Basis (linear algebra)

A basis B of a vector space V over a field F (such as the real numbers R or the complex numbers C) is a linearly independent subset of V that spans V. This means that a subset B of V is a basis if it satisfies the two following conditions:

linear independence

for every finite subset ${\displaystyle \{\mathbf {v} _{1},\dotsc ,\mathbf {v} _{m}\}}$ of B, if ${\displaystyle c_{1}\mathbf {v} _{1}+\cdots +c_{m}\mathbf {v} _{m}=\mathbf {0} }$ for some ${\displaystyle c_{1},\dotsc ,c_{m}}$ in F, then ${\displaystyle c_{1}=\cdots =c_{m}=0}$;

spanning property

for every vector v in V, one can choose ${\displaystyle a_{1},\dotsc ,a_{n}}$ in F and ${\displaystyle \mathbf {v} _{1},\dotsc ,\mathbf {v} _{n}}$ in B such that ${\displaystyle \mathbf {v} =a_{1}\mathbf {v} _{1}+\cdots +a_{n}\mathbf {v} _{n}}$.

The scalars ${\displaystyle a_{i}}$ are called the coordinates of the vector v with respect to the basis B, and by the first property they are uniquely determined.

A vector space that has a finite basis is called finite-dimensional. In this case, the finite subset can be taken as B itself to check for linear independence in the above definition.

It is often convenient or even necessary to have an ordering on the basis vectors, for example, when discussing orientation, or when one considers the scalar coefficients of a vector with respect to a basis without referring explicitly to the basis elements. In this case, the ordering is necessary for associating each coefficient to the corresponding basis element. This ordering can be done by numbering the basis elements. In order to emphasize that an order has been chosen, one speaks of an ordered basis, which is therefore not simply an unstructured set, but a sequence, an indexed family, or similar; see § Ordered bases and coordinates below.

This picture illustrates the standard basis in R². The blue and orange vectors are the elements of the basis; the green vector can be given in terms of the basis vectors, and so is linearly dependent upon them.

The set R² of the ordered pairs of real numbers is a vector space under the operations of component-wise addition

$${\displaystyle (a,b)+(c,d)=(a+c,b+d)}$$

and scalar multiplication

$${\displaystyle \lambda (a,b)=(\lambda a,\lambda b),}$$

where ${\displaystyle \lambda }$ is any real number. A simple basis of this vector space consists of the two vectors e₁ = (1, 0) and e₂ = (0, 1). These vectors form a basis (called the standard basis) because any vector v = (a, b) of R² may be uniquely written as

$${\displaystyle \mathbf {v} =a\mathbf {e} _{1}+b\mathbf {e} _{2}.}$$

Any other pair of linearly independent vectors of R², such as (1, 1) and (−1, 2), forms also a basis of R². More generally, if F is a field, the set ${\displaystyle F^{n}}$ of n-tuples of elements of F is a vector space for similarly defined addition and scalar multiplication. Let

$${\displaystyle \mathbf {e} _{i}=(0,\ldots ,0,1,0,\ldots ,0)}$$

be the n-tuple with all components equal to 0, except the ith, which is 1. Then ${\displaystyle \mathbf {e} _{1},\ldots ,\mathbf {e} _{n}}$ is a basis of ${\displaystyle F^{n},}$ which is called the standard basis of ${\displaystyle F^{n}.}$ A different flavor of example is given by polynomial rings. If F is a field, the collection F[X] of all polynomials in one indeterminate X with coefficients in F is an F-vector space. One basis for this space is the monomial basis B, consisting of all monomials:

$${\displaystyle B=\{1,X,X^{2},\ldots \}.}$$

Any set of polynomials such that there is exactly one polynomial of each degree (such as the Bernstein basis polynomials or Chebyshev polynomials) is also a basis. (Such a set of polynomials is called a polynomial sequence.) But there are also many bases for F[X] that are not of this form.

Many properties of finite bases result from the Steinitz exchange lemma, which states that, for any vector space V, given a finite spanning set S and a linearly independent set L of n elements of V, one may replace n well-chosen elements of S by the elements of L to get a spanning set containing L, having its other elements in S, and having the same number of elements as S.

Most properties resulting from the Steinitz exchange lemma remain true when there is no finite spanning set, but their proofs in the infinite case generally require the axiom of choice or a weaker form of it, such as the ultrafilter lemma.

