute a basis for Vλ

Proof: By Theorem 17.3.9 and Lemma 17.3.3,

Cn = Vλ₁ ⊕ • • • ⊕Vλk

and a basis for Cn is {D₁, • • • , Dr} where Dk is a basis for Vλk, ker (A−λkI)rk.

Let

S = ( B₁  • • •  Br )

where the Bi are the matrices described in the statement of the theorem. Then S⁻¹ must be of the form

where CiBi = Iri×ri. Also, if i ≠ j, then CiABj = 0 the last claim holding because A : Vλj →Vλj so the columns of ABj are linear combinations of the columns of vBj and each of these columns is orthogonal to the rows of Ci since CiBj = 0 if i ≠ j. Therefore,

and CrkABrk is an rk × rk matrix.

What about the eigenvalues of CrkABrk? The only eigenvalue of A restricted to Vλk is λk because if Ax = µx for some x ∈ Vλk and µ ≠ λk, then

contrary to the assumption that x ∈ Vλk. Suppose then that CrkABrkx = λx where x ≠ 0. Why is λ = λk? Let y = Brkx so y ∈ Vλk. Then

and so

Therefore, λ = λk because, as noted above, λk is the only eigenvalue of A restricted to Vλk. Now let Pk = CrkABrk.

The above theorem contains a result which is of suﬃcient importance to state as a corollary.

Corollary 17.4.3 Let A be an n × n matrix and let Dk denote a basis for the generalized eigenspace for λk. Then {D₁, • • • , Dr} is a basis for Cn.

More can be said. Recall Theorem 13.2.11 on Page 262. From this theorem, there exist unitary matrices, Uk such that U∗PkUk = Tk where Tk is an upper triangular matrix of the form

Now let U be the block diagonal matrix deﬁned by

By Theorem 17.4.2 there exists S such that

Therefore,

This proves most of the following corollary of Theorem 17.4.2.

Corollary 17.4.4 Let A be an n × n matrix. Then A is similar to an upper triangular, block diagonal matrix of the form

where Tk is an upper triangular matrix having only λk on the main diagonal. The diagonal blocks can be arranged in any order desired. If Tk is an mk × mk matrix, then

mk = dim (ker (A − λkI)rk

where the minimal polynomial of A is

Furthermore, mk is the multiplicity of λk as a zero of the characteristic polynomial of A.

Proof: The only thing which remains is the assertion that mk equals the multiplicity of λk as a zero of the characteristic polynomial. However, this is clear from the observation that since T is similar to A they have the same characteristic polynomial because

and the observation that since T is upper triangular, the characteristic polynomial of T is of the form

The above corollary has tremendous signiﬁcance especially if it is pushed even further resulting in the Jordan Canonical form. This form involves still more similarity transformations resulting in an especially revealing and simple form for each of the Tk, but the result of the above corollary is suﬃcient for most applications.

It is signiﬁcant because it enables one to obtain great understanding of powers of A by using the matrix T. From Corollary 17.4.4 there exists an n × n matrix S² such that

A = S⁻¹T S.

Therefore, A² = S⁻¹T SS⁻¹T S = S⁻¹T²S and continuing this way, it follows

Ak = S⁻¹TkS.

where T is given in the above corollary. Consider Tk. By block multiplication,

The matrix Ts is an ms × ms matrix which is of the form

which can be written in the form

Ts = D + N

for D a multiple of the identity and N an upper triangular matrix with zeros down the main diagonal. Therefore, by the Cayley Hamilton theorem, Nms = 0 because the characteristic equation for N is just λms = 0. Such a transformation is called nilpotent. You can see Nms = 0 directly also, without having to use the Cayley Hamilton theorem. Now since D is just a multiple of the identity, it follows

that DN = ND. Therefore, the usual binomial theorem may be applied and this yields the following equations for k ≥ ms.

the third equation holding because Nms = 0. Thus Tks is of the form

Lemma 17.4.5 Suppose T is of the form Ts described above in (17.4) where the constant α, on the main diagonal, is less than one in absolute value. Then

Proof: From (17.5), it follows that for large k, and j ≤ ms,

Therefore, letting C be the largest value of |(Nj)| for 0 ≤ j ≤ ms,

which converges to zero as k → ∞. This is most easily seen by applying the ratio test to the series

and then noting that if a series converges, then the kth term converges to zero.

17.5 The Matrix Of A Linear Transformation

If V is an n dimensional vector space and {v₁, • • • , vn} is a basis for V, there exists a linear map

deﬁned as

where

for ei the standard basis vectors for Fn consisting of

where the one is in the ith slot. It is clear that q deﬁned in this way, is one to one, onto, and linear. For v ∈ V, q⁻¹ (v) is a list of scalars called the components of v with respect to the basis {v₁, • • • , vn}.

Deﬁnition 17.5.1 Given a linear transformation L, mapping V to W, where

{v1, • • • , vn}

is a basis of V and {w₁, • • • , wm} is a basis for W, an m × n matrix A = (aij) is called the matrix of the transformation L with respect to the given choice of bases for V and W , if whenever v ∈ V, then multiplication of the components of v by (aij) yields the components of Lv.

The following diagram is descriptive of the deﬁnition. Here qV and qW are the maps deﬁned above with reference to the bases, {v₁, • • • , vn} and {w₁, • • • , wm} respectively.

Letting b ∈ Fn this requires

Now

for some choice of scalars cij because {w₁, • • • , wm} is a basis for W. Hence

It follows from the linear independence of {w₁, • • • , wm} that

for any choice of b ∈ Fn and consequently

aij = cij

where cij is deﬁned by (17.7). It may help to write (17.7) in the form

where C = (cij) , A = (aij) .

Example 17.5.2 Let

V ≡ { polynomials of degree 3 or less},

W ≡ { polynomials of degree 2 or less},

and L ≡ D where D is the differentiation operator. A basis for V is {1, x, x ², x³} and a basis for W is {1, x, x²}.

What is the matrix of this linear transformation with respect to this basis? Using (17.8),

( 0 1 2x 3x² ) = (1 x x2  ) C.

It follows from this that

Now consider the important case where V = Fn, W =Fm, and the basis chosen is the standard basis of vectors ei described above. Let L be a linear transformation from Fn to Fm and let A be the matrix of the transformation with respect to these bases. In this case the coordinate maps qV and qW are simply the identity map and the requirement that A is the matrix of the transformation amounts to

πi (Lb) = πi (Ab)

where πi denotes the map which takes a vector in Fm and returns the ith entry in the vector, the ith component of the vector with respect to the standard basis vectors. Thus, if the components of the vector in Fn with respect to the standard basis are (b₁, • • • , bn) ,

then

What about the situation where different pairs of bases are chosen for V and W ? How are the two matrices with respect to these choices related? Consider the following diagram which illustrates the situation.

In this diagram qi and pi are coordinate maps as described above. From the diagram,

where q⁻¹₁ and p⁻¹₂ p₂ are one to one, onto, and linear maps. Thus the effect of these maps is2 q 1 pidentical to multiplication by a suitable matrix.

Deﬁnition 17.5.3 In the special case where V = W and only one basis is used for V = W, this becomes

Letting S be the matrix of the linear transformation q− 1 ₁ with respect to the standard basis vectors in Fn,

When this occurs, A₁ is said to be similar to A₂ and A →S⁻¹AS is called a similarity transformation.

Here is some terminology.

Deﬁnition 17.5.4 Let S be a set. The symbol, ∼ is called an equivalence relation on S if it satisﬁes the following axioms.

1. x ∼ x  for all x ∈ S. (Reﬂexive)

2. If x ∼ y then y ∼ x. (Symmetric)

3. If x ∼ y and y ∼ z, then x ∼ z. (Transitive)

Deﬁnition 17.5.5 [x] denotes the set of all elements of S which are equivalent to x and [x] is called the equivalence class determined by x or just the equivalence class of x.

With the above deﬁnition one can prove the following simple theorem which you should do if you have not seen it.

Theorem 17.5.6 Let ∼ be an equivalence class deﬁned on a set, S and let H denote the set of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either x ∼ y and [x] = [y] or it is not true that x ∼y and [x] ∩ [y] = ∅.

Theorem 17.5.7 In the vector space of n × n matrices, deﬁne

A ∼ B

if there exists an invertible matrix S such that

A = S⁻¹BS.

Then ∼ is an equivalence relation and A ∼ B if and only if whenever V is an n dimensional vector space, there exists L ∈ L(V,V) and bases {v₁, • • • , vn} and {w₁, • • • , wn} such that A is the matrix of L with respect to {v₁, • • • , vn} and B is the matrix of L with respect to {w₁, • • • , wn}.

Proof: A ∼ A because S = I works in the deﬁnition. If A ∼ B, then B ∼ A, because

A = S⁻¹BS

implies

B = SAS⁻¹.

If A ∼ B and B ∼ C, then

A = S⁻¹BS, B = T⁻¹CT

and so

A = S⁻¹T⁻¹CTS = (T S)⁻¹CTS

which implies A ∼ C. This veriﬁes the first part of the conclusion.

Now let V be an n dimensional vector space, A ∼ B and pick a basis for V,

{v1, • • • , vn}.

Deﬁne L ∈ L(V,V) by

where A = (aij) . Then if B = (bij) , and S = (sij) is the matrix which provides the similarity transformation,

A = S⁻¹BS,

between A and B, it follows that

Now deﬁne

Then from (17.10),

and so

This proves the theorem because the if part of the conclusion was established earlier.

What if the linear transformation consists of multiplication by a matrix A and you want to find the matrix of this linear transformation with respect to another basis? Is there an easy way to do it? The answer is yes.

Proposition 17.5.8 Let A be an m × n matrix and let L be the linear transformation which is deﬁned by

In simple language, to find Lx, you multiply on the left of x by A. Then the matrix M of this linear transformation with respect to the bases {u₁, • • • , un} for Fn and {w₁, • • • , wm} for Fm is given by

M = ( w₁ • • • wm )⁻¹A ( u₁ • • • un )

where ( w₁ ... wm ) is the m × m matrix which has wj as its jth column.

Proof: Consider the following diagram.

Here the coordinate maps are deﬁned in the usual way. Thus

Therefore, qV can be considered the same as multiplication of a vector in Fn on the left by the matrix

( u₁ • • • un ) .

Similar considerations apply to qW. Thus it is desired to have the following for an arbitrary x ∈ Fn.

A ( u₁   ...   un ) x = ( w₁   ...   wn ) Mx

Therefore, the conclusion of the proposition follows.

Deﬁnition 17.5.9 An n × n matrix A, is diagonalizable if there exists an invertible n × n matrix S such that S⁻¹AS = D, where D is a diagonal matrix. Thus D has zero entries everywhere except on the main diagonal. Write diag (λ₁• • • , λn) to denote the diagonal matrix having the λi down the main diagonal.

Which matrices are diagonalizable?

Theorem 17.5.10 Let A be an n × n matrix. Then A is diagonalizable if and only if Fn has a basis of eigenvectors of A. In this case, S of Deﬁnition 17.5.9 consists of the n × n matrix whose columns are the eigenvectors of A and D = diag (λ₁, • • • , λn) .

