Added to Favorites

Related Searches

Definitions

Nearby Words

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

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 = { v_{1}, …, v_{n} } 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 a
_{1}, …, a_{n}∈ F, if a_{1}v_{1}+ … + a_{n}v_{n}= 0, then necessarily a_{1}= … = a_{n}= 0; and

- the spanning property,

- for every x in V it is possible to choose a
_{1}, …, a_{n}∈ F such that x = a_{1}v_{1}+ … + a_{n}v_{n}.

The numbers a_{i} 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) B ⊂ V is a basis, if

- every finite subset B
_{0}⊆ B obeys the independence property shown above; and - for every x in V it is possible to choose a
_{1}, …, a_{n}∈ F and v_{1}, …, v_{n}∈ B such that x = a_{1}v_{1}+ … + a_{n}v_{n}.

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.

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. 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 R^{n} {E1,...,En} where En is the n-th column of the identity matrix which consists of all ones in the main diagonal and zeros everywhere else. This is because the columns of the identity matrix are linerly independent can always span a vector set by expressing it as a linear combination.

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

In M_{22} {M_{1,1},M_{1,2},M_{2,1},M_{2,2}} where M_{22} is the set of all 2x2 matrices. and M_{m,n} is the 2x2 matrix with a 1 in the m,n position and zeros everywhere else. This again is a standard basis since it is linearly independent and spanning.

- Consider R
^{2}, 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 e_{1}= (1,0) and e_{2}= (0,1): suppose that v = (a, b) is a vector in R^{2}, 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 R^{2}(see the section Proving that a finite set is a basis further down). - More generally, the vectors e
_{1}, e_{2}, ..., e_{n}are linearly independent and generate R^{n}. Therefore, they form a basis for R^{n}and the dimension of R^{n}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, x
^{2}, ...) is a basis of R[x]. The dimension of R[x] is therefore equal to aleph-0.

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.

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.

This does not work for infinite-dimensional vector spaces.

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

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

- $$

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

$$

and a-b=0 ;

$$

Subtracting the first equation from the second, we obtain:a+2b=0. ,

- $$

3b=0 ;

$$

And from the first equation then:b=0. ,

- $a=0.\; ,$

Part II: To prove that these two vectors generate R^{2}, we have to let (a,b) be an arbitrary element of R^{2}, and show that there exist numbers x,y such that:

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

- $x-y=a\; ,$

- $x+2y=b.\; ,$

- $$

3y=b-a, ,

- $$

y=(b-a)/3, ,

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

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 R^{2} is 2, the two vectors already form a basis of R^{2} without needing any extension.

Simply compute the determinant

- $detbegin\{bmatrix\}1\&-11\&2end\{bmatrix\}=3neq0.$

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. 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 F^{n} to V.

Proof. The proof makes use of the fact that the standard basis of F^{n} is an ordered basis.

Suppose first that

- φ : F
^{n}→ V

- v
_{i}= φ(e_{i}) for 1 ≤ i ≤ n

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

- φ(x) = x
_{1}v_{1}+ x_{2}v_{2}+ ... + x_{n}v_{n},

These two constructions are clearly inverse to each other. Thus ordered bases for V are in 1-1 correspondence with linear isomorphisms F^{n} → V.

The inverse of the linear isomorphism φ determined by an ordered basis {v_{i}} equips V with coordinates: if, for a vector v ∈ V, φ^{-1}(v) = (a_{1}, a_{2},...,a_{n}) ∈ F^{n}, then the components a_{j} = a_{j}(v) are the coordinates of v in the sense that v = a_{1}(v) v_{1} + a_{2}(v) v_{2} + ... + a_{n}(v) v_{n}.

The maps sending a vector v to the components a_{j}(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.

In the context of infinite dimensional vector spaces over real or complex numbers, the notion Hamel basis (named after Georg Hamel, or algebraic basis) is used to refer to a basis as defined in this article. This is because there exist different notions of bases better suited to the infinite dimensional setting.

The most important alternatives are orthogonal bases on Hilbert spaces, Schauder bases and Markushevich bases on normed linear spaces.

The common feature of the other notions is that they permit for taking infinite linear combinations of the basic 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 clase 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 an easy consequence of 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 basis and there are infinite dimensional (non-complete) normed spaces which have countable Hamel basis. Consider $c\_\{00\}$, the space of the sequences $x=(x\_n)$ of real numbers which have only finitely many non-zero coordinates, with the norm $|x|=sup\_n\; |x\_n|.$ The standard basis is its countable Hamel basis.

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

- $int\_0^\{2pi\}\; left|f(x)right|^2,dxmath>$

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

- $lim\_\{nrightarrowinfty\}int\_0^\{2pi\}biggl|a\_0+sum\_\{k=1\}^n\; bigl(a\_kcos(kx)+b\_ksin(kx)bigr)-f(x)biggr|^2,dx=0$

for suitable (real or complex) coefficients a_{k}, b_{k}. But most 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 of little (if any) interest, whereas orthonormal bases of these spaces are essential in Fourier analysis.

- MIT Linear Algebra Lecture on Bases at Google Video, from MIT OpenCourseWare

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday September 30, 2008 at 10:41:55 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday September 30, 2008 at 10:41:55 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2014 Dictionary.com, LLC. All rights reserved.