If V is a vector space over a field F, then:

• If L is a linearly independent subset of a spanning set S ⊆ V, then there is a basis B such that
  $${\displaystyle L\subseteq B\subseteq S.}$$

  
• V has a basis (this is the preceding property with L being the empty set, and S = V).
• All bases of V have the same cardinality, which is called the dimension of V. This is the dimension theorem.
• A generating set S is a basis of V if and only if it is minimal, that is, no proper subset of S is also a generating set of V.
• A linearly independent set L is a basis if and only if it is maximal, that is, it is not a proper subset of any linearly independent set.

If V is a vector space of dimension n, then:

• A subset of V with n elements is a basis if and only if it is linearly independent.
• A subset of V with n elements is a basis if and only if it is a spanning set of V.

Let V be a vector space of finite dimension n over a field F, and

$${\displaystyle B=\{\mathbf {b} _{1},\ldots ,\mathbf {b} _{n}\}}$$

be a basis of V. By definition of a basis, every v in V may be written, in a unique way, as

$${\displaystyle \mathbf {v} =\lambda _{1}\mathbf {b} _{1}+\cdots +\lambda _{n}\mathbf {b} _{n},}$$

where the coefficients ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}}$ are scalars (that is, elements of F), which are called the coordinates of v over B. However, if one talks of the set of the coefficients, one loses the correspondence between coefficients and basis elements, and several vectors may have the same set of coefficients. For example, ${\displaystyle 3\mathbf {b} _{1}+2\mathbf {b} _{2}}$ and ${\displaystyle 2\mathbf {b} _{1}+3\mathbf {b} _{2}}$ have the same set of coefficients {2, 3}, and are different. It is therefore often convenient to work with an ordered basis; this is typically done by indexing the basis elements by the first natural numbers. Then, the coordinates of a vector form a sequence similarly indexed, and a vector is completely characterized by the sequence of coordinates. An ordered basis is also called a frame, a word commonly used, in various contexts, for referring to a sequence of data allowing defining coordinates.

Let, as usual, ${\displaystyle F^{n}}$ be the set of the n-tuples of elements of F. This set is an F-vector space, with addition and scalar multiplication defined component-wise. The map

$${\displaystyle \varphi$$ :(\lambda _{1},\ldots ,\lambda _{n})\mapsto \lambda _{1}\mathbf {b} _{1}+\cdots +\lambda _{n}\mathbf {b} _{n}}

is a linear isomorphism from the vector space ${\displaystyle F^{n}}$ onto V. In other words, ${\displaystyle F^{n}}$ is the coordinate space of V, and the n-tuple ${\displaystyle \varphi ^{-1}(\mathbf {v} )}$ is the coordinate vector of v.

The inverse image by ${\displaystyle \varphi }$ of ${\displaystyle \mathbf {b} _{i}}$ is the n-tuple ${\displaystyle \mathbf {e} _{i}}$ all of whose components are 0, except the ith that is 1. The ${\displaystyle \mathbf {e} _{i}}$ form an ordered basis of ${\displaystyle F^{n}}$, which is called its standard basis or canonical basis. The ordered basis B is the image by ${\displaystyle \varphi }$ of the canonical basis of ${\displaystyle F^{n}}$.

It follows from what precedes that every ordered basis is the image by a linear isomorphism of the canonical basis of ${\displaystyle F^{n}}$, and that every linear isomorphism from ${\displaystyle F^{n}}$ onto V may be defined as the isomorphism that maps the canonical basis of ${\displaystyle F^{n}}$ onto a given ordered basis of V. In other words it is equivalent to define an ordered basis of V, or a linear isomorphism from ${\displaystyle F^{n}}$ onto V.

Let V be a vector space of dimension n over a field F. Given two (ordered) bases ${\displaystyle B_{\text{old}}=(\mathbf {v} _{1},\ldots ,\mathbf {v} _{n})}$ and ${\displaystyle B_{\text{new}}=(\mathbf {w} _{1},\ldots ,\mathbf {w} _{n})}$ of V, it is often useful to express the coordinates of a vector x with respect to ${\displaystyle B_{\mathrm {old} }}$ in terms of the coordinates with respect to ${\displaystyle B_{\mathrm {new} }.}$ This can be done by the change-of-basis formula, that is described below. The subscripts "old" and "new" have been chosen because it is customary to refer to ${\displaystyle B_{\mathrm {old} }}$ and ${\displaystyle B_{\mathrm {new} }}$ as the old basis and the new basis, respectively. It is useful to describe the old coordinates in terms of the new ones, because, in general, one has expressions involving the old coordinates, and if one wants to obtain equivalent expressions in terms of the new coordinates; this is obtained by replacing the old coordinates by their expressions in terms of the new coordinates.

