Jordan normal form

Some textbooks have the ones on the subdiagonal; that is, immediately below the main diagonal instead of on the superdiagonal. The eigenvalues are still on the main diagonal.[10][11]

An n × n matrix A is diagonalizable if and only if the sum of the dimensions of the eigenspaces is n. Or, equivalently, if and only if A has n linearly independent eigenvectors. Not all matrices are diagonalizable; matrices that are not diagonalizable are called defective matrices. Consider the following matrix:

${\displaystyle A=\left[{\begin{array}{*{20}{r}}5&4&2&1\\[2pt]0&1&-1&-1\\[2pt]-1&-1&3&0\\[2pt]1&1&-1&2\end{array}}\right].}$

Including multiplicity, the eigenvalues of A are λ = 1, 2, 4, 4. The dimension of the eigenspace corresponding to the eigenvalue 4 is 1 (and not 2), so A is not diagonalizable. However, there is an invertible matrix P such that J = P⁻¹AP, where

${\displaystyle J={\begin{bmatrix}1&0&0&0\\[2pt]0&2&0&0\\[2pt]0&0&4&1\\[2pt]0&0&0&4\end{bmatrix}}.}$

The matrix ${\displaystyle J}$ is almost diagonal. This is the Jordan normal form of A. The section Example below fills in the details of the computation.

Consider the matrix ${\displaystyle A}$ from the example in the previous section. The Jordan normal form is obtained by some similarity transformation:

${\displaystyle P^{-1}AP=J;}$ that is, ${\displaystyle AP=PJ.}$

Let ${\displaystyle P}$ have column vectors ${\displaystyle p_{i}}$, ${\displaystyle i=1,\ldots ,4}$, then

${\displaystyle A{\begin{bmatrix}p_{1}&p_{2}&p_{3}&p_{4}\end{bmatrix}}={\begin{bmatrix}p_{1}&p_{2}&p_{3}&p_{4}\end{bmatrix}}{\begin{bmatrix}1&0&0&0\\0&2&0&0\\0&0&4&1\\0&0&0&4\end{bmatrix}}={\begin{bmatrix}p_{1}&2p_{2}&4p_{3}&p_{3}+4p_{4}\end{bmatrix}}.}$

We see that

${\displaystyle (A-1I)p_{1}=0}$

${\displaystyle (A-2I)p_{2}=0}$

${\displaystyle (A-4I)p_{3}=0}$

${\displaystyle (A-4I)p_{4}=p_{3}.}$

For ${\displaystyle i=1,2,3}$ we have ${\displaystyle p_{i}\in \operatorname {Ker} (A-\lambda _{i}I)}$, that is, ${\displaystyle p_{i}}$ is an eigenvector of ${\displaystyle A}$ corresponding to the eigenvalue ${\displaystyle \lambda _{i}}$. For ${\displaystyle i=4}$, multiplying both sides by ${\displaystyle (A-4I)}$ gives

${\displaystyle (A-4I)^{2}p_{4}=(A-4I)p_{3}.}$

But ${\displaystyle (A-4I)p_{3}=0}$, so

${\displaystyle (A-4I)^{2}p_{4}=0.}$

Thus, ${\displaystyle p_{4}\in \operatorname {Ker} (A-4I)^{2}.}$

Vectors such as ${\displaystyle p_{4}}$ are called generalized eigenvectors of A.

This example shows how to calculate the Jordan normal form of a given matrix.

Consider the matrix

${\displaystyle A={\begin{bmatrix}5&4&2&1\\0&1&-1&-1\\-1&-1&3&0\\1&1&-1&2\end{bmatrix}}}$

which is mentioned in the beginning of the article.

The characteristic polynomial of A is

${\displaystyle {\begin{aligned}\chi (\lambda )&=\det(\lambda I-A)\\&=\lambda ^{4}-11\lambda ^{3}+42\lambda ^{2}-64\lambda +32\\&=(\lambda -1)(\lambda -2)(\lambda -4)^{2}.\,\end{aligned}}}$

This shows that the eigenvalues are 1, 2, 4 and 4, according to algebraic multiplicity. The eigenspace corresponding to the eigenvalue 1 can be found by solving the equation Av = λ v. It is spanned by the column vector v = (−1, 1, 0, 0)^T. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by w = (1, −1, 0, 1)^T. Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional (even though this is a double eigenvalue) and is spanned by x = (1, 0, −1, 1)^T. So, the geometric multiplicity (that is, the dimension of the eigenspace of the given eigenvalue) of each of the three eigenvalues is one. Therefore, the two eigenvalues equal to 4 correspond to a single Jordan block, and the Jordan normal form of the matrix A is the direct sum

