Basis (linear algebra)

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Basis vector redirects here. For basis vector in the context of crystals, see crystal structure.

In linear algebra, a basis is a set of vectors that, in a linear combination, can represent every vector in a given vector space. In other words, a basis is a linearly independent spanning set.

1 Definition
2 Properties
3 Examples
4 Basis extension
5 Proving that a set is a basis
6 Example of alternative proofs
7 Ordered basis
8 Related notions
- 8.1 Example
9 See also

[edit] Definition

This picture illustrates the standard basis in R²

A basis B of a vector space V is a linearly independent subset of V that spans (or generates) V.

In more detail, suppose that B = { v₁, …, v_n } is a finite subset of a vector space V over a field F. Then B is a basis if it satisfies the following conditions:

the linear independence property,

for all a₁, …, a_n ∈ F, if a₁v₁ + … + a_nv_n = 0, then necessarily a₁ = … = a_n = 0; and

the spanning property,

for every x in V it is possible to choose a₁, …, a_n ∈ F such that x = a₁v₁ + … + a_nv_n.

A vector space that admits 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) B ⊂ V is a basis, if

every finite subset B₀ ⊆ B obeys the independence property shown above; and
for every x in V it is possible to choose a₁, …, a_n ∈ F and v₁, …, v_n ∈ B such that x = a₁v₁ + … + a_nv_n.

The axioms of a vector space do not permit us to meaningfully speak about an infinite sum of vectors. That is why the sums in the above definition are all finite. Settings that permit infinite linear combinations allow alternative definitions of the basis concept: see Related notions below.

When we want to describe the matrix of a linear transformation and in some other situations, it is convenient to list the basis vectors in a specific order. 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.

Another way to think about this is:

every ordered basis of a finite-dimensional vector space V corresponds to a linear isomorphism from f:Rⁿ → V, and vice versa.

Proof. Given such an isomorphism, the sequence f(e₁), ... , f(e_n), where e₁, ... e_n is the standard basis of Rⁿ, is an ordered basis of V. Conversely, given an ordered basis, the mapping defined by

f(a) = a₁v₁ + … + a_nv_n,

where a= a₁e₁ + … + a_ne_n is an element of Rⁿ, is necessarily a linear isomorphism.

[edit] 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 but no proper subset of B is.
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. For a given vector, the corresponding coefficients in the linear combination are the coordinates of the vector relative to the basis.

The theorem that every vector space has a basis is implied by the well-ordering theorem, or any other equivalent of the axiom of choice. (Proof: Well-order the elements of the vector space. Create the subset of all elements not linearly dependent on their predecessors. This is easily shown to be a basis). The converse is also true. All bases of a vector space have the same cardinality (number of elements), called the dimension of the vector space. The latter result is known as the dimension theorem, and requires the ultrafilter lemma, a strictly weaker form of the axiom of choice.

[edit] Examples

Consider R², the vector space of all co-ordinates (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 (see the section Proving that a set is a basis further down).

More generally, the vectors e₁, e₂, ..., e_n 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 e^t and e^2t. 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.

[edit] Basis extension

Between any linearly independent set and any generating set there is a basis. More formally: if L is a linearly independent set in the vector space V and G is a generating set of V containing L, then there exists a basis of V that contains L and is contained in G. In particular (taking G = V), any linearly independent set L can be "extended" to form a basis of V. These extensions are not unique.

[edit] Proving that a set is a basis

To prove that a set B is a basis for a (finite-dimensional) vector space V, it is sufficient to show that the number of elements in B equals the dimension of V, and one of the following:

B is linearly independent, or
span(B) = V.

[edit] 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.

[edit] From the definition of basis

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

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

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

Then:

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

and

$a-b=0 \;$

and

$a+2b=0. \,$

Subtracting the first equation from the second, we obtain:

$3b=0 \;$

$b=0. \,$

And from the first equation then:

$a=0. \,$

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 x,y such that:

$x(1,1)+y(-1,2)=(a,b). \,$

Then we have to solve the equations:

$x-y=a \,$

$x+2y=b. \,$

Subtracting the first equation from the second, we get:

$3y=b-a, \,$

and then

$y=(b-a)/3, \,$

and finally

$x=y+a=((b-a)/3)+a=(b+2a)/3. \,$

[edit] 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.

[edit] 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.

[edit] Ordered basis

A basis is just a set of vectors with no 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 {v_i} by the first n integers.

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 from the coordinate space Fⁿ, with its standard basis, to V. To see this, let

A : Fⁿ → V

be a linear isomorphism. Define an ordered basis {v_i} for V by

v_i = A(e_i) for 1 ≤ i ≤ n

where {e_i} is the standard basis for Fⁿ. Conversely, given any ordered basis {v_i} for V define a linear map A : Fⁿ → V by

$A(x) = \sum_{i=1}^n x_i v_i$

It is not hard to check that A is an isomorphism. Thus ordered bases for V are in 1-1 correspondence with linear isomorphisms Fⁿ → V.

[edit] Related notions

The phrase Hamel basis (or algebraic basis) is sometimes used to refer to a basis as defined in this article, where the number of terms in the linear combination a₁v₁ + … + a_nv_n is always finite.

In Hilbert spaces and other Banach spaces, there is a need to work with linear combinations of infinitely many vectors. In an infinite-dimensional Hilbert space, a set of vectors orthogonal to each other can never span the whole space via their finite linear combinations. What is called an orthonormal basis is a set of mutually orthogonal unit vectors that "span" the space via sometimes-infinite linear combinations. Except in the finite-dimensional case, this concept is not purely algebraic, and is distinct from a Hamel basis; it is also more generally useful. An orthonormal basis of an infinite-dimensional Hilbert space is therefore not a Hamel basis.

In topological vector spaces, quite generally, one may define infinite sums (infinite series) and express elements of the space as certain infinite linear combinations of other elements. To keep clear the distinction of bases using finite and infinite combination, the former ones are called Hamel bases and the latter ones Schauder bases, if the context requires it. The corresponding dimensions are also known as Hamel dimension and Schauder dimension.

[edit] Example

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

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

These functions are linearly independent, and every function that is quadratically integrable on that interval is an "infinite linear combination" of them. That means that

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

for suitable coefficients a_k, b_k. But most quadratically 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 of little if any interest; orthonormal bases of these spaces are important to Fourier analysis.

Category: Linear algebra