Proof: Suppose first that Fn has a basis of eigenvectors, {v₁, • • • , vn} where Avi = λivi. Then

let S denote the matrix (v₁ ... vn) and let S⁻¹ ≡  where uTiv j = δij ≡ 

exists because S has rank n. Then from block multiplication,

Next suppose A is diagonalizable so S⁻¹AS = D ≡ diag (λ₁, • • • , λn) . Then the columns of S form a basis because S⁻¹ is given to exist. It only remains to verify that these columns of A are eigenvectors. But letting S = (v₁ • • • vn) , AS = SD and so (Av₁ • • • Avn) = (λ₁v₁ • • • λnvn) which shows that Avi = λivi.

It makes sense to speak of the determinant of a linear transformation as described in the following corollary.

Corollary 17.5.11 Let L ∈ L(V, V) where V is an n dimensional vector space and let A be the matrix of this linear transformation with respect to a basis on V. Then it is possible to deﬁne

det (L) ≡ det (A) .

Proof: Each choice of basis for V determines a matrix for L with respect to the basis. If A and B are two such matrices, it follows from Theorem 17.5.7 that

A = S⁻¹BS

and so

det (A) = det (A) = det (S⁻¹) det (B) det (S)

But

1 = det (I) = det (S⁻¹S) = det (S) det (S⁻¹)

and so

det (A) = det (B)

Deﬁnition 17.5.12 Let A ∈ L(X, Y) where X and Y are finite dimensional vector spaces. Deﬁne rank (A) to equal the dimension of A (X) .

The following theorem explains how the rank of A is related to the rank of the matrix of A.

Theorem 17.5.13 Let A ∈ L(X, Y). Then rank (A) = rank (M) where M is the matrix of A taken with respect to a pair of bases for the vector spaces X, and Y.

Proof: Recall the diagram which describes what is meant by the matrix of A. Here the two bases are as indicated.

Let {z₁ ... zr} be a basis for A (X) . Then since the linear transformation, qY is one to one and onto, {qY⁻¹z₁, ... , qY⁻¹zr} is a linearly independent set of vectors in Fm. Let Aui = zi. Then

and so the dimension of M (Fn) ≥ r. Now if M (Fn) < r then there exists

But then there exists x ∈ Fn with Mx = y. Hence

a contradiction.

The following result is a summary of many concepts.

Theorem 17.5.14 Let L ∈ L(V, V) where V is a finite dimensional vector space. Then the following are equivalent.

1. L is one to one.

2. L maps a basis to a basis.

3. L is onto.

4. det (L)≠ 0

5. If Lv = 0 then v = 0.

Proof: Suppose first L is one to one and let {vi}ni=1 be a basis. Then if ∑ni=1 ciLvi = 0 it follows L (∑ni=1 civi) = 0 which means that since L (0) = 0, and L is one to one, it must be the case that ∑ni=1 civi = 0. Since {vi} is a basis, each ci = 0 which shows {Lvi} is a linearly independent set. Since there are n of these, it must be that this is a basis.

Now suppose 2.). Then letting {vi} be a basis, and y ∈ V, it follows from part 2.) that there are constants, {ci} such that y = ∑ni=1 ci Lvi = L (∑ni=1 civi) . Thus L is onto. It has been shown that 2.) implies 3.).

Now suppose 3.). Then the operation consisting of multiplication by the matrix of L, ML, must be onto. However, the vectors in Fn so obtained, consist of linear combinations of the columns of ML. Therefore, the column rank of ML is n. By Theorem 8.5.7 this equals the determinant rank and so det (ML) ≡ det (L)≠ 0.

Now assume 4.) If Lv = 0 for some v ≠ 0, it follows that MLx = 0 for some x ≠ 0. Therefore, the columns of ML are linearly dependent and so by Theorem 8.5.7, det (ML) = det (L) = 0 contrary to 4.). Therefore, 4.) implies 5.).

Now suppose 5.) and suppose Lv = Lw. Then L (v − w) = 0 and so by 5.), v − w = 0 showing that L is one to one.

Also it is important to note that composition of linear transformation corresponds to multiplication of the matrices. Consider the following diagram.

where A and B are two linear transformations, A ∈ L(X, Y) and B ∈ L(Y, Z) . Then B ◦ A ∈ L(X, Z) and so it has a matrix with respect to bases given on X and Z, the coordinate maps for these bases being qX and qZ respectively. Then

But this shows that MBMA plays the role of MB◦A, the matrix of B◦A. Hence the matrix of B◦A equals the product of the two matrices MA and MB. Of course it is interesting to note that although MB◦A must be unique, the matrices, MB and MA are not unique, depending on the basis chosen for Y .

Theorem 17.5.15 The matrix of the composition of linear transformations equals the product of the matrices of these linear transformations.

17.5.1 Some Geometrically Deﬁned Linear Transformations

This is a review of earlier material. If T is any linear transformation which maps Fn to Fm, there is always an m × n matrix A with the property that

| Ax = T x | (17.11)

for all x ∈ Fn. How does this relate to what is discussed above? In terms of the above diagram,

where

Thus those two maps are really just the identity map. Thus, to find the matrix of the linear transformation T with respect to the standard basis vectors,

T ek = Mek

In other words, the kth column of M equals Tek as noted earlier. All the earlier considerations apply. These considerations were just a specialization to the case of the standard basis vectors of this more general notion which was just presented.

17.5.2 Rotations About A Given Vector

As an application, I will consider the problem of rotating counter clockwise about a given unit vector which is possibly not one of the unit vectors in coordinate directions. First consider a pair of perpendicular unit vectors, u₁ and u₂ and the problem of rotating in the counterclockwise direction about u₃ where u₃ = u₁ × u₂ so that u₁, u₂, u₃ forms a right handed orthogonal coordinate system. Thus the vector u₃ is coming out of the page.

Let T denote the desired rotation. Then

T (au₁ + bu₂ + cu₃) = aT u₁ + bT u₂ + cT u₃

= (a cos θ − b sin θ) u₁ + (a sin θ + b cos θ) u₂ + cu₃.

Thus in terms of the basis {u₁, u₂, u₃}, the matrix of this transformation is

I want to write this transformation in terms of the usual basis vectors, {e₁, e₂, e₃}. From Proposition 17.5.8, if A is this matrix,

and so you can solve for A if you know the ui.

Suppose the unit vector about which the counterclockwise rotation takes place is (a, b, c). Then I obtain vectors, u₁ and u₂ such that {u₁, u₂, u₃} is a right handed orthogonal system with u₃ = (a, b, c) and then use the above result. It is of course somewhat arbitrary how this is accomplished. I will assume, however that |c| ̸ = 1 since otherwise you are looking at either clockwise or counterclockwise rotation about the positive z axis and this is a problem which has been dealt with earlier. (If c = −1, it amounts to clockwise rotation about the positive z axis while if c = 1, it is counterclockwise rotation about the positive z axis.) Then let u3= (a, b, c) and This one is perpendicular to u₃. If {u₁, u₂, u₃} is to be a right hand system it is necessary to have

Now recall that u₃ is a unit vector and so the above equals

=

Then from the above, A is given by

Of course the matrix is an orthogonal matrix so it is easy to take the inverse by simply taking the transpose. Then doing the computation and then some simpliﬁcation yields

| [picture] | (17.12)

With this, it is clear how to rotate clockwise about the unit vector (a, b, c) . Just rotate counter clockwise through an angle of −θ. Thus the matrix for this clockwise rotation is just

In deriving (17.12) it was assumed that c ≠ ±1 but even in this case, it gives the correct answer. Suppose for example that c = 1 so you are rotating in the counter clockwise direction about the positive z axis. Then a, b are both equal to zero and (17.12) reduces to the correct matrix for rotation about the positive z axis.

17.6 Exercises

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of π/3.

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of π/4.

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of π/12. Hint: Note that π/12 = π/3 − π/4.

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of 2π/3 and then reﬂects across the x axis.

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of π/3 and then reﬂects across the y axis.

Find the matrix with respect to the standard basis vectors for the linear transformation which rotates every vector in R² through an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4.

Let V be an inner product space and u ≠ 0. Show that the function Tu deﬁned by Tu (v) ≡ v − proju (v) is also a linear transformation. Here

Now show directly from the axioms of the inner product that

(Tuv, u) = 0

Let V be a finite dimensional inner product space, the field of scalars equal to either R or C. Verify that f given by fv ≡ (v, z) is in L(V,F). Next suppose f is an arbitrary element of L(V,F). Show the following.

(a) If f = 0, the zero mapping, then fv = (v, 0) for all v ∈ V .

(b) If f ≠ 0 then there exists z ≠ 0 satisfying (u, z) = 0 for all u ∈ ker (f) .

(c) Explain why f (y) z − f (z) y ∈ ker (f).

(d) Use part b. to show that there exists w such that f (y) = (y, w) for all y ∈ V .

(e) Show there is at most one such w.

You have now proved the Riesz representation theorem which states that every f ∈ L(V,F) is of the form

f (y) = (y, w)

for a unique w ∈ V.

↑Let A ∈ L(V, W ) where V, W are two finite dimensional inner product spaces, both having field of scalars equal to F which is either R or C. Let f ∈ L(V,F) be given by

f (y) ≡ (Ay, z)

where 〈 〉 now refers to the inner product in W. Use the above problem to verify that there exists a unique w ∈ V such that f (y) = (y, w), the inner product here being the one on V . Let A∗z ≡ w. Show that A∗ ∈ L(W, V ) and by construction,

(Ay, z) = (y,A*b)

In the case that V = Fn and W = Fm and A consists of multiplication on the left by an m × n matrix, give a description of A∗.

Let A be the linear transformation deﬁned on the vector space of smooth functions (Those which have all derivatives) given by Af = D² + 2D + 1. Find ker (A). Hint: First solve (D + 1) z = 0. Then solve (D + 1) y = z.

Let A be the linear transformation deﬁned on the vector space of smooth functions (Those which have all derivatives) given by Af = D²+ 5D + 4. Find ker (A). Note that you could first find ker (D + 4) where D is the differentiation operator and then consider ker (D + 1) (D + 4) = ker (A) and consider Sylvester’s theorem.

Suppose Ax = b has a solution where A is a linear transformation. Explain why the solution is unique precisely when Ax = 0 has only the trivial (zero) solution.

Verify the linear transformation determined by the matrix

maps R³ onto R² but the linear transformation determined by this matrix is not one to one.

Let L be the linear transformation taking polynomials of degree at most three to polynomials of degree at most three given by

D² + 2D + 1

where D is the differentiation operator. Find the matrix of this linear transformation relative to the basis {1, x, x², x3}. Find the matrix directly and then find the matrix with respect to the differential operator D + 1 and multiply this matrix by itself. You should get the same thing. Why?

Let L be the linear transformation taking polynomials of degree at most three to polynomials of degree at most three given by

D² + 5D + 4

where D is the differentiation operator. Find the matrix of this linear transformation relative to the bases {1, x, x², x3}. Find the matrix directly and then find the matrices with respect to the differential operators D + 1, D + 4 and multiply these two matrices. You should get the same thing. Why?

Show that if L ∈ L(V, W) (linear transformation) where V and W are vector spaces, then if Lyp = f for some yp ∈ V, then the general solution of Ly = f is of the form

ker (L) + yp.

Let L ∈ L(V, W ) where V, W are vector spaces, finite or inﬁnite dimensional, and deﬁne x ∼ y if x − y ∈ ker (L) . Show that ∼ is an equivalence relation. Next deﬁne addition and scalar multiplication on the space of equivalence classes as follows.