${\displaystyle J=J_{1}(1)\oplus J_{1}(2)\oplus J_{2}(4)={\begin{bmatrix}1&0&0&0\\0&2&0&0\\0&0&4&1\\0&0&0&4\end{bmatrix}}.}$

There are three Jordan chains. Two have length one: {v} and {w}, corresponding to the eigenvalues 1 and 2, respectively. There is one chain of length two corresponding to the eigenvalue 4. To find this chain, calculate

${\displaystyle \ker {(A-4I)}^{2}=\operatorname {span} \,\left\{{\begin{bmatrix}1\\0\\0\\0\end{bmatrix}},{\begin{bmatrix}1\\0\\-1\\1\end{bmatrix}}\right\}}$

where I is the 4 × 4 identity matrix. Pick a vector in the above span that is not in the kernel of A − 4I; for example, y = (1,0,0,0)^T. Now, (A − 4I)y = x and (A − 4I)x = 0, so {y, x} is a chain of length two corresponding to the eigenvalue 4.

The transition matrix P such that P⁻¹AP = J is formed by putting these vectors next to each other as follows

${\displaystyle P=\left[{\begin{array}{c|c|c|c}v&w&x&y\end{array}}\right]={\begin{bmatrix}-1&1&1&1\\1&-1&0&0\\0&0&-1&0\\0&1&1&0\end{bmatrix}}.}$

A computation shows that the equation P⁻¹AP = J indeed holds.

${\displaystyle P^{-1}AP=J={\begin{bmatrix}1&0&0&0\\0&2&0&0\\0&0&4&1\\0&0&0&4\end{bmatrix}}.}$

If we had interchanged the order in which the chain vectors appeared, that is, changing the order of v, w and {x, y} together, the Jordan blocks would be interchanged. However, the Jordan forms are equivalent Jordan forms.

We give a proof by induction that any complex-valued matrix A may be put in Jordan normal form. Since the underlying vector space can be shown[14] to be the direct sum of invariant subspaces associated with the eigenvalues, A can be assumed to have just one eigenvalue λ. The 1 × 1 case is trivial. Let A be an n × n matrix. The range of A − λ I, denoted by Ran(A − λ I), is an invariant subspace of A. Also, since λ is an eigenvalue of A, the dimension of Ran(A − λ I), r, is strictly less than n, so, by the inductive hypothesis, Ran(A − λ I) has a basis {p₁, …, p_r} composed of Jordan chains.

Next consider the kernel, that is, the subspace Ker(A − λ I). If

${\displaystyle \mathrm {Ran} (A-\lambda I)\cap \mathrm {Ker} (A-\lambda I)=\{0\},}$

the desired result follows immediately from the rank–nullity theorem. (This would be the case, for example, if A were Hermitian.)

Otherwise, if

${\displaystyle Q=\mathrm {Ran} (A-\lambda I)\cap \mathrm {Ker} (A-\lambda I)\neq \{0\},}$

let the dimension of Q be s ≤ r. Each vector in Q is an eigenvector, so Ran(A − λ I) must contain s Jordan chains corresponding to s linearly independent eigenvectors. Therefore the basis {p₁, ..., p_r} must contain s vectors, say {p_r−s+1, ..., p_r}, that are lead vectors of these Jordan chains. We can "extend the chains" by taking the preimages of these lead vectors. (This is the key step.) Let q_i be such that

${\displaystyle \;(A-\lambda I)q_{i}=p_{i}{\mbox{ for }}i=r-s+1,\ldots ,r.}$

The set {q_i}, being preimages of the linearly independent set {p_i} under A − λ I, is also linearly independent. Clearly no non-trivial linear combination of the q_i can lie in Ker(A − λ I), for {p_i}_{i=r−s+1, ..., r} is linearly independent. Furthermore, no non-trivial linear combination of the q_i can belong to Ran(A − λ I) because it would then be a linear combination of the basic vectors p₁, ..., p_r, and this linear combination would have a contribution of basic vectors not in Ker(A − λ I) because otherwise it would belong to Ker(A − λ I). The action of A − λ I on both linear combinations would then produce an equality of a non-trivial linear combination of lead vectors and such a linear combination of non-lead vectors, which would contradict the linear independence of (p₁, ..., p_r).