Typically, the new basis vectors are given by their coordinates over the old basis, that is,

$${\displaystyle \mathbf {w} _{j}=\sum _{i=1}^{n}a_{i,j}\mathbf {v} _{i}.}$$

If ${\displaystyle (x_{1},\ldots ,x_{n})}$ and ${\displaystyle (y_{1},\ldots ,y_{n})}$ are the coordinates of a vector x over the old and the new basis respectively, the change-of-basis formula is

$${\displaystyle x_{i}=\sum _{j=1}^{n}a_{i,j}y_{j},}$$

for i = 1, ..., n.

This formula may be concisely written in matrix notation. Let A be the matrix of the ${\displaystyle a_{i,j}}$, and

$${\displaystyle X={\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}}\quad {\text{and}}\quad Y={\begin{bmatrix}y_{1}\\\vdots \\y_{n}\end{bmatrix}}}$$

be the column vectors of the coordinates of v in the old and the new basis respectively, then the formula for changing coordinates is

$${\displaystyle X=AY.}$$

The formula can be proven by considering the decomposition of the vector x on the two bases: one has

$${\displaystyle \mathbf {x} =\sum _{i=1}^{n}x_{i}\mathbf {v} _{i},}$$

and

$${\displaystyle \mathbf {x} =\sum _{j=1}^{n}y_{j}\mathbf {w} _{j}=\sum _{j=1}^{n}y_{j}\sum _{i=1}^{n}a_{i,j}\mathbf {v} _{i}=\sum _{i=1}^{n}\left(\sum _{j=1}^{n}a_{i,j}y_{j}\right)\mathbf {v} _{i}.}$$

The change-of-basis formula results then from the uniqueness of the decomposition of a vector over a basis, here ${\displaystyle B_{\text{old}}}$; that is

$${\displaystyle x_{i}=\sum _{j=1}^{n}a_{i,j}y_{j},}$$

for i = 1, ..., n.

Let V be any vector space over some field F. Let X be the set of all linearly independent subsets of V.

The set X is nonempty since the empty set is an independent subset of V, and it is partially ordered by inclusion, which is denoted, as usual, by ⊆.

Let Y be a subset of X that is totally ordered by ⊆, and let L_Y be the union of all the elements of Y (which are themselves certain subsets of V).

Since (Y, ⊆) is totally ordered, every finite subset of L_Y is a subset of an element of Y, which is a linearly independent subset of V, and hence L_Y is linearly independent. Thus L_Y is an element of X. Therefore, L_Y is an upper bound for Y in (X, ⊆): it is an element of X, that contains every element of Y.

As X is nonempty, and every totally ordered subset of (X, ⊆) has an upper bound in X, Zorn's lemma asserts that X has a maximal element. In other words, there exists some element L_max of X satisfying the condition that whenever L_max ⊆ L for some element L of X, then L = L_max.

It remains to prove that L_max is a basis of V. Since L_max belongs to X, we already know that L_max is a linearly independent subset of V.

If there were some vector w of V that is not in the span of L_max, then w would not be an element of L_max either. Let L_w = L_max ∪ {w}. This set is an element of X, that is, it is a linearly independent subset of V (because w is not in the span of L_max, and L_max is independent). As L_max ⊆ L_w, and L_max ≠ L_w (because L_w contains the vector w that is not contained in L_max), this contradicts the maximality of L_max. Thus this shows that L_max spans V.

Hence L_max is linearly independent and spans V. It is thus a basis of V, and this proves that every vector space has a basis.

This proof relies on Zorn's lemma, which is equivalent to the axiom of choice. Conversely, it has been proved that if every vector space has a basis, then the axiom of choice is true.[9] Thus the two assertions are equivalent.

• Basis of a matroid
• Basis of a linear program
• Change of basis – Coordinate change in linear algebra
• Frame of a vector space
• Spherical basis – Basis used to express spherical tensors

• Instructional videos from Khan Academy
  • Introduction to bases of subspaces
  • Proof that any subspace basis has same number of elements
• "Linear combinations, span, and basis vectors". Essence of linear algebra. August 6, 2016. Archived from the original on 2021-11-17 – via YouTube.
• "Basis", Encyclopedia of Mathematics, EMS Press, 2001 [1994]