# Dual space

In mathematics, any vector space, V, has a corresponding dual vector space (or just dual space for short) consisting of all linear functionals on V together with a naturally induced linear structure. Dual vector spaces for finite-dimensional vector spaces can be used for studying tensors. When applied to vector spaces of functions (which are typically infinite-dimensional), dual spaces are employed for defining and studying concepts like measures, distributions, and Hilbert spaces. Consequently, the dual space is an important concept in the study of functional analysis.

There are two types of dual spaces: the algebraic dual space, and the continuous dual space. The algebraic dual space is defined for all vector spaces. When defined for a topological vector space there is a subspace of this dual space, corresponding to continuous linear functionals, which constitutes a continuous dual space.

## Algebraic dual space

Given any vector space V over a field F, the dual space V* is defined as the set of all linear maps φ: VF (linear functionals). The dual space V* itself becomes a vector space over F when equipped with the following addition and scalar multiplication:

{\begin{aligned}&(\varphi +\psi )(x)=\varphi (x)+\psi (x)\\&(a\varphi )(x)=a\left(\varphi (x)\right)\end{aligned}} for all φ and ψV*, xV, and aF. Elements of the algebraic dual space V* are sometimes called covectors or one-forms.

The pairing of a functional φ in the dual space V* and an element x of V is sometimes denoted by a bracket: φ(x) = [φ,x]  or φ(x) = ⟨φ,x. The pairing defines a nondegenerate bilinear mapping [·,·] : V* × VF.

### Finite-dimensional case

If V is finite-dimensional, then V has the same dimension as V. Given a basis {e1, ..., en} in V, it is possible to construct a specific basis in V, called the dual basis. This dual basis is a set {e1, ..., en} of linear functionals on V, defined by the relation

$\mathbf {e} ^{i}(c^{1}\mathbf {e} _{1}+\cdots +c^{n}\mathbf {e} _{n})=c^{i},\quad i=1,\ldots ,n$ for any choice of coefficients ciF. In particular, letting in turn each one of those coefficients be equal to one and the other coefficients zero, gives the system of equations

$\mathbf {e} ^{i}(\mathbf {e} _{j})=\delta _{j}^{i}$ where $\delta _{j}^{i}$ is the Kronecker delta symbol. For example if V is R2, and its basis chosen to be {e1 = (1, 0), e2 = (0, 1)}, then e1 and e2 are one-forms (functions that map a vector to a scalar) such that e1(e1) = 1, e1(e2) = 0, e2(e1) = 0, and e2(e2) = 1. (Note: The superscript here is the index, not an exponent).

In particular, if we interpret Rn as the 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. One way to see this is that a functional maps every n-vector x into a Real number y. Then, seeing this functional as a matrix M, and x,y as a nx1 matrix and an 1x1 matrix (trivially, a Real number)respectively, if we have Mx=y, then, by dimension reasons, M must be an 1xn matrix itself, i.e., M must be a row vector.

If V consists of the space of geometrical vectors (arrows) in the plane, then the level curves of an element of V form a family of parallel lines in V, 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 V 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, one needs only to determine which of the lines the vector lies on. Or, informally, one "counts" how many lines the vector crosses. More generally, if V is a vector space of any dimension, then the level sets of a linear functional in V are parallel hyperplanes in V, and the action of a linear functional on a vector can be visualized in terms of these hyperplanes.

### Infinite-dimensional case

If V is not finite-dimensional but has a basis eα indexed by an infinite set A, then the same construction as in the finite-dimensional case yields linearly independent elements eα (αA) of the dual space, but they will not form a basis.

Consider, for instance, the space R, whose elements are those sequences of real numbers which have only finitely many non-zero entries, which has a basis indexed by the natural numbers N: for iN, ei is the sequence consisting of all zeroes except in the ith position, which is 1. The dual space of R is RN, the space of all sequences of real numbers: such a sequence (an) is applied to an element (xn) of R to give the number ∑anxn, which is a finite sum because there are only finitely many nonzero xn. The dimension of R is countably infinite, whereas RN does not have a countable basis.

This observation generalizes to any infinite-dimensional vector space V over any field F: a choice of basis {eα : αA} identifies V with the space (FA)0 of functions f : A → F such that fα = f(α) is nonzero for only finitely many αA, where such a function f is identified with the vector

$\sum _{\alpha \in A}f_{\alpha }\mathbf {e} _{\alpha }$ in V (the sum is finite by the assumption on f, and any vV may be written in this way by the definition of the basis).