Finally, we can pick any linearly independent set {z₁, ..., z_t} whose projection spans

${\displaystyle \;\mathrm {Ker} (A-\lambda I)/Q.}$

Each z_i forms a Jordan chain of length 1. By construction, the union of the three sets {p₁, ..., p_r}, {q_{r−s +1}, ..., q_r}, and {z₁, ..., z_t} is linearly independent, and its members combine to form Jordan chains. Finally, by the rank–nullity theorem, the cardinality of the union is n. In other words, we have found a basis composed of Jordan chains, and this shows A can be put in Jordan normal form.

It can be shown that the Jordan normal form of a given matrix A is unique up to the order of the Jordan blocks.

Knowing the algebraic and geometric multiplicities of the eigenvalues is not sufficient to determine the Jordan normal form of A. Assuming the algebraic multiplicity m(λ) of an eigenvalue λ is known, the structure of the Jordan form can be ascertained by analyzing the ranks of the powers (A − λ I)^m(λ). To see this, suppose an n × n matrix A has only one eigenvalue λ. So m(λ) = n. The smallest integer k₁ such that

${\displaystyle (A-\lambda I)^{k_{1}}=0}$

is the size of the largest Jordan block in the Jordan form of A. (This number k₁ is also called the index of λ. See discussion in a following section.) The rank of

${\displaystyle (A-\lambda I)^{k_{1}-1}}$

is the number of Jordan blocks of size k₁. Similarly, the rank of

${\displaystyle (A-\lambda I)^{k_{1}-2}}$

is twice the number of Jordan blocks of size k₁ plus the number of Jordan blocks of size k₁−1. The general case is similar.

This can be used to show the uniqueness of the Jordan form. Let J₁ and J₂ be two Jordan normal forms of A. Then J₁ and J₂ are similar and have the same spectrum, including algebraic multiplicities of the eigenvalues. The procedure outlined in the previous paragraph can be used to determine the structure of these matrices. Since the rank of a matrix is preserved by similarity transformation, there is a bijection between the Jordan blocks of J₁ and J₂. This proves the uniqueness part of the statement.

Using the Jordan normal form, direct calculation gives a spectral mapping theorem for the polynomial functional calculus: Let A be an n × n matrix with eigenvalues λ₁, ..., λ_n, then for any polynomial p, p(A) has eigenvalues p(λ₁), ..., p(λ_n).

The characteristic polynomial of A is ${\displaystyle p_{A}(\lambda )=\det(\lambda I-A)}$. Similar matrices have the same characteristic polynomial. Therefore, ${\textstyle p_{A}(\lambda )=p_{J}(\lambda )=\prod _{i}(\lambda -\lambda _{i})^{m_{i}}}$, where ${\displaystyle \lambda _{i}}$ is the ith root of ${\textstyle p_{J}}$ and ${\displaystyle m_{i}}$ is its multiplicity, because this is clearly the characteristic polynomial of the Jordan form of A.

The Cayley–Hamilton theorem asserts that every matrix A satisfies its characteristic equation: if p is the characteristic polynomial of A, then ${\displaystyle p_{A}(A)=0}$. This can be shown via direct calculation in the Jordan form, since if ${\displaystyle \lambda _{i}}$ is an eigenvalue of multiplicity ${\displaystyle m}$, then its Jordan block ${\displaystyle J_{i}}$ clearly satisfies ${\displaystyle (J_{i}-\lambda _{i}I)^{m_{i}}=0}$. As the diagonal blocks do not affect each other, the ith diagonal block of ${\displaystyle (A-\lambda _{i}I)^{m_{i}}}$ is ${\displaystyle (J_{i}-\lambda _{i}I)^{m_{i}}=0}$; hence ${\textstyle p_{A}(A)=\prod _{i}(A-\lambda _{i}I)^{m_{i}}=0}$.

The Jordan form can be assumed to exist over a field extending the base field of the matrix, for instance over the splitting field of p; this field extension does not change the matrix p(A) in any way.

The minimal polynomial P of a square matrix A is the unique monic polynomial of least degree, m, such that P(A) = 0. Alternatively, the set of polynomials that annihilate a given A form an ideal I in C[x], the principal ideal domain of polynomials with complex coefficients. The monic element that generates I is precisely P.

Let λ₁, …, λ_q be the distinct eigenvalues of A, and s_i be the size of the largest Jordan block corresponding to λ_i. It is clear from the Jordan normal form that the minimal polynomial of A has degree Σs_i.

While the Jordan normal form determines the minimal polynomial, the converse is not true. This leads to the notion of elementary divisors. The elementary divisors of a square matrix A are the characteristic polynomials of its Jordan blocks. The factors of the minimal polynomial m are the elementary divisors of the largest degree corresponding to distinct eigenvalues.