[x] + [y] ≡ [x + y]

  α [x] = [αx]

Show that these are well deﬁned deﬁnitions and that the set of equivalence classes is a vector space with respect to these operations. The zero is [ker L] . Denote the resulting vector space by V/ ker (L) . Now suppose L is onto W. Deﬁne a mapping A : V/ ker (K) →W as follows.

A [x] ≡Lx

Show that A is well deﬁned, one to one and onto.

If V is a finite dimensional vector space and L ∈ L(V, V ) , show that the minimal polynomial for L equals the minimal polynomial of A where A is the n × n matrix of L with respect to some basis.

Let A be an n × n matrix. Describe a fairly simple method based on row operations for computing the minimal polynomial of A. Recall, that this is a monic polynomial p (λ) such that p (A) = 0 and it has smallest degree of all such monic polynomials. Hint: Consider I, A²,• • •. Regard each as a vector in Fn2 and consider taking the row reduced echelon form or something like this. You might also use the Cayley Hamilton theorem to note that you can stop the above sequence at An.

Let A be an n × n matrix which is non defective. That is, there exists a basis of eigenvectors. Show that if p (λ) is the minimal polynomial, then p (λ) has no repeated roots. Hint: First show that the minimal polynomial of A is the same as the minimal polynomial of the diagonal matrix

Where D (λ) is a diagonal matrix having λ down the main diagonal and in the above, the λi are distinct. Show that the minimal polynomial is ∏ri =1 (λ−λi) .

Show that if A is an n × n matrix and the minimal polynomial has no repeated roots, then A is non defective and there exists a basis of eigenvectors. Thus, from the above problem, a matrix may be diagonalized if and only if its minimal polynomial has no repeated roots. (It turns out this condition is something which is relatively easy to determine. You look at the polynomial and its derivative and ask whether these are relatively prime. The answer to this question can be determined using routine algorithms as discussed above in the section on polynomials and fields. Thus it is possible to determine whether an n × n matrix is defective.) Hint: You might want to use Theorem 17.3.1.

² The S here is written as S⁻¹ in the corollary.

The Jordan Canonical Form*

Recall Corollary 17.4.4. For convenience, this corollary is stated below.

Corollary A.0.1 Let A be an n × n matrix. Then A is similar to an upper triangular, block diagonal matrix of the form

where Tk is an upper triangular matrix having only λk on the main diagonal. The diagonal blocks can be arranged in any order desired. If Tk is an mk × mk matrix, then

mk = dim ker (A−λkI)rk.

where the minimal polynomial of A is

The Jordan Canonical form involves a further reduction in which the upper triangular matrices, Tk assume a particularly revealing and simple form.

Deﬁnition A.0.2 Jk (α) is a Jordan block if it is a k × k matrix of the form

In words, there is an unbroken string of ones down the super diagonal and the number, α filling every space on the main diagonal with zeros everywhere else. A matrix is strictly upper triangular if it is of the form

where there are zeroes on the main diagonal and below the main diagonal.

The Jordan canonical form involves each of the upper triangular matrices in the conclusion of Corollary 17.4.4 being a block diagonal matrix with the blocks being Jordan blocks in which the size of the blocks decreases from the upper left to the lower right. The idea is to show that every square matrix is similar to a unique such matrix which is in Jordan canonical form. It is assumed here that the field of scalars is C but everything which will be done below works just fine if the minimal polynomial can be completely factored into linear factors in the field of scalars.

Note that in the conclusion of Corollary 17.4.4 each of the triangular matrices is of the form αI + N where N is a strictly upper triangular matrix. The existence of the Jordan canonical form follows quickly from the following lemma.

Lemma A.0.3 Let N be an n × n matrix which is strictly upper triangular. Then there exists an invertible matrix S such that

where and

Proof: First note the only eigenvalue of N is 0. Let v₁ be an eigenvector. Then {v₁, v₂,· · ·, vr} is called a chain if Nvk+1 = vk for all k = 1, 2,· · ·,r and v₁ is an eigenvector so Nv₁ = 0. It will be called a maximal chain if there is no solution v, to the equation, Nv = vr.

Claim 1: The vectors in any chain are linearly independent and for

{v₁, v₂,· · ·, vr}

a chain based on v₁,

| N : span (v₁, v₂,· · ·, vr) span (v₁, v₂,· · ·, vr) . | (1.1)

Also if {v₁, v₂,· · ·, vr} is a chain, then r ≤ n.

Proof:First note that (1.1) is obvious because

It only remains to verify the vectors of a chain are independent. Suppose then

Do Nr−1 to it to conclude cr = 0. Next do Nr−2 to it to conclude cr−1 = 0 and continue this way. Now it is obvious r ≤n because the chain is independent. This proves the claim.

Consider the set of all chains based on eigenvectors. Since all have total length no larger than n it follows there exists one which has maximal length, If span (B₁) contains all eigenvectors of N, then stop. Otherwise, consider all chains based on eigenvectors not in span (B₁) and pick one, which is as long as possible. Thus r₂ ≤ r₁. If span (B₁, B₂) contains all eigenvectors of N, stop. Otherwise, consider all chains based on eigenvectors not in span (B₁, B₂) and pick one, such that is as large as possible. Continue this way. Thus rk ≥ rk + 1.

Claim 2: The above process terminates with a finite list of chains

{B₁,· · ·, Bs}

because for any k, {B₁,· · ·, Bk} is linearly independent.

Proof of Claim 2: The claim is true if k = 1. This follows from Claim 1. Suppose it is true for k − 1, k ≥ 2. Then {B₁,· · ·, Bk−1} is linearly independent. Suppose

where the wq come from {B₁,· · ·, Bk−1, Bk}. By induction, some of these wq must come from Bk. Let be the one for which i is as large as possible. Then do Ni−1 to both sides to obtain eigenvector upon which the chain Bk is based, is a linear combination of {B₁,· · ·, Bk−1} contrary to the construction. Since {B₁,· · ·, Bk} is linearly independent, the process terminates. This proves the claim.

Claim 3: Suppose Nw = 0. (w is an eigenvector) Then there exist scalars, ci such that

Recall that is the eigenvector in the ith chain on which this chain is based.

Proof of Claim 3: From the construction, w ∈ span (B₁,· · ·, Bs) since otherwise, it could serve as a base for another chain. Therefore,

Now apply N to both sides.

and so by Claim 2, = 0 if k ≥ 2. Therefore,

and this proves the claim.

It remains to verify that span (B₁,· · ·, Bs) = Fn. Suppose w ∉ span (B₁,· · ·, Bs) . By Claim 3 this implies w is not an eigenvector since all the eigenvectors are in span (B₁,· · ·, Bs) . Since Nn = 0, there exists a smallest integer, k ≥ 2 such that Nkw = 0 but Nk−1w ≠ 0. Then k ≤ min (r₁,· · ·, rs) because there exists a chain of length k based on the eigenvector Nk−1w, namely

Nk−1w, Nk−2w,Nk−3w,· · ·, w

and this chain must be no longer than the preceding chains because of the construction in which a longest possible chain was chosen at each step. Since Nk−1w is an eigenvector, it follows from Claim 3 that

Therefore,

and so,

which implies is an eigenvector and so by Claim 3 there exist di such that

and so

Continuing this way it follows that for each j < k, there exists a vector

zj ∈ span (B₁,· · ·, Bs)

such that

Nk−j (w−zj) = 0.

In particular, taking j = (k − 1) yields

N (w−zk−1) = 0

and now using Claim 3 again yields w ∈ span (B₁,· · ·, Bs), a contradiction. Therefore,

span (B₁,· · ·, Bs) = Fn

after all and so {B₁,· · ·, Bs} is a basis for Fn. 

Now consider the block matrix

S = ( B₁ · · · Bs )

where

Thus

where CiBi = Iri × ri and CiN Bj = 0 if i ≠ j. Let

Then

which equals an rk × rk matrix of the form

That is, it has ones down the super diagonal and zeros everywhere else. It follows

as claimed.

Now let the upper triangular matrices, Tk be given in the conclusion of Corollary 17.4.4. Thus, as noted earlier,

Tk = λkIrk × rk + Nk

where Nk is a strictly upper triangular matrix of the sort just discussed in Lemma A.0.3. Therefore, there exists Sk such that is of the form given in Lemma A.0.3. Now = and so is of the form

where i₁ ≥ i₂ ≥ · · · ≥ is and This proves the following corollary.

Corollary A.0.4 Suppose A is an upper triangular n × n matrix having α in every position on the main diagonal. Then there exists an invertible matrix S such that

where k₁ ≥ k₂ ≥ · · · ≥ kr ≥ 1 and

The next theorem is gives the existence of the Jordan canonical form.

Theorem A.0.5 Let A be an n × n matrix having eigenvalues λ₁,· · ·, λr where the multiplicity of λi as a zero of the characteristic polynomial equals mi. Then there exists an invertible matrix S such that

| [picture] | (1.2)

where J (λk) is an mk × mk matrix of the form

| [picture] | (1.3)

where k₁ ≥ k₂ ≥ · · · ≥ kr ≥ 1 and

Proof: From Corollary 17.4.4, there exists S such that S⁻¹AS is of the form

where Tk is an upper triangular mk × mk matrix having only λk on the main diagonal. By Corollary A.0.4 There exist matrices, Sk such that where J (λk) is described in (1.3). Now let M be the block diagonal matrix given by

It follows that M⁻¹S⁻¹ASM = M⁻¹T M and this is of the desired form.

What about the uniqueness of the Jordan canonical form? Obviously if you change the order of the eigenvalues, you get a different Jordan canonical form but it turns out that if the order of the eigenvalues is the same, then the Jordan canonical form is unique. In fact, it is the same for any two similar matrices.

Theorem A.0.6 Let A and B be two similar matrices. Let JA and JB be Jordan forms of A and B respectively, made up of the blocks JA (λi) and JB (λi) respectively. Then JA and JB are identical except possibly for the order of the J (λi) where the λi are deﬁned above.

Proof: First note that for λi an eigenvalue, the matrices JA (λi) and JB (λi) are both of size mi × mi because the two matrices A and B, being similar, have exactly the same characteristic equation and the size of a block equals the algebraic multiplicity of the eigenvalue as a zero of the characteristic equation. It is only necessary to worry about the number and size of the Jordan blocks making up JA (λi) and JB (λi) . Let the eigenvalues of A and B be {λ₁,· · ·, λr}. Consider the two sequences of numbers {rank (A − λI)m} and {rank (B − λI)m}. Since A and B are similar, these two sequences coincide. (Why?) Also, for the same reason, {rank (JA − λI)m} coincides with {rank (JB − λI)m}. Now pick λk an eigenvalue and consider {rank (JA − λkI)m} and {rank (JB − λkI)m}. Then

and a similar formula holds for JB − λkI. Here

and

and it suﬃces to verify that li = ki for all i. As noted above,∑ki = ∑li. Now from the above formulas,

which shows rank (J A(0)m) = rank (JB (0)m) for all m. However,

with a similar formula holding for JA (0)mand rank (JB (0)m) = rank (Jli (0)m) , similar for rank (JA (0)m) . In going from m to m + 1,

rank (Jli(0)m) − 1 = rank Jli (0)m+1)

until m = li at which time there is no further change. Therefore, p = r since otherwise, there would exist a discrepancy right away in going from m = 1 to m = 2. Now suppose the sequence {li} is not equal to the sequence, {ki}. Then lr−b ≠ kr − b for some b a nonnegative integer taken to be a small as possible. Say lr−b > kr−b. Then, letting m = kr−b,

and in going to m + 1 a discrepancy must occur because the sum on the right will contribute less to the decrease in rank than the sum on the left.

