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Basis (linear algebra)
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In linear algebra, a basis is a set of vectors that, in a linear combination, can represent every vector in a given vector space, and such that no element of the set can be represented as a linear combination of the others. In other words, a basis is a linearly independent spanning set.
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 = 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:
linear independence property,
for all a1, …, an ? F, if a1v1 + … + anvn = 0, then necessarily a1 = … = an = 0; and
spanning property,
for every x in V it is possible to choose a1, …, an ? F 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) 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, …, an ? F and v1, …, vn ? B 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.
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 (see the section Proving that a finite 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.
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.
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 finite 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.
This does not work for infinite-dimensional vector spaces.
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: To prove that they are linearly independent, suppose that there are numbers a,b such that:
Then:
- and and
Subtracting the first equation from the second, we obtain:
- so
And from the first equation then:
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:
Then we have to solve the equations:
Subtracting the first equation from the second, we get:
- 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 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
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 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 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 f 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 : Fn ? V
is a linear isomorphism. Define an ordered basis for V by
- vi = φ(ei) for 1 = i = n
where 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 f 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 Fn ? V.
The inverse of the linear isomorphism f determined by an ordered basis equips V with coordinates: if, for a vector v ? V, f-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 f-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 a1v1 + … + anvn 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 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.
Banach SpacesAs a result of the Baire category theorem if a Banach space has infinite Hamel dimension, it must have an uncountable Hamel basis. The Banach space could not be written as a countable union of the finite spanning sets. Thus any Banach space X of cardinality strictly greater than the continuum ( ) must have an uncountable Hamel basis. An example of such a space is the set of bounded functions from [0,1] to equipped with the uniform norm. This is a Banach space that contains the subset (see characteristic function). This last subset has cardinality , so the Banach space has cardinality strictly greater than and hence has infinite Hamel dimension. From the existence of such a Banach space X, one can take the guaranteed infinite Hamel basis, normalize and then construct a linear bijection T from X onto itself so that the graph of T is not closed (even though the image of T being X must be closed). Such T could not be bounded and would have to be unbounded on say a countable sub-collection of the Hamel basis. This is an instance in which the closed graph theorem is not applicable as T would be unbounded.
ExampleIn the study of Fourier series, one learns that the functions ? are an "orthogonal basis" of the (real or complex) vector space of all (real or complex valued) functions on the interval [0, 2p] that are square-integrable on this interval, i.e., functions f satisfying
The functions ? are linearly independent, and every function f that is square-integrable on [0, 2p] is an "infinite linear combination" of them, in the sense that
for suitable (real or complex) coefficients ak, bk. 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.
See also
External links- at Google Video, from MIT OpenCourseWare
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