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# Basis (linear algebra)

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 Title: Basis (linear algebra) Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Basis (linear algebra) The same vector can be represented in two different bases (purple and red arrows).
Basis vector redirects here. For basis vector in the context of crystals, see crystal structure. For a more general concept in physics, see frame of reference.

A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set.

Given a basis of a vector space V, every element of V can be expressed uniquely as a linear combination of basis vectors, whose coefficients are referred to as vector coordinates or components. A vector space can have several distinct sets of basis vectors; however each such set has the same number of elements, with this number being the dimension of the vector space.

## Definition 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.

A basis B of a vector space V over a field F is a linearly independent subset of V that spans V.

In more detail, suppose that B = { v1, …, vn } is a finite subset of a vector space V over a field F (such as the real or complex numbers R or C). Then B is a basis if it satisfies the following conditions:

• the linear independence property,
for all a1, …, anF, if a1v1 + … + anvn = 0, then necessarily a1 = … = an = 0; and
• the spanning property,
for every x in V it is possible to choose a1, …, anF such that x = a1v1 + … + anvn.

The numbers ai are called the coordinates of the vector x 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. To deal with infinite-dimensional spaces, we must generalize the above definition to include infinite basis sets. We therefore say that a set (finite or infinite) BV is a basis, if

• every finite subset B0B obeys the independence property shown above; and
• for every x in V it is possible to choose a1, …, anF and v1, …, vnB such that x = a1v1 + … + anvn.

The sums in the above definition are all finite because without additional structure the axioms of a vector space do not permit us to meaningfully speak about an infinite sum of vectors. Settings that permit infinite linear combinations allow alternative definitions of the basis concept: see Related notions below.

It is often convenient to list the basis vectors in a specific order, for example, when considering the transformation matrix of a linear map with respect to a basis. We then speak of an ordered basis, which we define to be a sequence (rather than a set) of linearly independent vectors that span V: see Ordered bases and coordinates below.

## Properties

Again, B denotes a subset of a vector space V. Then, B is a basis if and only if any of the following equivalent conditions are met:

• B is a minimal generating set of V, i.e., it is a generating set and no proper subset of B is also a generating set.
• B is a maximal set of linearly independent vectors, i.e., it is a linearly independent set but no other linearly independent set contains it as a proper subset.
• Every vector in V can be expressed as a linear combination of vectors in B in a unique way. If the basis is ordered (see Ordered bases and coordinates below) then the coefficients in this linear combination provide coordinates of the vector relative to the basis.

Every vector space has a basis. The proof of this requires the axiom of choice. All bases of a vector space have the same cardinality (number of elements), called the dimension of the vector space. This result is known as the dimension theorem, and requires the ultrafilter lemma, a strictly weaker form of the axiom of choice.

Also many vector sets can be attributed a standard basis which comprises both spanning and linearly independent vectors.

Standard bases for example:

In Rn, {e1, ..., en}, where ei is the ith column of the identity matrix.

In P2, where P2 is the set of all polynomials of degree at most 2, {1, x, x2} is the standard basis.

In M22, {M1,1, M1,2, M2,1, M2,2}, where M22 is the set of all 2×2 matrices. and Mm,n is the 2×2 matrix with a 1 in the m,n position and zeros everywhere else.

## Change of basis

Given a vector space V over a field F and suppose that {v1, ..., vn} and {α1, ..., αn} are two bases for V. By definition, if ξ is a vector in V then ξ = x1α1 + ... + xnαn for a unique choice of scalars x1, ..., xn in F called the coordinates of ξ relative to the ordered basis {α1, ..., αn}. The vector x = (x1, ..., xn) in Fn is called the coordinate tuple of ξ (relative to this basis). The unique linear map φ : FnV with φ(vj) = αj for j = 1, ..., n is called the coordinate isomorphism for V and the basis {α1, ..., αn}. Thus φ(x) = ξ if and only if ξ = x1α1 + ... + xnαn.

A set of vectors can be represented by a matrix of which each column consists of the components of the corresponding vector of the set. As a basis is a set of vectors, a basis can be given by a matrix of this kind. The change of basis of any object of the space is related to this matrix. For example, coordinate tuples change with its inverse.

## Examples

• Consider R2, the vector space of all coordinates (a, b) where both a and b are real numbers. Then a very natural and simple basis is simply the vectors e1 = (1,0) and e2 = (0,1): suppose that v = (a, b) is a vector in R2, then v = a(1,0) + b(0,1). But any two linearly independent vectors, like (1,1) and (−1,2), will also form a basis of R2.
• More generally, the vectors e1, e2, ..., en are linearly independent and generate Rn. Therefore, they form a basis for Rn and the dimension of Rn is n. This basis is called the standard basis.
• Let V be the real vector space generated by the functions et and e2t. These two functions are linearly independent, so they form a basis for V.
• Let R[x] denote the vector space of real polynomials; then (1, x, x2, ...) is a basis of R[x]. The dimension of R[x] is therefore equal to aleph-0.