The dual space of V may then be identified with the space FA of all functions from A to F: a linear functional T on V is uniquely determined by the values θα = T(eα) it takes on the basis of V, and any function θ : A → F (with θ(α) = θα) defines a linear functional T on V by

$T{\biggl (}\sum _{\alpha \in A}f_{\alpha }\mathbf {e} _{\alpha }{\biggr )}=\sum _{\alpha \in A}f_{\alpha }T(e_{\alpha })=\sum _{\alpha \in A}f_{\alpha }\theta _{\alpha }.$ Again the sum is finite because fα is nonzero for only finitely many α.

Note that (FA)0 may be identified (essentially by definition) with the direct sum of infinitely many copies of F (viewed as a 1-dimensional vector space over itself) indexed by A, i.e., there are linear isomorphisms

$V\cong (F^{A})_{0}\cong \bigoplus _{\alpha \in A}{F}.$ On the other hand FA is (again by definition), the direct product of infinitely many copies of F indexed by A, and so the identification

$V^{*}\cong {\biggl (}\bigoplus _{\alpha \in A}F{\biggr )}^{*}\cong \prod _{\alpha \in A}F^{*}\cong \prod _{\alpha \in A}F\cong F^{A}$ is a special case of a general result relating direct sums (of modules) to direct products.

Thus if the basis is infinite, then the algebraic dual space is always of larger dimension than the original vector space. This is in marked 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.

### Bilinear products and dual spaces

If V is finite-dimensional, then V is isomorphic to V*. But there is in general no natural isomorphism between these two spaces. Any bilinear form ⟨•,•⟩ on V gives a mapping of V into its dual space via

$v\mapsto \langle v,\cdot \rangle$ where the right hand side is defined as the functional on V taking each wV to ⟨v,w⟩. In other words, the bilinear form determines a linear mapping

$\Phi _{\langle \cdot ,\cdot \rangle }:V\to V^{*}$ defined by

$[\Phi _{\langle \cdot ,\cdot \rangle }(v),w]=\langle v,w\rangle .$ If the bilinear form is nondegenerate, then this is an isomorphism onto a subspace of V*. If V is finite-dimensional, then this is an isomorphism onto all of V*. Conversely, any isomorphism Φ from V to a subspace of V* (resp., all of V*) defines a unique nondegenerate bilinear form ⟨•,•⟩Φ on V by

$\langle v,w\rangle _{\Phi }=(\Phi (v))(w)=[\Phi (v),w].\,$ Thus there is a one-to-one correspondence between isomorphisms of V to subspaces of (resp., all of) V* and nondegenerate bilinear forms on V.

If the vector space V 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 V with the complex conjugate of the dual space

$\Phi _{\langle \cdot ,\cdot \rangle }:V\to {\overline {V^{*}}}.$ The conjugate space V* can be identified with the set of all additive complex-valued functionals f : VC such that

$f(\alpha v)={\overline {\alpha }}f(v).$ ### 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 vV, φV*. This map Ψ is always injective; it is an isomorphism if and only if V is finite-dimensional. Indeed, the isomorphism of a finite-dimensional vector space with its double dual is an archetypal example of a natural isomorphism. Note that infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals.

### Transpose of a linear map

{{#invoke:main|main}} If f : V → W is a linear map, then the transpose (or dual) f * : W* → V* is defined by

$f^{*}(\varphi )=\varphi \circ f\,$ for every φW*. The resulting functional f *(φ) in V* is called the pullback of φ along f.

The following identity holds for all φW* and vV:

$[f^{*}(\varphi ),\,v]=[\varphi ,\,f(v)],$ where the bracket [•,•] on the left is the duality pairing of V with its dual space, and that on the right is the duality pairing of W with its dual. This identity characterizes the transpose, and is formally similar to the definition of the adjoint.

The assignment ff * 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 if and only if 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 transpose 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 the transpose matrix AT with respect to the dual bases of W* and V*, hence the name. Alternatively, as f is represented by A acting on the left on column vectors, f * is represented by the same matrix acting on the right on row vectors. These points of view are related by the canonical inner product on Rn, which identifies the space of column vectors with the dual space of row vectors.

### Quotient spaces and annihilators

Let S be a subset of V. The annihilator of S in V*, denoted here So, is the collection of linear functionals fV* such that [f, s] = 0 for all sS. That is, So consists of all linear functionals f : V → F such that the restriction to S vanishes: f|S = 0.