The degree of an elementary divisor is the size of the corresponding Jordan block, therefore the dimension of the corresponding invariant subspace. If all elementary divisors are linear, A is diagonalizable.

The Jordan form of a n × n matrix A is block diagonal, and therefore gives a decomposition of the n dimensional Euclidean space into invariant subspaces of A. Every Jordan block J_i corresponds to an invariant subspace X_i. Symbolically, we put

${\displaystyle \mathbb {C} ^{n}=\bigoplus _{i=1}^{k}X_{i}}$

where each X_i is the span of the corresponding Jordan chain, and k is the number of Jordan chains.

One can also obtain a slightly different decomposition via the Jordan form. Given an eigenvalue λ_i, the size of its largest corresponding Jordan block s_i is called the index of λ_i and denoted by ν(λ_i). (Therefore, the degree of the minimal polynomial is the sum of all indices.) Define a subspace Y_i by

${\displaystyle Y_{i}=\operatorname {Ker} (\lambda _{i}I-A)^{\nu (\lambda _{i})}.}$

This gives the decomposition

${\displaystyle \mathbb {C} ^{n}=\bigoplus _{i=1}^{l}Y_{i}}$

where l is the number of distinct eigenvalues of A. Intuitively, we glob together the Jordan block invariant subspaces corresponding to the same eigenvalue. In the extreme case where A is a multiple of the identity matrix we have k = n and l = 1.

The projection onto Y_i and along all the other Y_j ( j ≠ i ) is called the spectral projection of A at λ_i and is usually denoted by P(λ_i ; A). Spectral projections are mutually orthogonal in the sense that P(λ_i ; A) P(λ_j ; A) = 0 if i ≠ j. Also they commute with A and their sum is the identity matrix. Replacing every λ_i in the Jordan matrix J by one and zeroing all other entries gives P(λ_i ; J), moreover if U J U⁻¹ is the similarity transformation such that A = U J U⁻¹ then P(λ_i ; A) = U P(λ_i ; J) U⁻¹. They are not confined to finite dimensions. See below for their application to compact operators, and in holomorphic functional calculus for a more general discussion.

Comparing the two decompositions, notice that, in general, l ≤ k. When A is normal, the subspaces X_i's in the first decomposition are one-dimensional and mutually orthogonal. This is the spectral theorem for normal operators. The second decomposition generalizes more easily for general compact operators on Banach spaces.

It might be of interest here to note some properties of the index, ν(λ). More generally, for a complex number λ, its index can be defined as the least non-negative integer ν(λ) such that

${\displaystyle \mathrm {Ker} (\lambda -A)^{\nu (\lambda )}=\operatorname {Ker} (\lambda -A)^{m},\;\forall m\geq \nu (\lambda ).}$

So ν(λ) > 0 if and only if λ is an eigenvalue of A. In the finite-dimensional case, ν(λ) ≤ the algebraic multiplicity of λ.

The Jordan form is used to find a normal form of matrices up to conjugacy such that normal matrices make up an algebraic variety of a low fixed degree in the ambient matrix space.

Sets of representatives of matrix conjugacy classes for Jordan normal form or rational canonical forms in general do not constitute linear or affine subspaces in the ambient matrix spaces.

Vladimir Arnold posed[16] a problem: Find a canonical form of matrices over a field for which the set of representatives of matrix conjugacy classes is a union of affine linear subspaces (flats). In other words, map the set of matrix conjugacy classes injectively back into the initial set of matrices so that the image of this embedding—the set of all normal matrices, has the lowest possible degree—it is a union of shifted linear subspaces.

It was solved for algebraically closed fields by Peteris Daugulis.[17] The construction of a uniquely defined plane normal form of a matrix starts by considering its Jordan normal form.

Let X be a Banach space, L(X) be the bounded operators on X, and σ(T) denote the spectrum of T ∈ L(X). The holomorphic functional calculus is defined as follows:

Fix a bounded operator T. Consider the family Hol(T) of complex functions that is holomorphic on some open set G containing σ(T). Let Γ = {γ_i} be a finite collection of Jordan curves such that σ(T) lies in the inside of Γ, we define f(T) by

${\displaystyle f(T)={\frac {1}{2\pi i}}\int _{\Gamma }f(z)(z-T)^{-1}dz.}$

The open set G could vary with f and need not be connected. The integral is defined as the limit of the Riemann sums, as in the scalar case. Although the integral makes sense for continuous f, we restrict to holomorphic functions to apply the machinery from classical function theory (for example, the Cauchy integral formula). The assumption that σ(T) lie in the inside of Γ ensures f(T) is well defined; it does not depend on the choice of Γ. The functional calculus is the mapping Φ from Hol(T) to L(X) given by

${\displaystyle \;\Phi (f)=f(T).}$

We will require the following properties of this functional calculus:

1. Φ extends the polynomial functional calculus.
2. The spectral mapping theorem holds: σ(f(T)) = f(σ(T)).
3. Φ is an algebra homomorphism.