The Fundamental Theorem Of Algebra

The fundamental theorem of algebra states that every non constant polynomial having coeﬃcients in C has a zero in C. If C is replaced by R, this is not true because of the example, x² + 1 = 0. This theorem is a very remarkable result and notwithstanding its title, all the best proofs of it depend on either analysis or topology. It was first mostly proved by Gauss in 1797. The first complete proof was given by Argand in 1806. The proof given here follows Rudin [13]. See also Hardy [7] for a similar proof, more discussion and references. The best proof is found in the theory of complex analysis. Recall De Moivre’s theorem from trigonometry which is listed here for convenience.

Theorem B.0.7 Let r > 0 be given. Then if n is a positive integer,

[r (cos t + i sin t)]n = rn (cos nt + i sin nt) .

Recall that this theorem is the basis for proving the following corollary from trigonometry, also listed here for convenience.

Corollary B.0.8 Let z be a non zero complex number and let k be a positive integer. Then there are always exactly k kth roots of z in C.

Lemma B.0.9 Let ak ∈ C for k = 1,· · ·, n and let  Then p is continuous.

Proof:

|azn − awn| ≤ |a| |z − w| zn⁻¹ + zn⁻²w + · · · + wn⁻¹ .

Then for |z − w| < 1, the triangle inequality implies |w|< 1 +|z| and so if |z − w|< 1,

|azn − awn| ≤ |a| |z − w|n (1 + |z|)

If ε > 0 is given, let

It follows from the above inequality that for |z − w| < δ, |azn − awn| < ε. The function of the lemma is just the sum of functions of this sort and so it follows that it is also continuous.

Theorem B.0.10 (Fundamental theorem of Algebra) Let p (z) be a nonconstant polynomial. Then there exists z ∈ C such that p (z) = 0.

Proof: Suppose not. Then

where an ≠ = 0, n >  0.  Then

and so

| [picture] | (2.1)

Now let

λ ≡ inf {|p (z)| : z ∈ C} .

By (2.1), there exists an R > 0 such that if |z| > R, it follows that |p (z)| > λ + 1. Therefore,

λ ≡ inf {|p (z)| : z ∈ C} = inf {|p (z)| : |z| ≤ R} .

The set {z : |z| ≤ R} is a closed and bounded set and so this infimum is achieved at some point w with |w| ≤ R. A contradiction is obtained if |p (w)| = 0 so assume |p (w)| > 0. Then consider

It follows q (z) is of the form

q (z) = 1 + ckzk + · · · + cnzn

where ck ≠ 0, because q (0) = 1. It is also true that |q (z)| ≥ 1 by the assumption that |p (w)| is the smallest value of |p (z)| . Now let θ ∈ C be a complex number with |θ| = 1 and

ckwk = −|w|k|ck| .

If

and if w = 0,  = 1 will work. Now let ηk =  and let t be a small positive number.

q (tηw) ≡ 1 − tk |w|k |ck| + · · · + cntn (ηw)n

which is of the form

1 − tk |w|k |ck| + tk (g (t,w))

where limt→ 0 g (t,w) = 0. Letting t be small enough,

|g (t,w)| < |w|k |ck| /2

and so for such t,

|q (tηw)| < 1 − tk |w|k |ck| + tk |w|k |ck| /2 < 1,

a contradiction to |q (z)| ≥ 1.

Bibliography

[1] Apostol T. Calculus Volume II Second edition, Wiley 1969.

[2] Baker, Roger, Linear Algebra, Rinton Press 2001.

[3] Davis H. and Snider A., Vector Analysis Wm. C. Brown 1995.

[4] Edwards C.H. Advanced Calculus of several Variables, Dover 1994.

[5] Golub, G. and Van Loan, C., Matrix Computations, Johns Hopkins University Press, 1996.

[6] Gurtin M. An introduction to continuum mechanics, Academic press 1981.

[7] Hardy G. A Course Of Pure Mathematics, Tenth edition, Cambridge University Press 1992.

[8] Horn R. and Johnson C. matrix Analysis, Cambridge University Press, 1985.

[9] Jacobsen N. Basic Algebra Freeman 1974.

[10] Karlin S. and Taylor H. A First Course in Stochastic Processes, Academic Press, 1975.

[11] Kuttler K. Linear Algebra On web page. Linear Algebra

[12] Nobel B. and Daniel J. Applied Linear Algebra, Prentice Hall, 1977.

[13] Rudin W. Principles of Mathematical Analysis, McGraw Hill, 1976.

[14] Salas S. and Hille E., Calculus One and Several Variables, Wiley 1990.

[15] Strang Gilbert, Linear Algebra and its Applications, Harcourt Brace Jovanovich 1980.

[16] Wilkinson, J.H., The Algebraic Eigenvalue Problem, Clarendon Press Oxford 1965.

[17] Yosida K., Functional Analysis, Springer Verlag, 1978.

Answers To Selected Exercises

C.1 Exercises 21

1 Let z = 5 + i9. Find z⁻¹.
(5 + i9)⁻¹ =

2 Let z = 2+i7 and let w = 3−i8. Find zw, z+w,z², and w/z.
62 + 5i, 5−i,−45 + 28i, and

4 Graph the complex cube roots of 8 in the complex plane. Do the same for the four fourth roots of 16. The cube roots are the solutions to z³ + 8 = 0, Solution is:
The fourth roots are the solutions to z⁴ + 16 = 0, Solution is:

When you graph these, you will have three equally spaced points on the circle of radius 2 for the cube roots and you will have four equally spaced points on the circle of radius 2 for the fourth roots. Here are pictures which should result.

5 If z is a complex number, show there exists ω a complex number with |ω| = 1 and ωz = |z|.
If z = 0, let ω = 1. If z ≠ 0, let ω =

7 You already know formulas for cos (x + y) and sin (x + y) these were used to prove De Moivre’s theorem.
Now using De Moivre’s theorem, derive a formula for sin (5x) and one for cos (5x).
sin (5x) = 5 cos⁴ x sin x − 10 cos² x sin³ x + sin⁵ x cos (5x) = cos⁵ x − 10 cos³ x sin² x + 5 cos x sin⁴ x

9 Factor x³ + 8 as a product of linear factors.
x³ + 8 = 0, Solution is: and so this polynomial equals

10 Write x³+27 in the form (x + 3) (x² + ax + b) where x² + ax + b cannot be factored any more using only real numbers.
x³ + 27 = (x + 3) (x² − 3x + 9)

12 Factor x⁴+16 as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers.
. 

You can use the information in the preceding problem. Note that has real coefficients.

13 If z, w are complex numbers prove and then show by induction that Also verify that In words this says the conjugate of a product equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates.

which is the same thing. 

Thus it holds for a product of two complex numbers. Now suppose you have that it is true for the product of n complex numbers. Then and now, by induction this equals

As to sums, this is even easier.

14 Suppose p (x) = a nxn + an−1xn−1 +· · · + a₁x + a₀ where all the ak are real numbers. Suppose also that p (z) = 0 for some z ∈ C. Show it follows that p () = 0 also.
You just use the above problem. If p (z) = 0, then you have

15 Show that 1 + i, 2 + i are the only two zeros to

p (x) = x₂ − (3 + 2i) x + (1 + 3i)

so the zeros do not necessarily come in conjugate pairs if the coefficients are not real.

(x − (1 + i)) (x − (2 + i)) = x₂ − (3 + 2i) x + 1 + 3i

16 I claim that 1 = −1. Here is why.

This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?
Something is wrong. There is no single

17 De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents, not just integers. 1 = 1(1/4) = (cos 2π + i sin 2π)1/4 = cos (π/2) + i sin (π/2) = i. Therefore, squaring both sides it follows 1 = −1 as in the previous problem. What does this tell you about De Moivre’s theorem? Is there a profound difference between raising numbers to integer powers and raising numbers to non integer powers?

It doesn’t work. This is because there are four fourth roots of 1.

18 Here is another question: If n is an integer, is it always true that (cos − i sin )n = cos (n)−i sin (n)? Explain.

Yes, this is true.

(cos − i sin )n = (cos (−) + i sin (−))n
= cos (−n) + i sin (−n)
= cos (n) − i sin (n)

19 Suppose you have any polynomial in cos and sin . By this I mean an expression of the form where aαβ ∈ C. Can this always be written in the form ? Explain. Yes it can. It follows from the identities for the sine and cosine of the sum and difference of angles that

Now cos = 1 cos + 0 sin and sin = 0 cos + 1 sin . Suppose that whenever k ≤ n,

for some numbers aj , bj . Then

Now use the above identities to write all products as sums of sines and cosines of (j − 1) , j, (j + 1) . Then adjusting the constants, it follows

You can do something similar with sinn () and with products of the form

cosα sinβ .

20 Suppose p (x) = anxn + an−1xn−1 +· · · + a₁x + a₀ is a polynomial and it has n zeros,

z₁, z₂,· · ·, zn

listed according to multiplicity. (z is a root of multiplicity m if the polynomial f (x) = (x − z)m divides p (x) but (x − z) f (x) does not.) Show that

p (x) = an (x − z₁) (x − z₂) · · · (x − zn) .

p (x) = (x − z₁) q (x)+r (x) where r (x) is a nonzero constant or equal to 0. However, r (z₁) = 0 and so r (x) = 0. Now do to q (x) what was done to p (x) and continue until the degree of the resulting q (x) equals 0. Then you have the above factorization.

21 Give the solutions to the following quadratic equations having real coefficients.

(a) x² − 2x + 2 = 0, Solution is: 1 + i, 1 − i
(b) 3x²+x+3 = 0, Solution is:
(c) x² − 6x + 13 = 0, Solution is: 3 + 2i, 3 − 2i
(d) x²+4x+9 = 0, Solution is:
(e) 4x² + 4x + 5 = 0, Solution is:

22 Give the solutions to the following quadratic equations having complex coefficients. Note how the solutions do not come in conjugate pairs as they do when the equation has real coefficients.

(a) x² + 2x + 1 + i = 0, Solution is : x = -1 +

(b) 4x²+4ix−5 = 0, Solution is :

(c) 4x² +(4 + 4i) x+1+2i = 0, Solution is :

(d) x²−4ix−5 = 0, Solution is : x = −1+2i, x = 1 + 2i

(e) 3x² + (1 − i) x + 3i = 0, Solution is :

23 Prove the fundamental theorem of algebra for quadratic polynomials having coefficients in C.

This is pretty easy because you can simply write the quadratic formula. Finding the square roots of complex numbers is easy from the above presentation. Hence, every quadratic polynomial has two roots in C. Note that the two square roots in the quadratic formula are on opposite sides of the unit circle so one is −1 times the other.

C.2 Exercises 36

= 9.56 degrees

Will need 68. 966 gallons of gas. Therefore, it will not make it.

It takes 2 hours

miles. He ends up 1/3 mile down stream.

miles. He ends up 1/3 mile down stream.

(−3, 2,−5) .

(3, 0, 0).

C.3 Exercises 47

1 This formula says that
u·v = |u| |v| where is the included angle between the two vectors. Thus 

and equality holds if and only if = 0 or π. This means that the two vectors either point in the same direction or opposite directions. Hence one is a multiple of the other.

3 −0.197 39 = cos , Thus = 1. 769 5 radians.

4 −0.444 44 = cos , = 2. 031 3 radians.

