In mathematics, every vector space has a corresponding dual vector space (or just dual space for short) consisting of all linear forms on together with the vector space structure of pointwise addition and scalar multiplication by constants.
The dual space as defined above is defined for all vector spaces, and to avoid ambiguity may also be called the . When defined for a topological vector space, there is a subspace of the dual space, corresponding to continuous linear functionals, called the continuous dual space.
Dual vector spaces find application in many branches of mathematics that use vector spaces, such as in tensor analysis with finite-dimensional vector spaces. When applied to vector spaces of functions (which are typically infinite-dimensional), dual spaces are used to describe measures, distributions, and Hilbert spaces. Consequently, the dual space is an important concept in functional analysis.
Early terms for dual include polarer Raum [Hahn 1927], espace conjugué, adjoint space [Alaoglu 1940], and transponierter Raum [Schauder 1930] and [Banach 1932]. The term dual is due to Bourbaki.
Algebraic dual space
Given any vector space over a field , the (algebraic) dual space (alternatively denoted by p. 19, §3.1 or p. 20, §13)For used in this way, see An Introduction to Manifolds (). This notation is sometimes used when is reserved for some other meaning. For instance, in the above text, is frequently used to denote the codifferential of , so that represents the pullback of the form .
uses to denote the algebraic dual of . However, other authors use for the continuous dual, while reserving for the algebraic dual (). is defined as the set of all linear maps (linear functionals). Since linear maps are vector space homomorphisms, the dual space may be denoted . The dual space itself becomes a vector space over when equipped with an addition and scalar multiplication satisfying:
for all , , and . For example, if we express the vector space as the set of vectors (where and are real numbers), the function
is an element of , since it is -linear and maps vectors in to elements of .
Elements of the algebraic dual space are sometimes called covectors, one-forms, or linear forms.
The pairing of a functional in the dual space and an element of is sometimes denoted by a bracket: p. 21, §14 or . This pairing defines a nondegenerate bilinear mappingIn many areas, such as quantum mechanics, is reserved for a sesquilinear form defined on . called the natural pairing.
Dual set
Given a vector space and a basis on that space, one can define a linearly independent set in called the dual set. Each vector in corresponds to a unique vector in the dual set. This correspondence yields an injection .
If is finite-dimensional, the dual set is a basis, called the dual basis, and the injection is an isomorphism.
Finite-dimensional case If is finite-dimensional and has a basis, in , the dual basis is a set of linear functionals on , defined by the relation
for any choice of coefficients . In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations
where is the Kronecker delta symbol. This property is referred to as the bi-orthogonality property.
Consider the basis of V. Let be defined as the following:
.
These are a basis of because:
The are linear functionals, which map such as and to scalars and . Then also, and . Therefore, for . Suppose . Applying this functional on the basis vectors of successively, lead us to (The functional applied in results in ). Therefore, is linearly independent on . Lastly, consider . Then so . So generates . Hence, it is a basis of .
For example, if is , let its basis be chosen as . The basis vectors are not orthogonal to each other. Then, and are one-forms (functions that map a vector to a scalar) such that , , , and . (Note: The superscript here is the index, not an exponent.) This system of equations can be expressed using matrix notation as
Solving for the unknown values in the first matrix shows the dual basis to be . Because and are functionals, they can be rewritten as and .
In general, when is , if is a matrix whose columns are the basis vectors and is a matrix whose columns are the dual basis vectors, then
where is the identity matrix of order . The biorthogonality property of these two basis sets allows any point to be represented as
even when the basis vectors are not orthogonal to each other. Strictly speaking, the above statement only makes sense once the inner product and the corresponding duality pairing are introduced, as described below in .
In particular, can be interpreted as the space of columns of real numbers, its dual space is typically written as the space of rows of real numbers. Such a row acts on as a linear functional by ordinary matrix multiplication. This is because a functional maps every -vector into a real number . Then, seeing this functional as a matrix , and as an matrix, and a matrix (trivially, a real number) respectively, if then, by dimension reasons, must be a matrix; that is, must be a row vector.
If consists of the space of geometrical vectors in the plane, then the level curves of an element of form a family of parallel lines in , because the range is 1-dimensional, so that every point in the range is a multiple of any one nonzero element. So an element of can be intuitively thought of as a particular family of parallel lines covering the plane. To compute the value of a functional on a given vector, it suffices to determine which of the lines the vector lies on. Informally, this "counts" how many lines the vector crosses. More generally, if is a vector space of any dimension, then the level sets of a linear functional in are parallel hyperplanes in , and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.