The annihilator of a subset is itself a vector space. In particular, o = V* is all of V* (vacuously), whereas Vo = 0 is the zero subspace. Furthermore, the assignment of an annihilator to a subset of V reverses inclusions, so that if STV, then

$0\subset T^{o}\subset S^{o}\subset V^{*}.$ Moreover, if A and B are two subsets of V, then

$(A\cap B)^{o}\supseteq A^{o}+B^{o},$ and equality holds provided V is finite-dimensional. If Ai is any family of subsets of V indexed by i belonging to some index set I, then

$\left(\bigcup _{i\in I}A_{i}\right)^{o}=\bigcap _{i\in I}A_{i}^{o}.$ In particular if A and B are subspaces of V, it follows that

$(A+B)^{o}=A^{o}\cap B^{o}.\,$ If V is finite-dimensional, and W is a vector subspace, then

$W^{oo}=W\,$ after identifying W with its image in the second dual space under the double duality isomorphism VV**. Thus, in particular, forming the annihilator is a Galois connection on the lattice of subsets of a finite-dimensional vector space.

If W is a subspace of V then the quotient space V/W is a vector space in its own right, and so has a dual. By the first isomorphism theorem, a functional f : V → F factors through V/W if and only if W is in the kernel of f. There is thus an isomorphism

$(V/W)^{*}\cong W^{o}.$ As a particular consequence, if V is a direct sum of two subspaces A and B, then V* is a direct sum of Ao and Bo.

## Continuous dual space

When dealing with topological vector spaces, one is typically only interested in the continuous linear functionals from the space into the base field ${\mathbb {F} }={\mathbb {C} }$ (or ${\mathbb {R} }$ ). This gives rise to the notion of the "continuous dual space" or "topological dual" which is a linear subspace of the algebraic dual space $V^{*}$ , denoted by $V'$ . 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 maps. Nevertheless in the theory of topological vector spaces the terms "continuous dual space" and "topological dual space" are rarely used, as a rule they are replaced by "dual space", since there is no serious need to consider discontinuous maps in this field.

For a topological vector space $V$ its continuous dual space, or topological dual space, or just dual space (in the sense of the theory of topological vector spaces) $V'$ is defined as the space of all continuous linear functionals $\varphi :V\to {\mathbb {F} }$ .

$\|\varphi \|_{A}=\sup _{x\in A}|\varphi (x)|,$ $\forall A\in {\mathcal {A}}\qquad \|\varphi _{i}-\varphi \|_{A}=\sup _{x\in A}|\varphi _{i}(x)-\varphi (x)|{\underset {i\to \infty }{\longrightarrow }}0.$ Usually (but not necessarily) the class ${\mathcal {A}}$ is supposed to satisfy the following conditions:

$\forall x\in V\qquad \exists A\in {\mathcal {A}}\qquad x\in A,$ $\forall A,B\in {\mathcal {A}}\qquad \exists C\in {\mathcal {A}}\qquad A\cup B\subseteq C,$ $\forall A\in {\mathcal {A}}\qquad \forall \lambda \in {\mathbb {F} }\qquad \lambda \cdot A\in {\mathcal {A}},$ If these requirements are fulfilled then the corresponding topology on $V'$ is Hausdorff and the sets

$U_{A}=\{\varphi \in V':\quad ||\varphi ||_{A}<1\},\qquad A\in {\mathcal {A}},$ form its local base.

Here are the three most important special cases.

$\|\varphi \|=\sup _{\|x\|\leq 1}|\varphi (x)|.$ Each of these three choices of topology on $V'$ leads to a variant of reflexivity property for topological vector spaces.

### Examples

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

$\|\mathbf {a} \|_{p}=\left(\sum _{n=0}^{\infty }|a_{n}|^{p}\right)^{1/p}$ is finite. Define the number q by 1/p + 1/q = 1. Then the continuous dual of  p is naturally identified with  q: given an element φ ∈ ( p)′, the corresponding element of  q is the sequence (φ(en)) where en denotes the sequence whose n-th term is 1 and all others are zero. Conversely, given an element a = (an) ∈  q, the corresponding continuous linear functional φ on  p is defined by φ(b) = ∑n anbn for all b = (bn) ∈  p (see Hölder's inequality).

In a similar manner, the continuous dual of  1 is naturally identified with  ∞ (the space of bounded sequences). Furthermore, the continuous duals of the Banach spaces c (consisting of all convergent sequences, with the supremum norm) and c0 (the sequences converging to zero) are both naturally identified with  1.