5

7

8 It makes no sense.

9

11

13 20 300 = 4242. 6

15 (−4, 3,−4)·(0, 1, 0) × 10 = 30

17

19 (1, 2, 3, 4)· (2, 0, 1, 3) = 17

21

C.4 Exercises 56

1 If a 0, then the condition says that |a × u|=|a| sin = 0 for all angles . Hence a = 0 after all.

3

5

7 113

9 It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.

12 a × b × c is meaningless.

13

14

15 [v, w, z] [u, v, w]

16 0

18 u·u = c, Therefore, u′ · u + u · u′ = 0 so u′ · u = 0.

C.5 Exercises 72

1

2

5

6 No solution exists.

8 It appears that the solution exists but is not unique.

9 It appears that there is a unique solution.

11 There might be a solution. If so, there are inﬁnitely many.

12 No. Consider x + y + z = 2 and x + y + z = 1.

13 These can have a solution. For example, x + y = 1, 2x + 2y = 2, 3x + 3y = 3 even has an inﬁnite set of solutions.

14 h = 4

15 Any h will work.

16 Any h will work.

17 If h 2 there will be a unique solution for any k. If h = 2 and k 4, there are no solutions. If h = 2 and k = 4, then there are infinitely many solutions.

19 There is no solution.

20

22

23 z = t, y = 4t, x = 2− 4t.

24

25 x5 = t, x3 = s. Then the other variables are given

26 [x = 1− 2t, z = 1, y = t]

27 [x = 2− 4t, y = −8t, z = t]

25 [x = −1, y = 2, z = −1]

29 [x = 2, y = 4, z = 5]

30 [x = 1, y = 2, z = −5]

31 [x = −1, y = −5, z = 4]

32 [x = 2t + 1, y = 4t, z = t]

33 [x = 1, y = 5, z = 3]

34 [x = 4, y = −4, z = −2]

35 These are not legitimate row operations. They do not preserve the solution set of the system.

36 {g = 60, I = 90, b = 200, s = 50}

37 [w = 15, x = 15, y = 20, z = 10] .

C.6 Exercises 92

1.

Not possible, , Not possible, Not

possible.

2.

Not possible,

4 , Not possible, ( −14 13 2 ), Not possible,

7

8

9 [k = 4]

10 There is no possible choice of k which will make these matrices commute.

11 To get −A, just replace every entry of A with its additive inverse. The 0 matrix is the one which has all zeros in it.

12

13 If A = −AT, then aii = −aii and so each aii = 0. 

14

= iω₂u₃−iω₃u₂−jω₁u₃ + jω₃u₁ + kω₁u₂−kω₂u₁.

In terms of matrices, this is

The matrix is of the form

for suitable ωi since it is

skew symmetric. Now you note that multiplication on the left by this matrix gives the same thing as the cross product.

15

16 0′ = 0′ + 0 = 0.

17 0A = (0 + 0) A = 0A+0A. Now add −(0A) to both sides. Then 0 = 0A.

18 A + (−1) A = (1 + (−1)) A = 0A = 0. Therefore, from the uniqueness of the additive inverse, it follows that −A = (−1) A.

19

20

24 Explain why if AB = AC and A⁻¹ exists, then B = C. 

Because you can multiply on the left by A⁻¹.

28

29

30

31 does not exist. The row reduced echelon form of this

matrix

is

32

33

34

35 row echelon form:

There is no inverse

36

37  

38 Write in the form A

where A is an appropriate matrix.

39

42

43 Multiply on the left of both sides by A−1.

44 Multiply on both sides on the left by A⁻¹. Thus 0 = A⁻¹0 =

A⁻¹ (Ax) =(A⁻¹A)x = Ix = x

45 A⁻¹ = A⁻¹I = A⁻¹ (AB)

=(A⁻¹A)B = IB = B.

46 You just need to show that (A⁻¹)T acts like the inverse of AT because from uniqueness in the above problem, this will imply it is the inverse. But from properties of the transpose,

Hence (A⁻¹)T =(AT)⁻¹ and this last matrix exists.

47

51

C.7 Exercises 111

2 6

3 2

4 6

5 −4

6 −6

7 −32

8 63

9 211

11 It does not change the determinant. This was just taking the transpose.

13 The determinant is unchanged. It was just the first row added to the second.

15 In this case the two columns were switched so the determinant of the second is −1 times the determinant of the first.

17 det (aA) = det (aIA) = det (aI) det (A) = an det (A) . The matrix which has a down the main diagonal has determinant equal to an.

19 This is not true at all. Consider

20 It must be 0 because 0 = det (0) = det (Ak) = (det (A))k.

21 det (A) = 1, or −1.

23 If A = S⁻¹BS, then SAS⁻¹ = B and so if A ∼ B, then B ∼ A. It is obvious that A ∼ A because you can let S = I. Say A ∼ B and B ∼ C. Then A = P⁻¹BP and B = Q⁻¹CQ. Therefore, 

and so

25 Say M = S⁻¹N S. Then

31

33

38 Show that if det (A) ≠ 0 for A an n × n matrix, it follows that if Ax = 0, then x = 0.

If det (A) ≠ 0, then A⁻¹ exists and so you could multiply on both sides on the left by A⁻¹ and obtain that x = 0.

39 Suppose A, B are n × n matrices and that A B = I. Show that then BA = I. Hint: You might do something like this: First explain why det (A) , det (B) are both nonzero. Then (AB) A = A and then show BA (BA − I) = 0. From this use what is given to conclude A (BA − I) = 0. Then use Problem 38.

You have 1 = det (A) det (B). Hence both A and B have inverses. Letting x be given,

and so it follows from the above problem that

(BA − I) x = 0.

Since x is arbitrary, it follows that BA = I.

40

41

C.8 Exercises 158

4 a. is not, b. is, and so is c.

5

7 It is because you cannot have more than min (m, n) nonzero rows in the row reduced echelon form. Recall that the number of pivot columns is the same as the number of nonzero rows from the description of this row reduced echelon form.

8 is a basis

9 is a basis

11 Yes. This is a subspace. It is closed with respect to vector addition and scalar multiplication.

13 Yes, this is a subspace.

15 This is a subspace.

17 Not a subspace.

19

21 If AB = I, then B must be one to one. Otherwise there exists x ≠ 0 such that Bx = 0. But then you would have

x = I x = ABx = A0 = 0

In particular, the columns of B are linearly independent. Therefore, B is also onto. Also,

(BA − I) Bx = B (AB) x−Bx = 0

Since B is onto, it follows that BA − I maps every vector to 0 and so this matrix is 0. Thus BA = I.

23 These vectors are not linearly independent. They are linearly dependent. In fact −1 times the first added to 2 times the second is the third.

25 These cannot be linearly independent because there are 4 of them. You can have at most three. However, it might be that they span R³.

, row echelon form:

. The dimension of the span of these vectors

 is 2 so they do not span R³ .

27 These vectors are not linearly independent and so they are not a basis. The remaining question is whether they span.

, row echelon form:

. The dimension of the span of these vectors is 

2 so they do not span R³ .

31 Yes it is. It is the span of the vectors

Since these two vectors are a linearly independent set, the given subspace has dimension 2.

33 It is a subspace and it equals the span of the vectors which form the columns of the following matrix.

, row echelon form:

. It follows that the dimension of this subspace 

equals 3. A basis .

37 This is obvious. If x, y ∈ V ∩ W, then for scalars α, β, the linear combination αx + βy must be in both V and W since they are both subspaces.

39 Let {x₁, · · · , xk} be a basis for V ∩ W. Then there is a basis for V and W which are respectively

{x₁, · · · , xk, yk₊₁, · · · , yp} ,

{x₁, · · · , xk, zk₊₁, · · · , zq}

It follows that you must have

k + p − k + q − k ≤ n

and so you must have

p + q − n ≤ k

41 No. There must then be inﬁnitely many solutions. If the system is Ax = b, then there are inﬁnitely many solutions to Ax = 0 and so the solutions to Ax = b are a particular solution to Ax = b added to the solutions to Ax = 0 of which there are infinitely many.

43 Yes. It has a unique solution.

45 a. Inﬁnite solution set. b. This surely can’t happen. If you add in another column, the rank does not get smaller. c. You can’t have the rank equal 4 if you only have two columns. d. In this case, there is no solution to the system of equations represented by the augmented matrix. e. In this case, there is a unique solution since the columns of A are independent.The columns are independent. Therefore, A is one to one.

49 Suppose ABx = 0. Then Bx ∈ ker (A) ∩ B (Fp) and so showing that

Consider B (Fp) ∩ ker (A) and let a basis be

{w₁,· · ·, wk}.

Then each wi is of the form Bzi = wi. Therefore, {z₁,· · ·, zk} is linearly independent and ABzi = 0. Now let {u₁,· · ·, ur} be a basis for ker (B) . If ABx = 0, then Bx ∈ ker (A) ∩ B (Fp) and so

which implies

and so it is of the form

It follows that if ABx = 0 so that x ∈ ker (AB) , then

x ∈ span (z₁, · · · , zk, u₁, · · · , ur) .

Therefore,

dim (ker (A B))
≤ k + r = dim (B (Fp) ∩ ker (A))
+dim (ker (B))
≤ dim (ker (A)) + dim (ker (B))

51 If det (A − λI) = 0 then (A − λI) ⁻¹ does not exist and so the columns are not independent which means that for some x ≠ 0, (A − λI) x = 0.

53 Since A₁ is not one to one, it follows there exists x ≠ 0 such that A₁x = 0. Hence Ax = 0 although x ≠ 0 so it follows that A is not one to one. From another point of view, if A were one to one, then ker (A) ⊥ = Rn and so by the Fredholm alternative, AT would be onto Rn. However, AT has only m columns so this cannot take place.

54 That (ATA)T = ATA follows from the properties of the transpose. Therefore,

Suppose ATAx = 0. Then (AT Ax, x) = (Ax,Ax) and so Ax = 0. Therefore,

(ATb, x)= (b,Ax) = (b, 0) = 0

It follows that ATb ∈ (ker((ATA)T))⊥ and so there exists a solution x to the equation

ATAx = ATb

by the Fredholm alternative.

55

|b−Ay|²
= |b−Ax+Ax − Ay|²
= |b−Ax|² +|Ax − Ay|²
+2 (b−Ax,A (x − y))
= |b−Ax|² + |Ax − Ay|²
+2(ATb−ATAx, (x − y))
= |b−Ax|² +|Ax−Ay|²

and so, Ax is closest to b out of all vectors Ay.

56 The dimension of Fn2 is n². Therefore, there exist F scalars ck such that

Let p (λ) be the monic polynomial having smallest degree such that p (A) = 0. If q (A) = 0 then from the Euclidean algorithm,

q (λ) = p (λ) l (λ) + r (λ)

where the degree of r (λ) is less than the degree of p (λ) or else r (λ) equals 0. However, if it is not zero, you could plug in A and obtain

0 = q (A) = 0 + r (A)

and this would contradict the deﬁnition of p (λ) as being the polynomial having smallest degree which sends A to 0. Hence q (λ) = p (λ) l (λ) .

C.9 Exercises 175

3

4

5

6

7

8

9

16

17

19

21

25

26

Now, why does this matrix preserve distance? For short, call it Q and note that QTQ = I. Then

|x|² = (QT Qx, x) = (Qx,Qx) = |Qx|²

and so Q preserves distances.