In the finite-dimensional case, σ(T) = {λ_i} is a finite discrete set in the complex plane. Let e_i be the function that is 1 in some open neighborhood of λ_i and 0 elsewhere. By property 3 of the functional calculus, the operator

${\displaystyle e_{i}(T)}$

is a projection. Moreover, let ν_i be the index of λ_i and

${\displaystyle f(z)=(z-\lambda _{i})^{\nu _{i}}.}$

The spectral mapping theorem tells us

${\displaystyle f(T)e_{i}(T)=(T-\lambda _{i})^{\nu _{i}}e_{i}(T)}$

has spectrum {0}. By property 1, f(T) can be directly computed in the Jordan form, and by inspection, we see that the operator f(T)e_i(T) is the zero matrix.

By property 3, f(T) e_i(T) = e_i(T) f(T). So e_i(T) is precisely the projection onto the subspace

${\displaystyle \operatorname {Ran} e_{i}(T)=\mathrm {Ker} (T-\lambda _{i})^{\nu _{i}}.}$

The relation

${\displaystyle \sum _{i}e_{i}=1}$

implies

${\displaystyle \mathbb {C} ^{n}=\bigoplus _{i}\;\operatorname {Ran} e_{i}(T)=\bigoplus _{i}\;\mathrm {Ker} (T-\lambda _{i})^{\nu _{i}}}$

where the index i runs through the distinct eigenvalues of T. This is the invariant subspace decomposition

${\displaystyle \mathbb {C} ^{n}=\bigoplus _{i}Y_{i}}$

given in a previous section. Each e_i(T) is the projection onto the subspace spanned by the Jordan chains corresponding to λ_i and along the subspaces spanned by the Jordan chains corresponding to λ_j for j ≠ i. In other words, e_i(T) = P(λ_i;T). This explicit identification of the operators e_i(T) in turn gives an explicit form of holomorphic functional calculus for matrices:

For all f ∈ Hol(T),

${\displaystyle f(T)=\sum _{\lambda _{i}\in \sigma (T)}\sum _{k=0}^{\nu _{i}-1}{\frac {f^{(k)}}{k!}}(T-\lambda _{i})^{k}e_{i}(T).}$

Notice that the expression of f(T) is a finite sum because, on each neighborhood of λ_i, we have chosen the Taylor series expansion of f centered at λ_i.

Let T be a bounded operator λ be an isolated point of σ(T). (As stated above, when T is compact, every point in its spectrum is an isolated point, except possibly the limit point 0.)

The point λ is called a pole of operator T with order ν if the resolvent function R_T defined by

${\displaystyle R_{T}(\lambda )=(\lambda -T)^{-1}}$

has a pole of order ν at λ.

We will show that, in the finite-dimensional case, the order of an eigenvalue coincides with its index. The result also holds for compact operators.

Consider the annular region A centered at the eigenvalue λ with sufficiently small radius ε such that the intersection of the open disc B_ε(λ) and σ(T) is {λ}. The resolvent function R_T is holomorphic on A. Extending a result from classical function theory, R_T has a Laurent series representation on A:

${\displaystyle R_{T}(z)=\sum _{-\infty }^{\infty }a_{m}(\lambda -z)^{m}}$

where

${\displaystyle a_{-m}=-{\frac {1}{2\pi i}}\int _{C}(\lambda -z)^{m-1}(z-T)^{-1}dz}$ and C is a small circle centered at λ.

By the previous discussion on the functional calculus,

${\displaystyle a_{-m}=-(\lambda -T)^{m-1}e_{\lambda }(T)}$ where ${\displaystyle e_{\lambda }}$ is 1 on ${\displaystyle B_{\epsilon }(\lambda )}$ and 0 elsewhere.

But we have shown that the smallest positive integer m such that

${\displaystyle a_{-m}\neq 0}$ and ${\displaystyle a_{-l}=0\;\;\forall \;l\geq m}$

is precisely the index of λ, ν(λ). In other words, the function R_T has a pole of order ν(λ) at λ.