basis vector
- Basis vector redirects here. For basis vector in the context of crystals, see crystal structure.
Definition
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}.
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}.
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 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.
Examples
- Consider R^{2}, 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 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 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.
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.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.
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 R^{2}.From the definition of basis
We have to prove that these two vectors are linearly independent and that they generate R^{2}.Part I: To prove that they are linearly independent, suppose that there are numbers a,b such that:
- and and
- so
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:
- and then
- and finally
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 R^{2} is 2, the two vectors already form a basis of R^{2} without needing any extension.By the invertible matrix theorem
Simply compute the determinantOrdered bases and coordinates
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.
Related notions
The phrase Hamel basis (named after Georg Hamel, 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 a_{1}v_{1} + … + a_{n}v_{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.
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 (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 satisfyingThe 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
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.
External links
- MIT Linear Algebra Lecture on Bases at Google Video, from MIT OpenCourseWare
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