27 From the above problem this preserves distances and QT = Q. Now do it to x.

and so

Qx − Qy= y − x
Qx+Qy= x + y

Hence, adding these yields Qx = y and then subtracting them gives Qy = x.

28 Linear transformations take 0 to 0. Also

Ta (u + v) ≠ Tau + Tav.

31 

33 

35 

37 

38 

39 

41 

43 

48

51 If x, y ∈ ker (A) then

A (ax+by) = aAx + bAy = a0 + b0 = 0

and so ker (A) is closed under linear combinations. Hence it is a subspace.

C.10 Exercises 190

11 An LU factorization of the coefficient matrix is

First solve

which yields u = 1, v = 2. w = 6. Next solve.

This yields z = 6, y = -16, x = 27.

22 You would have QRx = b and so then you would have Rx = QTb. Now R is a upper triangular and so the solution of this problem is fairly simple.

C.11 Exercises 214

1 The minimum is −11/2 and it occurs when x₁ = x₃ = x₆ = 0 and x₂ = 7/2, x₄ = 13/2, x₆ = −11/2. The maximum is 7 and it occurs when x₁ = 7, x₂ = 0, x₃ = 0, x₄ = 3, x₅ = 5, x₆ = 0.

2 Maximize and minimize the following if possible. All variables are nonnegative

(a) The maximum is 7 when x₁ = 7 and x₁, x₃ = 0.
  The minimum is −7 and it happens when x₁ = 0, x₂ = 7/2, x₃ = 0.

(b) The minimum is −21 and it occurs when x₁ = x₂ = 0, x₃ = 7.
  The maximum is 7 and it occurs when x₁ = 7, x₂ = 0, x₃ = 0.

(c) The minimum is 0 and it occurs when x₁ = x₂ = 0, x₃ = 1.
  The maximum is 14 and it happens when x₁ = 7, x₂ = x₃ = 0.

(d) The maximum is 7 and it happens when x₂ = 7/2, x₃ = x₁ = 0.
  The minimum is 0 when x₁ = x₂ = 0, x₃ = 1.

4 Find solutions if possible.

(a) There is no solution to these inequalities with x₁, x₂ ≥ 0.

(b) A solution is x₁ = 8/5, x₂ = x₃ = 0.

(c) No solution to these inequalities for which all the variables are nonnegative.

(d) There is a solution when x₂ = 2, x₃ = 0, x₁ = 0. 

(e) There is no solution.

C.12 Exercises 239

3 If it did have λ ∈ R as an eigenvalue, then there would exist a vector x such that Ax = λx for λ a real number. Therefore, Ax and x would need to be parallel. However, this doesn’t happen because A rotates the vectors.

5 Amx = λmx for any integer. In the case of −1,

A⁻¹λx = AA⁻¹x = x

so A⁻¹x = λ⁻¹x. Thus the eigenvalues of A⁻¹ are just λ⁻¹ where λ is an eigenvalue of A.

7 Let x be the eigenvector. Then Amx = λmx,Amx = Ax = λx and so

λm = λ

Hence if λ0, then

λm-1 = 1

and so |λ| = 1.

9 eigenvectors:  2. This is a

defective matrix.

11 eigenvectors:   -1,

This matrix is not defective because, even though λ = 1 is a repeated eigenvalue, it has a 2 dimensional eigenspace.

13 eigenvectors:   

This matrix is not defective.

15 eigenvectors:

This matrix is defective. In this case, there is only one eigenvalue, −1 of multiplicity 3 but the dimension of the eigenspace is only 2.

18 eigenvectors: 0 This one is defective.

19 eigenvectors: 1

This if defective.

21 eigenvectors:

23 eigenvectors:

2 + 2i This matrix is not defective.

25 eigenvectors:

This is not defective.

27 The characteristic polynomial is of degree three and it has real coefficients. Therefore, there is a real root and two distinct complex roots. It follows that A cannot be defective because it has three distinct eigenvalues. 

29 eigenvectors:

31 eigenvectors:

33 eigenvectors:

35 In terms of percentages in the various locations,

37 Obviously A cannot be onto because the range of A has dimension 1 and the dimension of this space should be 3 if the matrix is onto. Therefore, A cannot be invertible. Its row reduced echelon form cannot be I since if it were, A would be onto. Aw = w so it has an eigenvalue equal to 1. Now suppose Ax = λx. Thus, from the Cauchy Schwarz inequality,

39 Since the vectors are linearly independent, the matrix S has an inverse. Denoting this inverse by

it follows by deﬁnition that

wTi xj = δij.

Therefore,

S⁻¹MS = S⁻¹ (Mx₁, · · · ,Mxn)

41 The diagonally dominant condition implies that none of the Gerschgorin disks contain 0. Therefore, 0 is not an eigenvalue. Hence A is one to one, hence invertible.

43 First note that (AB)∗= B∗A∗. Say Mx = λx, x ≠ 0. Then

Hence λ =

46 Suppose A is skew symmetric. Then what about iA?

(iA)∗= −iA∗= −iAT = iA

and so iA is self adjoint. Hence it has all real eigenvalues. Therefore, the eigenvalues of A are all of the form iλ where λ is real. Now what about the eigenvectors? You need

Ax = iλx

where λ ≠ 0 is real and A is real. Then

ARe (x) = iλ Re (x)

The left has all real entries and the right has all pure imaginary entries. Hence Re (x) = 0 and so x has all imaginary entries.

C.13 Exercises 276

1 a. orthogonal and transformation, b. symmetric, c. skew symmetric.

4 ||Ux||² = (Ux, Ux)
= (UTUx, x) = (Ix, x) = ||x||²
Next suppose distance is preserved by U. Then

But since U preserves distances, it is also the case that

(U (x + y) , U (x + y))
= ||x||²+||y||² + 2 (x, y)

Hence

(x, y) = (UTUx, y)

and so

((UTU −I)x, y) = 0

Since y is arbitrary, it follows that UTU − I = 0. Thus U is orthogonal.

5

7 If A is symmetric, then A = UTDU for some D a diagonal matrix in which all the diagonal entries are non zero. Hence A⁻¹ = U⁻¹D⁻¹U−T. Now

and so A⁻¹ = QD⁻¹QT, where Q is orthogonal. Is this thing on the right symmetric? Take its transpose. This is QD⁻¹QT which is the same thing, so it appears that a symmetric matrix must have symmetric inverse. Now consider raising it to a power.

Ak = UTDkU

and the right side is clearly symmetric.

9 Yes.

11 eigenvectors:

12 eigenvectors:

13 eigenvectors:

15

and so λ = . This shows that all eigenvalues are real. It follows all the eigenvectors are real. Why? Because A is real. , so is an eigenvector. Hence it can be assumed all eigenvectors are real.

Now let x, y, µ and λ be given as above.

and so

(λ − µ) x·y = 0.

Since λ ≠ µ, it follows x · y = 0.

17

and so λ = . This shows that all eigenvalues are real. Now let x, y, µ and λ be given as above.

and so

(λ−µ) x·y = 0.

Since λ ≠ µ, it follows x · y = 0.

19 Certainly not.

29 eigenvectors:

The matrix U has these as its columns.

34 eigenvectors:

37 If A is given by the formula, then

AT = UTDTU = UTDU = A

Next suppose A = AT. Then by the theorems on symmetric matrices, there exists an orthogonal matrix U such that

UAUT = D

for D diagonal. Hence

A = UTDU

39 There exists U unitary such that A = U∗T U such that T is uppser triangular. Thus A and T are similar. Hence they have the same determinant. Therefore, det (A) = det (T) , but det (T) equals the product of the entries on the main diagonal which are the eigenvalues of A.

43

44 y = −0.125x² + 1. 425x + 0.925

46 Find an orthonormal basis for the spans of the following sets of vectors.

47

49

51 It satisfies the properties of an inner product. Note that

so

The product is obviously linear in the first argument. If (A,A)F = 0, then

52 From the singular value decomposition,

Then

53 

C.14 Exercises 296

1 

2 

5 

6 

7 It indicates that they are no good for doing it.

C.15 Exercises 322

1 The actual largest eigenvalue is 8 with corresponding eigenvector

4 The largest eigenvalue is −16 and an eigenvector is

5 Eigenvalue near −1.35 : λ = −1. 341,

Eigenvalue near 1.5: λ = 1. 679 0,

Eigenvalue near 6.5: λ = 6. 662,

8 Eigenvalue near −1 : λ = −0.703 69,

Eigenvalue near .25 : λ = 0.189 11,

Eigenvalue near 7.5 : λ = 7. 514 6,

10

12 From the bottom line, a lower bound is −10 From the second line, an upper bound is 12.

C.16 Exercises 326

1 The hint is a good suggestion. Pick the first thing in S. By the Archimedean property, S ≠ . That is km > n for all k sufficiently large. Call this first thing q + 1. Thus n−(q + 1) m < 0 but n−qm ≥ 0. Then

n−qm < m

and so

0 ≤ r ≡n−qm < m.

2 First note that either m or −m is in S so S is a nonempty set of positive integers. By well ordering, there is a smallest element of S, called p = x₀m + y₀n. Either p divides m or it does not. If p does not divide m, then by the above problem,

m = pq + r

where 0 < r < p. Thus

m = (x₀m + y₀n) q + r

and so, solving for r,

r = m (1 − x₀) + (−y₀q) n ∈ S.

However, this is a contradiction because p was the smallest element of S. Thus p|m. Similarly p|n.Now suppose q divides both m and n. Then m = qx and n = qy for integers, x and y. Therefore,

showing q|p. Therefore, p = (m, n) .

3 Suppose r is the greatest common divisor of p and m. Then if r ≠ 1, it must equal p because it must divide p. Hence there exist integers x, y such that

p = xp + ym

which requires that p must divide m which is assumed not to happen. Hence r = 1 and so the two numbers are relatively prime.

4 The only substantive issue is why Zp is a field. Let [x] ∈ Zp where [x] ≠ [0]. Thus x is not a multiple of p. Then from the above problem, x and p are relatively prime. Hence from another of the above problems, there exist integers a, b such that

1 = ap + bx

Then

[1−bx] = [ap] = 0

and it follows that

[b] [x] = [1]

so [b] = [x]⁻¹.

C.17 Exercises 345

1 No. (1, 0, 0, 0) ∈ M but 10 (1, 0, 0, 0) ∉ M.

3 If not, you could add in a vector not in their span and obtain 6 vectors which are linearly independent. This cannot occur thanks to the exchange theorem.

10 For each x ∈ [a, b] , let fx (x) = 1 and fx (y) = 0 if y ≠ x. Then these vectors are obviously linearly independent.

12 A field also has multiplication. However, you can consider the elements of the field as vectors and then it satisﬁes all the vector space axioms. When you multiply a number (vector) in R by a scalar in Q you get something inR. All the axioms for a vector space are now obvious. For example, if α ∈ Q and x, y ∈ R,

α (x + y) = αx + αy

from the distributive law on R.

13 Simply let f (i) be the ith component of a vector x ∈ Fn. Thus a typical thing in Fn is (f (1) ,· · ·, f (n)).

14 Say for some n, ∑nk=1 ckek = 0, the zero function. Then pick i,

Since i was arbitrary, this shows these vectors are linearly independent.

15 Say

Then taking derivatives you have

This must hold when each equation is evaluated at x where you can pick the x at which the above determinant is nonzero. Therefore, this is a system of n equations in n variables, the ci and the coefficient matrix is invertible. Therefore, each ci = 0.

19 Which are linearly independent?

(a) These are linearly independent.