## Extending to a basis

Let S be a subset of a vector space V. To extend S to a basis means to find a basis B that contains S as a subset. This can be done if and only if S is linearly independent. Almost always, there is more than one such B, except in rather special circumstances (i.e. S is already a basis, or S is empty and V has two elements).

A similar question is when does a subset S contain a basis. This occurs if and only if S spans V. In this case, S will usually contain several different bases.

## Example of alternative proofs

Often, a mathematical result can be proven in more than one way. Here, using three different proofs, we show that the vectors (1,1) and (−1,2) form a basis for R2.

### From the definition of basis

We have to prove that these two vectors are linearly independent and that they generate R2.

Part I: If two vectors v,w are linearly independent, then av + bw = 0 (a and b scalars) implies a = 0, b = 0.

To prove that they are linearly independent, suppose that there are numbers a,b such that:

a(1,1)+b(-1,2)=(0,0) \,

(i.e., they are linearly dependent). Then:

(a-b,a+2b)=(0,0) \,
and
a-b=0 \;
and
a+2b=0. \,

Subtracting the first equation from the second, we obtain:

3b=0 \;
so
b=0. \,

Adding this equation to the first equation then:

a=0. \,

Hence we have linear independence.

Part II: To prove that these two vectors generate R2, we have to let (a,b) be an arbitrary element of R2, and show that there exist numbers r,s ∈ R such that:

r(1,1)+s(-1,2)=(a,b). \,

Then we have to solve the equations:

r-s=a \,
r+2s=b. \,

Subtracting the first equation from the second, we get:

3s=b-a, \,
and then
s=(b-a)/3, \,
and finally
r=s+a=((b-a)/3)+a=(b+2a)/3. \,

### By the dimension theorem

Since (−1,2) is clearly not a multiple of (1,1) and since (1,1) is not the zero vector, these two vectors are linearly independent. Since the dimension of R2 is 2, the two vectors already form a basis of R2 without needing any extension.

### By the invertible matrix theorem

Simply compute the determinant

\det\begin{bmatrix}1&-1\\1&2\end{bmatrix}=3\neq0.

Since the above matrix has a nonzero determinant, its columns form a basis of R2. See: invertible matrix.

## Ordered bases and coordinates

A basis is just a linearly independent set of vectors with or without a given ordering. For many purposes it is convenient to work with an ordered basis. For example, when working with a coordinate representation of a vector it is customary to speak of the "first" or "second" coordinate, which makes sense only if an ordering is specified for the basis. For finite-dimensional vector spaces one typically indexes a basis {vi} by the first n integers. An ordered basis is also called a frame.

Suppose V is an n-dimensional vector space over a field F. A choice of an ordered basis for V is equivalent to a choice of a linear isomorphism φ from the coordinate space Fn to V.

Proof. The proof makes use of the fact that the standard basis of Fn is an ordered basis.

Suppose first that

φ : FnV

is a linear isomorphism. Define an ordered basis {vi} for V by

vi = φ(ei) for 1 ≤ in

where {ei} is the standard basis for Fn.

Conversely, given an ordered basis, consider the map defined by

φ(x) = x1v1 + x2v2 + ... + xnvn,

where x = x1e1 + x2e2 + ... + xnen is an element of Fn. It is not hard to check that φ is a linear isomorphism.

These two constructions are clearly inverse to each other. Thus ordered bases for V are in 1-1 correspondence with linear isomorphisms FnV.

The inverse of the linear isomorphism φ determined by an ordered basis {vi} equips V with coordinates: if, for a vector vV, φ−1(v) = (a1, a2,...,an) ∈ Fn, then the components aj = aj(v) are the coordinates of v in the sense that v = a1(v) v1 + a2(v) v2 + ... + an(v) vn.

The maps sending a vector v to the components aj(v) are linear maps from V to F, because of φ−1 is linear. Hence they are linear functionals. They form a basis for the dual space of V, called the dual basis.

## Related notions

### Analysis

In the context of infinite-dimensional vector spaces over the real or complex numbers, the term Hamel basis (named after orthogonal bases on Hilbert spaces, Schauder bases and Markushevich bases on normed linear spaces. The term Hamel basis is also commonly used to mean a basis for the real numbers R as a vector space over the field Q of rational numbers. (In this case, the dimension of R over Q is uncountable, specifically the continuum, the cardinal number 20.)

The common feature of the other notions is that they permit the taking of infinite linear combinations of the basis vectors in order to generate the space. This, of course, requires that infinite sums are meaningfully defined on these spaces, as is the case for topological vector spaces – a large class of vector spaces including e.g. Hilbert spaces, Banach spaces or Fréchet spaces.

