In mathematics, the existence of a dual vector space reflects in an abstract way the relationship between row vectors (1×n) and column vectors (n×1). The construction can also take place for infinite-dimensional spaces and gives rise to important ways of looking at measures, distributions, and Hilbert space. The use of the dual space in some fashion is thus characteristic of functional analysis. It is also inherent in the Fourier transform.

Algebraic dual space

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Given any vector space V over some field F, we define the dual space V* to be the set of all linear functionals on V, i.e., scalar-valued linear transformations on V (in this context, a "scalar" is a member of the base-field F). V* itself becomes a vector space over F under the following definition of addition and scalar multiplication:

$(\backslash phi\; +\; \backslash psi\; )(\; x\; )\; =\; \backslash phi\; (\; x\; )\; +\; \backslash psi\; (\; x\; )\; \backslash ,$

$(\; a\; \backslash phi\; )\; (\; x\; )\; =\; a\; \backslash phi\; (\; x\; )\; \backslash ,$

for all φ, ψ in V*, a in F and x in V. In the language of tensors, elements of V are sometimes called covariant vectors, and elements of V*, contravariant vectors, covectors or one-forms.

Examples

If the dimension of V is finite, then V* has the same dimension as V; if {e₁,...,e_n} is a basis for V, then the associated dual basis {e¹,...,eⁿ} of V* is given by

$$

e^i (e_j)= \left\{\begin{matrix} 1, & \mbox{if }i = j \\ 0, & \mbox{if } i \ne j \end{matrix}\right.

In the case of R², its basis is B={e₁=(1,0),e₂=(0,1)}.Then, e¹ is a one-form (function which maps a vector to a scalar) such that e¹(e₁)=1, and e¹(e₂)=0. Similarity for e². (Note: The superscript here is an index, not an exponent.)

Concretely, if we interpret Rⁿ as space of columns of n real numbers, its dual space is typically written as the space of rows of n real numbers. Such a row acts on Rⁿ as a linear functional by ordinary matrix multiplication.

If V consists of the space of geometrical vectors (arrows) in the plane, then the elements of the dual V* can be intuitively represented as collections of parallel lines. Such a collection of lines can be applied to a vector to yield a number in the following way: one counts how many of the lines the vector crosses.

If V is infinite-dimensional, then the above construction of eⁱ does not produce a basis for V* and the dimension of V* is greater than that of V. Consider for instance the space R^(ω), whose elements are those sequences of real numbers which have only finitely many non-zero entries (dimension is countably infinite). The dual of this space is R^ω, the space of all sequences of real numbers (dimension is uncountably infinite). Such a sequence (a_n) is applied to an element (x_n) of R^(ω) to give the number ∑_na_nx_n.

Bilinear products and dual spaces

As we saw above, if V is finite-dimensional, then V is isomorphic to V*, but the isomorphism is not natural and depends on the basis of V we started out with. In fact, any isomorphism Φ from V to V* defines a unique non-degenerate bilinear form on V by

$\backslash langle\; v,w\; \backslash rangle\; =\; (\backslash Phi\; (v))(w)\; \backslash ,$

and conversely every such non-degenerate bilinear product on a finite-dimensional space gives rise to an isomorphism from V to V*.

Injection into the double-dual

There is a natural homomorphism Ψ from V into the double dual V**, defined by (Ψ(v))(φ) = φ(v) for all v in V, φ in V*. This map Ψ is always injective; it is an isomorphism if and only if V is finite-dimensional.

Pullback of a linear map

If f: V→W is a linear map, we may define its pullback f*: W*→V* by

$f^*\; (\backslash phi\; )\; =\; \backslash phi\; \backslash circ\; f\; \backslash ,$

where φ is an element of W*.

The assignment $f\; \backslash mapsto\; \backslash ,\; f^*$ produces an injective linear map between the space of linear operators from V to W and the space of linear operators from W* to V*; this homomorphism is an isomorphism iff W is finite-dimensional. If V = W 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 (fg)*=g*f*. In the language of category theory, taking the dual of vector spaces and the pullback of linear maps is therefore a contravariant functor from the category of vector spaces over F to itself. Note that one can identify (f*)* with f using the natural injection into the double dual.

If the linear map f is represented by the matrix A with respect to two bases of V and W, then f* is represented by same matrix acting by multiplication on the right on row vectors. Using the canonical inner product on Rⁿ, one may identify the space with its dual, in which case the matrix can be represented by the transposed matrix ^tA.

Structure of the dual space

The structure of the algebraic dual space is simply related to the structure of the vector space. If the space is finite dimensional then the space and its dual are isomorphic, while if the space is infinite dimensional then the dual space always has larger dimension.

Given a basis {e_α} for V indexed by A, one may construct the linearly independent set of dual vectors {σ^α}, as defined above. If V is infinite-dimensional however, the dual vectors do not form a basis for V*; the span of {σ^α} consists of all finite linear combinations of the dual vectors, but any infinite ordered tuple of dual vectors (thought of informally as an infinite sum) defines an element of the dual space. Because every vector of the vector space may be written as a finite linear combination of basis vectors {e_α}, an infinite tuple of dual vectors evaluates to nonzero scalars only finitely many times.