(b) These are also linearly independent.

21 This is obvious because when you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is {1,√2}. By deﬁnition, the span of these gives the collection of vectors. Are they independent? Say a + b√2 = 0 where a, b are rational numbers. If a ≠ 0, then b√2 = −a which can’t happen since a is rational. If b ≠ 0, then −a = b√2 which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both a, b = 0 and so this is a basis.

29 Consider the claim about ln σ.

1eln(σ) + (−1) σe⁰ = 0

The equation shown does hold from the deﬁnition of ln σ. However, if ln σ were algebraic, then eln σ, e⁰ would be linearly dependent with field of scalars equal to the algebraic numbers, contrary to the Lindemann Weierstrass theorem. The other instances are similar. In the case of cos σ, you could use the identity

contradicting independence of

eiσ, e−iσ, e⁰.

C.18 Exercises 363

1 I will show one of these. Verify that Examples 16.5.1-16.5.4 are each inner product spaces. First consider Example 16.5.1. All of the axioms of the inner product are obvious except one, the one which says that if ⟨f⟩, ⟨f⟩ = 0 then f = 0. This one depends on continuity of the functions. Suppose then that it is not true. In other words, ⟨f, f⟩ = 0 and yet f ≠ 0. Then for some x ∈I, f (x) ≠ 0. By continuity, there exists δ > 0 such that if y ∈ I ∩ (x−δ, x + δ) ≡Iδ, then

|f (y)−f (x)|< |f (x)|/2

It follows that for y ∈ Iδ,

|f (y)|> |f (x)| − |i (x) /2| = |f (x)|/2.

Hence

a contradiction. Note that min p > 0 because p is a continuous function deﬁned on a closed and bounded interval and so it achieves its minimum by the extreme value theorem of calculus.

2

where u = Σk ukvk and w =Σk wkvk.

5 It might be the case that ⟨z, z⟩ = 0 and yet z ≠ 0. Just let z = (z₁,· · ·, zn) where exactly p of the zi equal 1 but the remaining are equal to 0. Then ⟨z, z⟩ would reduce to 0 in the integers mod p. Another problem is the failure to have an order onZp. Consider first Z₂. Is 1 positive or negative? If it is positive, then 1 + 1 would need to be positive. But 1 + 1 = 0 in this case. If 1 is negative, then −1 is positive, but −1 is equal to 1. Thus 1 would be both positive and negative. You can consider the general case where p > 2 also. Simply take a ≠ 1. If a is positive, then consider a, a², a³· · ·. These would all have to be positive. However, eventually a repeat will take place. Thus an = am m < n, and so am (ak − 1) = 0 where k = n−m. Since am ≠ 0, it follows that ak = 1 for a suitable k. It follows that the sequence of powers of a must include each of {1, 2,· · ·, p− 1} and all these would therefore, be positive. However, 1 + (p− 1) = 0 contradicting the assertion that Zp can be ordered. So what would you mean by saying ⟨z, z⟩ ≥ 0? The Cauchy Schwarz inequality would not even apply.

7 In an inner product space, an open ball is the set Let δ = r− |z−x|. Then if y ∈B (z, δ) ,

|y − x| ≤ |y − z| +|z − x|< δ +|z − x|

= r− |z − x| +|z − x| = r

and so B (z, δ) ⊆B (x,r).

8

9 Let y go with λ and z go with µ.

z (p (x) y′)′ + (λq (x) + r (x)) yz = 0
y (p (x) z′)′ + (µq (x) + r (i)) zy = 0

Subtract.

z (p (x) y′)′ −y (p (x) z′)′ + (λ−µ) q (x) yz = 0

Now integrate from a to b. First note that

and so what you get is

Look at the stuff on the top line. From the assumptions on the boundary conditions,

C₁y (a) + C₂y′ (a) = 0
C₁z (a) + C₂z′ (a) = 0

and so

y (a) z′ (a)−y′ (a) z (a) = 0

Similarly,

y (b) z′ (b)−y′ (b) z (b) = 0

Hence, that stuff on the top line equals zero and so the orthogonality condition holds.

11

12

13

15

16

You know that t → Dn (t) is periodic of period 2π. Therefore, if f (y) = 1,

However, it follows directly from computation that Snf (x) = 1. Just take the integral of the sum which deﬁnes Dn.

17 From Lemma 16.5.11 and Theorem 16.5.12

for all w ∈ span Therefore,

Now if ⟨u, v⟩ = 0, then you can see right away from the deﬁnition that

|u + v|² = |u|² +|v|²

Applying this to

the above equals

the last step following because of similar reasoning to the above and the assumption that the uk are orthonormal. It follows the sum ∑∞k=1|⟨y, uk⟩| converges and so lim k→∞⟨y, uk⟩ = 0 because if a series converges, then the kth term must converge to 0.

18 Let f be any piecewise continuous function which is bounded on [−π, π] . Show, using the above problem, that

Let the inner product space consist of piecewise continuous bounded functions with the inner product deﬁned by

Then, from the above problem and the fact shown earlier that

form an orthonormal set of vectors in this inner product space, it follows that

without loss of generality, assume that f has real values. Then the above limit reduces to having both the real and imaginary parts converge to 0. This implies the thing which was desired. Note also that if α ∈ [−1, 1] , then

19 From the deﬁnition of Dn,

Now observe that Dn is an even function. Therefore, the formula equals

Now note that because

and Dn is even. Therefore,

From the formula for Dn (y) given earlier, this is dominated by an expression of the form

for a suitable constant C. The above is equal to

and the expression equals a bounded continuous function on [0, π] except at 0 where it is undeﬁned. This follows from elementary calculus. Therefore, changing the function at this single point does not change the integral and so we can consider this as a continuous bounded function deﬁned on [0, π] . Also, from the assumptions on f,

is equal to a piecewise continuous function on [0, π] except at the point 0. Therefore, the above integral converges to 0 by the previous problem. This shows that the Fourier series generally tries to converge to the midpoint of the jump.

20

You could also find the Fourier series for x² instead of x and get

because the periodic extension of this function is continuous. Let x = 0

This is one of those calculus problems where you show it converges absolutely by the comparison test with a p series. However, here is what it converges to.

and so

21 Consider for t ∈ [0, 1] the following.

|y − (x + t (w − x))|²

where w ∈K and x ∈K. It equals

f (t) = |y − x|² + t²|w − x|²−2t Re ⟨y − x, w − x⟩

Suppose x is the point of K which is closest to y. Then f′ (0) ≥ 0. However,

f′ (0) = −2 Re⟨y − x, w − x⟩.

Therefore, if x is closest to y,

Re ⟨y−x, w−x⟩ ≤ 0.

Next suppose this condition holds. Then you have

|y− (x + t (w − x))|² ≥
|y − x|² + t²|w − x|² ≥ |y − x|²

By convexity of K, a generic point of K is of the form x+t (w−x) for w ∈K. Hence x is the closest point.

22

Of course the same reasoning yields

23 Let xk be a minimizing sequence. The connection between xk and ck ∈ Fk is obvious because the{uk} are orthonormal. That is,

|xn − xm| = |cn − cm|Fp .

where x ∑jcjuj. Use the parallelogram identity.

Hence

Now the right hand side converges to 0 since {xk} is a minimizing sequence. Therefore, {xk} is a Cauchy sequence in U. Hence the sequence of component vectors{ck} is a Cauchy sequence in Fn and so it converges thanks to completeness of F. It follows that {xk} also must converge to some x. Then since K is closed, it follows that x ∈K. Hence

λ = |x−y|.

24

⟨P x − P y, y − P y⟩ ≤ 0
⟨P y − P x, x − P x⟩ ≤ 0

Thus

⟨P x−P y, x−P x⟩ ≥ 0

Hence

⟨P x − P y, x − P x⟩ − ⟨P x − P y, y − P y⟩ ≥ 0

and so

25 Let be a basis for V and if x ∈V, let xi be the components of x relative to this basis. Thus the xi are deﬁned according to

Then decree that {ui} is an orthonormal basis. It follows

Now letting {xk} be a sequence of vectors of V let {xk} denote the sequence of component vectors in Fn. One direction is easy, saying that  If this is not so, then there exists a sequence of vectors {xk} such that

dividing both sides by it can be assumed that Hence But from the triangle inequality,

Therefore, since this is a contradition to each It follows that there exists such that for all x,

Now consider the other direction. If it is not true, then there exists a sequence {xk} such that

Dividing both sides by |xk|, it can be assumed that |xk| = |xk| = 1. Hence, by compactness of the closed unit ball in Fn, there exists a further subsequence, still denoted by k such that xk → a ∈ Fn and it also follows that |a|Fn = 1. Also the above inequality implies Therefore,

which is a contradiction to the uj being linearly independent. Therefore, there exists such that for all x,

Now if you have any other norm on this finite dimensional vector space, say |||·|||, then from what was just shown, there exist scalars δi and ∆i all positive, such that

 It follows that

Hence

In other words, any two norms on a finite dimensional vector space are equivalent norms. What this means is that every consideration which depends on analysis or topology is exactly the same for any two norms. What might change are geometric properties of the norms.

C.19 Exercises 388

1

3

5

8 Let 

(a) If f = 0, the zero mapping, then fv = ⟨v, 0⟩ for all v ∈V .

so

(b) If f ≠ 0 then there exists z ≠ 0 satisfying ⟨u, z⟩ = 0 for all u ∈ ker (f) . ker (f) is a subspace and so there exists z₁ ∈ ker (f) . Then there exists a closest point of ker (f) to z₁ called x. Then let z = z1 − x. Thus ⟨u, z⟩ = 0 for all u ∈ ker (f).

(c) f (f (y) z−f (z) y) = f (y) f (z)−f (z) f (y) = 0.

(d)

and so

so  appears to work.

(e) If w₁, w₂ both work, then for every y, 

Now let y = w₁ −w₂ and so w₁ = w₂.

9 It is required to show that A∗ is linear.

Since y is arbitrary, this shows that A∗ is linear. In case A is an m × n matrix as described,

11 The two operators D + 1 and D + 4 commute and are each one to one on the kernel of the other. Also, it is obvious that ker (D + a) consists of functions of the form Ce−at. Therefore, ker (D + 1) (D + 4) consists of functions of the form

where C₁, C₂ are arbitrary constants. In other words, a basis for ker (D + 1) (D + 4) is {e−t, e−4t}.

14

17 It is obvious that x ∼ x. If x ∼ y, then y ∼ x is also clear. If x ∼ y and y ∼ z, then

z −x = z−y + y−x

and by assumption, both z−y and y−x ∈ ker (L) which is a subspace. Therefore, z − x ∈ ker (L) also and so ∼ is an equivalence relation. Are the operations well deﬁned? If [x] = [x′] , [y] = [y′] , is it true that [x + y] = [y′ + x′]? Of course. x′ + y′−(x + y) = (x′ −x)+(y′ −y) ∈ ker (L) because ker (L) is a subspace. Similar reasoning applies to the case of scalar multiplication. Now why is A well deﬁned? If [x] = [x′] , is Lx = Lx′? Of course this is so. x−x′ ∈ ker (L) by assumption. Therefore, Lx = Lx′. It is clear also that A is linear. If A [x] = 0, then Lx = 0 and so x ∈ ker (L) and so [x] = 0. Therefore, A is one to one. It is obviously onto L (V ) = W.