The preference of other types of bases for infinite-dimensional spaces is justified by the fact that the Hamel basis becomes "too big" in Banach spaces: If X is an infinite-dimensional normed vector space which is complete (i.e. X is a Banach space), then any Hamel basis of X is necessarily uncountable. This is a consequence of the Baire category theorem. The completeness as well as infinite dimension are crucial assumptions in the previous claim. Indeed, finite-dimensional spaces have by definition finite bases and there are infinite-dimensional (non-complete) normed spaces which have countable Hamel bases. Consider c_{00}, the space of the sequences x=(x_n) of real numbers which have only finitely many non-zero elements, with the norm \|x\|=\sup_n |x_n|. Its standard basis, consisting of the sequences having only one non-zero element, which is equal to 1, is a countable Hamel basis.

#### Example

In the study of Fourier series, one learns that the functions {1} ∪ { sin(nx), cos(nx) : n = 1, 2, 3, ... } are an "orthogonal basis" of the (real or complex) vector space of all (real or complex valued) functions on the interval [0, 2π] that are square-integrable on this interval, i.e., functions f satisfying

\int_0^{2\pi} \left|f(x)\right|^2\,dx<\infty.

The functions {1} ∪ { sin(nx), cos(nx) : n = 1, 2, 3, ... } are linearly independent, and every function f that is square-integrable on [0, 2π] is an "infinite linear combination" of them, in the sense that

\lim_{n\rightarrow\infty}\int_0^{2\pi}\biggl|a_0+\sum_{k=1}^n \bigl(a_k\cos(kx)+b_k\sin(kx)\bigr)-f(x)\biggr|^2\,dx=0

for suitable (real or complex) coefficients ak, bk. But many square-integrable functions cannot be represented as finite linear combinations of these basis functions, which therefore do not comprise a Hamel basis. Every Hamel basis of this space is much bigger than this merely countably infinite set of functions. Hamel bases of spaces of this kind are typically not useful, whereas orthonormal bases of these spaces are essential in Fourier analysis.

### Affine geometry

The related notions of an affine space, projective space, convex set, and cone have related notions of affine basis (a basis for an n-dimensional affine space is n+1 points in general linear position), projective basis (essentially the same as an affine basis, this is n+1 points in general linear position, here in projective space), convex basis (the vertices of a polytope), and cone basis (points on the edges of a polygonal cone); see also a Hilbert basis (linear programming).

## Proof that every vector space has a basis

Let V be any vector space over some field F. Every vector space must contain at least one element: the zero vector 0.

Note that if V = {0}, then the empty set is a basis for V. Now we consider the case where V contains at least one nonzero element, say v.

Define the set X as all linear independent subsets of V. Note that since V contains the nonzero element v, the singleton subset L = {v} of V is necessarily linearly independent.

Hence the set X contains at least the subset L = {v}, and so X is nonempty.

We let X be partially ordered by inclusion: If L1 and L2 belong to X, we say that L1 ≤ L2 when L1 ⊂ L2. It is easy to check that (X, ≤) satisfies the definition of a partially ordered set.

We now note that if Y is a subset of X that is totally ordered by ≤, then the union LY of all the elements of Y (which are themselves certain subsets of V) is an upper bound for Y. To show this, it is necessary to verify both that a) LY belong to X, and that b) every element L of Y satisfies L ≤ LY. Both a) and b) are easy to check.

Now we apply Zorn's lemma, which asserts that because X is nonempty, and every totally ordered subset of the partially ordered set (X, ≤) has an upper bound, it follows that X has a maximal element. (In other words, there exists some element Lmax of X satisfying the condition that whenever Lmax ≤ L for some element L of X, then L = Lmax.)

Finally we claim that Lmax is a basis for V. Since Lmax belong to X, we already know that Lmax is a linearly independent subset of V.

Now suppose Lmax does not span V. Then there exists some vector w of V that cannot be expressed as a linearly combination of elements of Lmax (with coefficients in the field F). Note that such a vector w cannot be an element of Lmax.

Now consider the subset Lw of V defined by Lw = Lmax ∪ {w}. It is easy to see that a) Lmax ≤ Lw (since Lmax is a subset of Lw), and that b) Lmax ≠ Lw (because Lw contains the vector w that is not contained in Lmax).

But the combination of a) and b) above contradict the fact that Lmax is a maximal element of X, which we have already proved. This contradiction shows that the assumption that Lmax does not span V was not true.

Hence Lmax does span V. Since we also know that Lmax is linearly independent over the field F, this verifies that Lmax is a basis for V. Which proves that the arbitrary vector space V has a basis.

Note: This proof relies on Zorn's lemma, which is logically equivalent to the Axiom of Choice. It turns out that, conversely, the assumption that every vector space has a basis can be used to prove the Axiom of Choice. Thus the two assertions are logically equivalent.