More explicitly, any infinite tuple (f_ασ^α) may be thought of as the infinite sum

$f=\backslash sum\_\{\backslash alpha\; \backslash in\; A\}f\_\backslash alpha\backslash sigma^\backslash alpha$

which satisfies

$f(\backslash mathbf\{e\}\_n)\; =\; f\_n.$

So f acts on an arbitrary vector

$\backslash mathbf\{v\}=\backslash sum\_\{i=1\}^n\; v^i\backslash mathbf\{e\}\_i$

in V by

$f(\backslash mathbf\{v\})\; =\; \backslash sum\_\{i=1\}^n\; v^if(\backslash mathbf\{e\}\_i)\; =\; \backslash sum\_\{i=1\}^n\; v^i\; f\_i.$

This dual vector f is linearly independent of the dual vectors {σ^α} unless A is finite. The dual space is the span of all such tuples. The idea of a dual vector as an infinite sum should not be taken too literally; in general infinite sums are defined in terms of a limit, which only makes sense in a topological space, and even then not all sums will be convergent. A basis for the dual space is a set of vectors such that every dual vector can be written as a finite linear combination. The existence of such a basis requires the axiom of choice, and cannot be exhibited explicitly.

This can be understand more rigorously, if perhaps more abstractly as follows. For any vector space V over F, we can find a basis. If that basis has cardinality α (thus α is the dimension of the vector space), then we may find a basis indexed by α. Since any field may be viewed as a one dimensional vector space over itself, we may construct the vector space direct sum of copies of F and the existence of the basis is equivalent to the existence of an isomorphism

$V\backslash cong\backslash bigoplus\_\backslash alpha\; \backslash mathbb\{F\}.$

Thus this isomorphism is nothing other than the equivalent statement that any vector can be uniquely written as a sum of finitely many basis vectors, which is simply the definition of a basis. Note that the isomorphism is not canonical; it depends on the particular choice of basis.

A property of the direct sum is that the operation of passing to the dual turns direct sums into direct products. That is,

$\backslash left(\backslash bigoplus\_\backslash alpha\backslash mathbb\{F\}\backslash right)^*=\backslash prod\_\backslash alpha\backslash mathbb\{F\}^*=\backslash prod\_\backslash alpha\backslash mathbb\{F\},$

and here in the second equation we use the fact that any field F, viewed as a vector space over itself, is canonically isomorphic to its dual space. Thus we see that

$V^*\backslash cong\; \backslash prod\_\backslash alpha\backslash mathbb\{F\}.$

Recall that the vector space direct sum is the set of tuples which are only nonzero finitely many times, while the vector space direct product is the set of all tuples (tuples which may be nonzero infinitely often). If α is infinite, then there are always more vectors in the dual space than the vector space. This is in marked contrast to the case of the continuous dual space, discussed below, which may be isomorphic to the vector space even for infinite-dimensional spaces. On the other hand, if α is finite, then all tuples are nonzero only finitely often, so the direct sum and direct product coincide; any finite dimensional vector space is isomorphic to its dual space, though usually not canonically so.

Continuous dual space

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When dealing with topological vector spaces, one is typically only interested in the continuous linear functionals from the space into the base field. This gives rise to the notion of the continuous dual space which is a linear subspace of the algebraic dual space. The continuous dual of a vector space V is denoted V′. When the context is clear, the continuous dual may just be called the dual.

The continuous dual V′ of a normed vector space V (e.g., a Banach space or a Hilbert space) forms a normed vector space. The norm ||φ|| of a continuous linear functional on V is defined by

$\backslash |\backslash phi\; \backslash |\; =\; \backslash sup\; \backslash \{\; |\backslash phi\; (\; x\; )|\; :\; \backslash |x\backslash |\; \backslash le\; 1\; \backslash \}$

This turns the continuous dual into a normed vector space, indeed into a Banach space so long as the underlying field is complete which is often included in the definition of the normed vector space. In other words, the dual of a normed space over a complete field is necessarily complete.

For any finite-dimensional normed vector space or topological vector space, such as Euclidean n-space, the continuous dual and the algebraic dual coincide. This is however false for any infinite-dimensional normed space, as shown by the example of discontinuous linear map.

Examples

Let 1 < p < ∞ be a real number and consider the Banach space L^p of all sequences a = (a_n) for which

$\backslash |\backslash mathbf\{a\}\backslash |\_p\; =\; \backslash left\; (\; \backslash sum\_\{n=0\}^\backslash infty\; |a\_n|^p\; \backslash right)\; ^\{1/p\}$

is finite. Define the number q by 1/p + 1/q = 1. Then the continuous dual of L^p is naturally identified with L^q: given an element φ ∈ (L^p)', the corresponding element of L^q is the sequence (φ(e_n)) where e_n denotes the sequence whose n-th term is 1 and all others are zero. Conversely, given an element a = (a_n) ∈ L^q, the corresponding continuous linear functional φ on L^p is defined by φ(b) = ∑_n a_n b_n for all b = (b_n) ∈ L^p (see Hölder's inequality).

In a similar manner, the continuous dual of L¹ is naturally identified with L^∞. Furthermore, the continuous duals of the Banach spaces c (consisting of all convergent sequences, with the supremums norm) and c₀ (the sequences converging to zero) are both naturally identified with L¹.

Further properties

If V is a Hilbert space, then its continuous dual is a Hilbert space which is anti-isomorphic to V. This is the content of the Riesz representation theorem, and gives rise to the bra-ket notation used by physicists in the mathematical formulation of quantum mechanics.

In analogy with the case of the algebraic double dual, there is always a naturally defined injective continuous linear operator Ψ : V → V″ from V into its continuous double dual V″. This map is in fact an isometry, meaning ||Ψ(x)|| = ||x|| for all x in V. Spaces for which the map Ψ is a bijection are called reflexive.

The continuous dual can be used to define a new topology on V, called the weak topology.

If the dual of V is separable, then so is the space V itself. The converse is not true; the space l₁ is separable, but its dual is l_∞, which is not separable.

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