19 An easy way to do this is to “unravel” the powers of the matrix making vectors in Fn2 and then making these the columns of a n²×n matrix. Look for linear relationships between the columns by obtaining the row reduced echelon form and using Lemma 8.2.5. As an example, consider the following matrix.

Lets find its minimal polynomial. We have the following powers

By the Cayley Hamilton theorem, I won’t need to consider any higher powers than this. Now I will unravel each and make them the columns of a matrix.

Next you can do row operations and obtain the row reduced echelon form for this matrix and then look for linear relationships.

From this and Lemma 8.2.5, you see that for A denoting the matrix,

A³ = 4A² − 5A + 2I

and so the minimal polynomial is

λ³ − 4λ² + 5λ− 2

No smaller degree polynomial can work either. Since it is of degree 3, this is also the characteristic polynomial. Note how we got this without expanding any determinants or solving any polynomial equations. If you factor this polynomial, you get λ³ − 4λ² + 5λ − 2 = (λ− 2) (λ− 1)² so this is an easy problem, but you see that this procedure for finding the minimal polynomial will work even when you can’t factor the characteristic polynomial.

20 If two matrices are similar, then they must have the same minimal polynomial. This is obvious from the fact that for p (λ) any polynomial and A = S⁻¹BS,

p (A) = S⁻¹p (B) S

Next you can do row operations and obtain the row reduced echelon form for this matrix and then look So what is the minimal polynomial of the diagonal matrix shown? It is obviously   Thus there are no repeated roots.

21 Show that if A is an n × n matrix and the minimal polynomial has no repeated roots, then A is non defective and there exists a basis of eigenvectors. Thus, from the above problem, a matrix may be diagonalized if and only if its minimal polynomial has no repeated roots. It turns out this condition is something which is relatively easy to determine. Hint: You might want to use Theorem 17.3.1. If A has a minimal polynomial which has no repeated roots, say

then from the material on decomposing into direct sums of generalized eigenspaces, you have

and by deﬁnition, the basis vectors for ker (A−λ₂I) are all eigenvectors. Thus Fn has a basis of eigenvectors and is therefore diagonalizable or non defective.

Index

∩ , 11

∪, 11

σ(A), 375

Abel’s formula, 116

Abelian group, 325

adjoint, 262, 266

adjugate, 106, 128

algebraic multiplicity, 223

algebraic number

minimal polynomial, 343

algebraic numbers, 342

field, 344

angle between vectors, 41

area

parallelogram, 51

area of a parallelogram, 49

augmented matrix, 63

axioms for a norm, 47

back substitution, 63

barallelepiped

volume, 53

bases, 149

basic feasible solution, 196

basic variables, 70, 196

basis, 149, 327

any two same size, 330

basis of eigenvectors

diagonalizable, 228

bijective, 369

block matrix, 257

block multiplication, 257

box product, 53

Cartesian coordinates, 24

Cauchy Schwarz inequality, 39, 45

Cayley Hamilton theorem, 129, 244, 282

characteristic equation, 218

characteristic polynomial, 128

characteristic value, 218

classical adjoint, 106

codomain, 12

cofactor, 102, 126

cofactor matrix, 102

column rank, 141

column space, 141

companion matrix, 310, 323

complex conjugate, 16

complex eigenvalues, 232

shifted inverse power method, 309

complex numbers, 15, 16

complex numbers

arithmetic, 16

roots, 18

triangle inequality, 17

component, 34, 47

component of a force, 43

component of force,⁴³

components of a matrix, 78

components of a vector, 26

composition of linear transformations, 386

condition number, 296

conformable,⁸¹

consistent, 72

Coordinates, 23

Cramer’s rule, 109, 128

cross product, 48, 49

area of parallelogram, 49

coordinate description, 50

distributive law, 51, 52, 54

geometric description, 49

cross product

coordinate description, 50

distributive law, 51

geometric description, 49

parallelepiped, 53

De Moivre’s theorem, 18

defective, 224

defective eigenvalue, 224

derivative, 298

determinant, 121

alternating property, 123

cofactor, 100

cofactor expansion, 125

expanding along row or column, 100

expansion along row (column), 126

linear transformation, 384

matrix inverse formula, 106, 127

minor, 99

product, 104, 124

product of eigenvalues, 281

row operations, 103

transpose, 123

determinant rank

row rank, 157

diagonal matrix, 228, 243

diagonalizable, 227, 228, 243, 255, 384

differential equations

first order systems, 348

dimension, 149

dimension of vector space, 330

direct sum, 371

distance formula, 27

Dolittle’s method, 183

domain, 12

dot product, 39

properties, 39

dynamical system, 244

echelon form, 64, 65

eigenspace, 220

eigenvalue, 218, 375

existence, 373

eigenvalues, 129

eigenvector, 218

Einstein summation convention, 55

elementary matrices, 131

elementary matrix

inverse, 136

properties, 136

elementary operations, 62

empty set, 12

entries of a matrix, 78

equivalence class, 338, 382

equivalence relation, 338, 382

exchange theorem, 327

field extensions, 341

Field of scalars, 325

force, 32

Fourier coeﬃcients, 358

Fredholm alternative, 155, 268

free variables, 70, 196

Frobinius norm, 282

function, 12

functions, 12

fundamental theorem of algebra, 20, 399

Gauss Elimination, 71

Gauss elimination, 63, 64

Gauss Jordan method for inverses, 89

Gauss Seidel, 288

Gauss Seidel method, 288

general solution, 175

solution space, 174

generalized eigenspace, 375

direct sum, 373

geometric multiplicity, 223

Gerschgorin’s theorem, 238

Gram Schmidt process, 261, 354

Grammian matrix, 355

greatest common divisor, 333

Hermitian, 264

homogeneous coordinates, 178

homogeneous syster, 174

homomorphism, 165

Householder matrix, 188

householder matrix, 177

inconsistent, 69, 72

independent set

extending to a basis, 151

independent set of vectors

extending to form a basis, 151

injective, 12, 369

inner produc

strange examplet, 47

inner product, 39, 45

axioms, 350

Cauchy Schwarz inequality, 352

inner product

properties, 45

integers mod a prime, 326

intersection, 11

intervals

notation, 11

inverse

left inverse, 128

right inverse, 128

inverses and determinants, 108, 127

invertible, 87

irreducible, 333

isomorphism, 165

Jacobi, 286

Jacobi method, 286

Jordan block, 391

joule, 44

ker, 173

kernel, 152, 173

Kroneker delta, 54

Kroneker symbol, 87

Laplace expansion, 101, 125

leading entry, 64

least square approximation, 265

linear combination, 124, 137

linear independence

enlarging to form a basis, 331

equivalent conditions, 145

linear relationships, 138

finding them, 147

linear transformation, 165, 369

matrix, 167, 173

linear transformations

commuting, 372

dimension, 369

linear trnsformation

rotation, 166

linearly dependent, 327

linearly independent, 144, 327

linearly independent sets, 148

LU deomposition

non existence, 181

LU factorization

by inspection, 181

justiﬁcation, 184

multipliers, 182

solving systems, 183

main diagonal, 102

Markov matrices, 235

matrices

more columns than rows, 139

multiplication, 81

one to one, onto, 157

similar, 227

matrix, 77

composition of linear transformations, 386

identity, 86

inverse, 87

invertible, product of elementary matrices, 158 

left inverse, 128

left inverse, right inverse, 172

lower triangular, 102, 128

main diagonal, 228

one to one, onto, 172

raising to a power, 229

right inverse, 128

right inverse left inverse and inverse, 140

rotation, 167

rotation about given vector, 169

self adjoint, 243, 278

symmetric, 243

transpose, 85

upper triangular, 102, 128

matrix exponential, 232

matrix inverse

finding it, 89

matrix multiplication

ij entry, 83

properties, 84

matrix of linear transformation, 380

mean square approximation, 359

migration matrix, 235

minimal polynomial, 164, 374

computation, 390

uniqueness, 374

minimization and orthogonality, 357

minor, 102, 126

monic, 333

monic polynomial, 374

multipliers, 185

Newton, 35

nilpotent, 113, 378

non defective

minimal polynomial, 390

nondefective, 255

nondefective eigenvalue, 224

normed linear space, 353

normed vector space, 353 

null space, 152, 173

nullity, 154

one to one, 12, 166

rank, 163

onto, 12, 167

open ball, 27

operator norm, 293

orthogonal matrix, 113, 177, 188, 247

switching two unit vectors, 188

orthogonal projection, 357

orthogonality and minimization, 265

orthonormal, 248, 260

independent, 354

orthonormal set, 353

p norms, 297

parallelepiped, 53

parallelogram identity, 367

particular solution, 173

partitioned matrix, 257

permutation matrices, 131

permutation symbol, 54

reduction identity, 55

perp, 155

perpendicular, 42

pivot, 69

pivot column, 65, 138

pivot columns, 65

pivot position, 65

pivot positions, 65

PLU factorization, 186

points and vectors, 23

polar form complex number, 18

polarization identity, 367

polynomial, 332

degree, 332

divides, 333

equal, 332

Euclidean algorithm, 332

greatest common divisor, 333

greatest common divisor description, 333

greatest common divisor, uniqueness, 333

irreducible, 333

irreducible factorization, 334

relatively prime, 333

root, 332

polynomials

canceling, 335

factorization, 336

position vector, 25, 26, 33

power method, 299

preserving distance, 269

principle directions, 234

product of matrices

composition of linear transformations, 386

projection, 43, 97

projection of a vector, 43

projections

matrix, 171

QR decomposition, 261

QR factorization, 188, 189, 261

thin, 261

range, 12

rank

column determinant and row, 142

finding the rank, 143

linear transformation, 385

rank and singular values, 273

rank of a matrix, 141, 156

Rayleigh quotient, 310

reﬂection

across a given vector, 177

regression line, 267

regular Sturm Liouville problem, 364

relations

graph, 13

resultant, 34

right handed system, 48

right inverse, 89

right polar factorization, 269

rotations

about given vector, 386

row equivalent, 139

row operations, 64, 103, 131

row rank, 141

row reduced echelon form, 65, 137

existence, 137

uniqueness, 139

row space, 141

scalar product, 39

scalars, 24, 77

scaling factor, 299

Schur’s theorem, 262

set notation, 11

shifted inverse power method, 301

complex eigenvalues, 309

sign function, 119

similar matrices, 381

similarity

block diagonal matrix, 375

upper triangular block diagonal, 377

similarity and equivalence, 382

similarity relation, 227

similarity transformation, 381

simplex tableau, 196, 197

simultaneous corrections, 286

singular value decomposition, 272

singular values, 272

skew lines, 60

skew symmetric, 86, 93

slack variables, 196, 198

slope, 14, 15

solution space, 173

span, 124, 137, 327

spanning sets, 148

spectrum, 218

speed, 35

standard position, 33

strictly upper triangular, 391

Sturm Liouville problem, 364

subspace, 147, 327

has a basis, 332

surjective, 12, 369

Sylvester’s theorem, 370

symmetric, 86, 93

symmetric matrix, 249

trace, 282

sum of eigenvalues, 282

triangle inequality, 30, 40, 47, 353

complex numbers, 17

union, 11

unitary, 262

upper Hessenberg form, 317

variation of constants formula, 350

variational inequality, 367

vector addition

geometric meaning, 26

vector space, 325

dimension, 330

vector space axioms, 24

vectors, 23, 32, 79

column vector, 79

row vector, 79

velocity, 35

work, 43

Wronskian, 116, 355

Wronskian alternative, 349

zero matrix, 78