Infinite-dimensional case
If is not finite-dimensional but has a basisSeveral assertions in this article require the axiom of choice for their justification. The axiom of choice is needed to show that an arbitrary vector space has a basis: in particular it is needed to show that has a basis. It is also needed to show that the dual of an infinite-dimensional vector space is nonzero, and hence that the natural map from to its double dual is injective. indexed by an infinite set , then the same construction as in the finite-dimensional case yields linearly independent elements () of the dual space, but they will not form a basis.
For instance, consider the space , whose elements are those sequences of real numbers that contain only finitely many non-zero entries, which has a basis indexed by the natural numbers . For , is the sequence consisting of all zeroes except in the -th position, which is 1. The dual space of is (isomorphic to) , the space of all sequences of real numbers: each real sequence defines a function where the element of is sent to the number
which is a finite sum because there are only finitely many nonzero . The dimension of is countably infinite, whereas does not have a countable basis.
This observation generalizes to any infinite-dimensional vector space over any field : a choice of basis identifies with the space of functions such that is nonzero for only finitely many , where such a function is identified with the vector
in (the sum is finite by the assumption on , and any may be written uniquely in this way by the definition of the basis).
The dual space of may then be identified with the space of all functions from to : a linear functional on is uniquely determined by the values it takes on the basis of , and any function (with ) defines a linear functional on by
Again, the sum is finite because is nonzero for only finitely many .
The set may be identified (essentially by definition) with the direct sum of infinitely many copies of (viewed as a 1-dimensional vector space over itself) indexed by , i.e. there are linear isomorphisms
On the other hand, is (again by definition), the direct product of infinitely many copies of indexed by , and so the identification
is a special case of a general result relating direct sums (of modules) to direct products.
If a vector space is not finite-dimensional, then its (algebraic) dual space is always of larger dimension (as a cardinal number) than the original vector space. This is in contrast to the case of the continuous dual space, discussed below, which may be isomorphic to the original vector space even if the latter is infinite-dimensional.
The proof of this inequality between dimensions results from the following.
If is an infinite-dimensional -vector space, the arithmetical properties of cardinal numbers implies that
where cardinalities are denoted as absolute values. For proving that it suffices to prove that which can be done with an argument similar to Cantor's diagonal argument. The exact dimension of the dual is given by the Erdős–Kaplansky theorem.
Bilinear products and dual spaces
If is finite-dimensional, then is isomorphic to . But there is in general no natural isomorphism between these two spaces. Any bilinear form on gives a mapping of into its dual space via
where the right hand side is defined as the functional on taking each to . In other words, the bilinear form determines a linear mapping
defined by
If the bilinear form is nondegenerate, then this is an isomorphism onto a subspace of . If is finite-dimensional, then this is an isomorphism onto all of . Conversely, any isomorphism from to a subspace of (resp., all of if is finite dimensional) defines a unique nondegenerate bilinear form on by
Thus there is a one-to-one correspondence between isomorphisms of to a subspace of (resp., all of) and nondegenerate bilinear forms on .
If the vector space is over the complex field, then sometimes it is more natural to consider sesquilinear forms instead of bilinear forms. In that case, a given sesquilinear form determines an isomorphism of with the complex conjugate of the dual space
The conjugate of the dual space can be identified with the set of all additive complex-valued functionals such that
Injection into the double-dual There is a natural homomorphism from into the double dual , defined by for all . In other words, if is the evaluation map defined by , then is defined as the map . This map is always injective; and it is always an isomorphism if is finite-dimensional. pp. 25, 28 Indeed, the isomorphism of a finite-dimensional vector space with its double dual is an archetypal example of a natural isomorphism. Infinite-dimensional Hilbert spaces are not isomorphic to their algebraic double duals, but instead to their continuous double duals.
Transpose of a linear map
If is a linear map, then the transpose (or dual) is defined by
for every . The resulting functional in is called the pullback of along .
The following identity holds for all and :
where the bracket on the left is the natural pairing of with its dual space, and that on the right is the natural pairing of with its dual. This identity characterizes the transpose, §44 and is formally similar to the definition of the adjoint.
The assignment produces an injective linear map between the space of linear operators from to and the space of linear operators from to ; this homomorphism is an isomorphism if and only if is finite-dimensional. If then the space of linear maps is actually an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that . In the language of category theory, taking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over to itself. It is possible to identify with using the natural injection into the double dual.