By the Riesz representation theorem, the continuous dual of a Hilbert space is again a Hilbert space which is anti-isomorphic to the original space. This gives rise to the bra–ket notation used by physicists in the mathematical formulation of quantum mechanics.

### Transpose of a continuous linear map

If T : V → W is a continuous linear map between two topological vector spaces, then the (continuous) transpose T′ : W′ → V′ is defined by the same formula as before:

$T'(\varphi )=\varphi \circ T,\quad \varphi \in W'.\,$ The resulting functional T′(φ) is in V′. The assignment T → T′ produces a linear map between the space of continuous linear maps from V to W and the space of linear maps from W′ to V′. When T and U are composable continuous linear maps, then

$(U\circ T)'=T'\circ U'.\,$ When V and W are normed spaces, the norm of the transpose in L(W′, V′) is equal to that of T in L(V, W). Several properties of transposition depend upon the Hahn–Banach theorem. For example, the bounded linear map T has dense range if and only if the transpose T′ is injective.

When T is a compact linear map between two Banach spaces V and W, then the transpose T′ is compact. This can be proved using the Arzelà–Ascoli theorem.

When V is a Hilbert space, there is an antilinear isomorphism iV from V onto its continuous dual V′. For every bounded linear map T on V, the transpose and the adjoint operators are linked by

$i_{V}\circ T^{*}=T'\circ i_{V}.\,$ When T is a continuous linear map between two topological vector spaces V and W, then the transpose T′ is continuous when W′ and V′ are equipped with"compatible" topologies: for example when, for X = V and X = W, both duals X′ have the strong topology β(X′, X) of uniform convergence on bounded sets of X, or both have the weak-∗ topology σ(X′, X) of pointwise convergence on X. The transpose T′ is continuous from β(W′, W) to β(V′, V), or from σ(W′, W) to σ(V′, V).

### Annihilators

Assume that W is a closed linear subspace of a normed space V, and consider the annihilator of W in V′,

$W^{\perp }=\{\varphi \in V':W\subset \ker \varphi \}.\,$ Then, the dual of the quotient V / W can be identified with W, and the dual of W can be identified with the quotient V′ / W. Indeed, let P denote the canonical surjection from V onto the quotient V / W; then, the transpose P′ is an isometric isomorphism from (V / W )′ into V′, with range equal to W. If j denotes the injection map from W into V, then the kernel of the transpose j′ is the annihilator of W:

$\ker(j')=W^{\perp }$ and it follows from the Hahn–Banach theorem that j′ induces an isometric isomorphism V′ / WW′.

### Further properties

If the dual of a normed space V is separable, then so is the space V itself. The converse is not true: for example the space  1 is separable, but its dual is  ∞ is not.

### Topologies on the dual

The topology of V and the topology of real or complex numbers can be used to induce on V′ a dual space topology.

### Double dual This is a natural transformation of vector addition from a vector space to its double dual. <x1, x2> denotes the ordered pair of two vectors. The addition + sends x1 and x2 to x1 + x2. The addition +' induced by the transformation can be defined as (Ψ(x1) +' Ψ(x2))(φ) = φ(x1 + x2) = φ(x) for any φ in the dual space.

In analogy with the case of the algebraic double dual, there is always a naturally defined continuous linear operator Ψ : VV′′ from a normed space V into its continuous double dual V′′, defined by

$\Psi (x)(\varphi )=\varphi (x),\quad x\in V,\ \varphi \in V'.\,$ As a consequence of the Hahn–Banach theorem, this map is in fact an isometry, meaning for all x in V. Normed spaces for which the map Ψ is a bijection are called reflexive.

When V is a topological vector space, one can still define Ψ(x) by the same formula, for every xV, however several difficulties arise. First, when V is not locally convex, the continuous dual may be equal to {0} and the map Ψ trivial. However, if V is Hausdorff and locally convex, the map Ψ is injective from V to the algebraic dual V′* of the continuous dual, again as a consequence of the Hahn–Banach theorem.

Second, even in the locally convex setting, several natural vector space topologies can be defined on the continuous dual V′, so that the continuous double dual V′′ is not uniquely defined as a set. Saying that Ψ maps from V to V′′, or in other words, that Ψ(x) is continuous on V′ for every xV, is a reasonable minimal requirement on the topology of V′, namely that the evaluation mappings

$\varphi \in V'\mapsto \varphi (x),\quad x\in V,\,$ be continuous for the chosen topology on V′. Further, there is still a choice of a topology on V′′, and continuity of Ψ depends upon this choice. As a consequence, defining reflexivity in this framework is more involved than in the normed case.
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