Basis (linear algebra)

(Redirected from Basis of a vector space)

In linear algebra, a basis is a minimum set of vectors that, when combined, can address every vector in a given space. More precisely, a basis for a vector space V is a set of linearly independent vectors that span all of V.

A subset B of V is a basis for V if it satisfies any of the following equivalent conditions:

1. B is both a set of linearly independent vectors and a generating set of V.
2. B is a minimal generating set of V, i.e. it is a generating set but no proper subset of B is.
3. B is a maximal set of linearly independent vectors, i.e. it is a linearly independent set but no proper superset is.
4. every vector in V can be expressed as a linear combination of vectors in B in a unique way.

Recall that a set B is a generating set of V if every vector in V is a linear combination of vectors in B. This definition includes a finiteness condition: a linear combination is always a finite sum of the form a1v1 + ... + anvn.

Importantly, one can show that every vector space has a basis. For spaces that cannot be finitely generated, Zorn's lemma is needed for the proof. Also, 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 for vector spaces.

 Contents

Examples

Example 0: Suppose scalars { 0, 1, 2, ... 8, 9, }. Then a set of vectors { 1, 10, 100, ... } form a basis for a set of natural numbers. For example, a number 123 would be constructed as follows:

123 = 100(1) + 10(2) + 1(3) + 1000(0) + 10000(0) + ... where 100, 10, 1 and the rest of numbers followed by zeros are basis vectors, and 1, 2, 3 and 0 are scalars.

Examples of proofs

Example I: Show that the vectors (1,1) and (-1,2) form a basis for R2.

Proof: We have to prove that these 2 vectors are both 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:

[itex] a(1,1)+b(-1,2)=(0,0). \,[itex]

Then:

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

Subtracting the first equation from the second, we obtain:

[itex]
3b=0 \;[itex]
so
[itex] b=0. \,[itex]

And from the first equation then:

[itex] a=0. \,[itex]

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:

[itex] x(1,1)+y(-1,2)=(a,b). \,[itex]

Then we have to solve the equations:

[itex] x-y=a \,[itex]
[itex] x+2y=b. \,[itex]

Subtracting the first equation from the second, we get:

[itex]
3y=b-a, \,[itex]
and then
[itex]
y=(b-a)/3, \,[itex]
and finally
[itex] x=y+a=((b-a)/3)+a. \,[itex]

Example II: It is easy to show that 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.

Example III: Let W be the real vector space generated by the functions et, e2t. The two functions are linearly independent, and therefore form a basis for W.

Example IV: 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.

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.

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 a ordering is specified for the basis. For finite-dimensional vector spaces one typically indexes a basis {vi</sup>} by the first n integers and orders the basis elements so that v1 < v2 < … .

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 : FnV

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

vi = A(ei) for 1 ≤ in

where {ei} is the standard basis for Fn. Conversely, given any ordered basis {vi} for V define a linear map A : FnV by

[itex]A(x) = \sum_{i=1}^n x_i v_i[itex]

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

Related notions

The phrase Hamel basis is sometimes used to refer to a basis as defined above, in which the fact that all linear combinations are finite is crucial. A set B is a Hamel basis of a vector space V if every member of V is a linear combination of just finitely many members of B.

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, vector space bases are called Hamel bases if the context requires it, and the vector space dimension is also known as Hamel dimension.

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

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

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

[itex]\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[itex]

for suitable coefficients ak, bk. 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.

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