Dual space

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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. Dual vector spaces defined on finite-dimensional vector spaces can be used for defining tensors which are studied in tensor algebra. When applied to vector spaces of functions (which typically are 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.

[edit] Algebraic dual space

Given any vector space, V, over some field, F, the dual space, V*, is defined as the set of all linear functionals on V, i.e., scalar-valued linear maps 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:

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

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

for all $φ,ψ$ in V*, a in F and x in V. 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, such as

$\phi(x)=[\phi,x],\quad\text{or}\,\,\, \phi(x)=\langle\phi, x\rangle.$

For the former notation, see (Halmos 1974). For the latter, see (Misner, Thorne & Wheeler 1973). The pairing defines a nondegenerate bilinear mapping,^[1]

$[\cdot,\cdot] : V^* \times V \to F.$

[edit] The finite dimensional case

If V is finite-dimensional, then V* has the same dimension as V. Given a basis of V, it is possible to give a basis of V*, called the dual basis. In detail, if {e₁,...,e_n} is a basis for V, then the associated dual basis of V* is an n-tuple {e¹,...,eⁿ} of linear functionals on V defined by the relation

$\mathbf{e}^i(c_1 \mathbf{e}_1+\cdots+c_n\mathbf{e}_n) = c_i$

for any choice of coefficients c_i (and hence any vector in V, since the e_i are assumed to be a basis). In particular, taking c_j=1 and every other coefficient zero gives the relation

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

the Kronecker delta. In the case of R², its basis is B={e₁=(1,0),e₂=(0,1)}. Then, e¹, and e² are one-forms (functions which 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 an index, not an exponent.)

Concretely, 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.

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. So an element of V* can be intuitively thought of as a particular family 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 (Misner, Thorne & Wheeler 1973, §2.5).

[edit] The infinite dimensional case

If V is not finite-dimensional but has a Hamel basis^[2] e_α indexed by an infinite set A, then the same construction as in the finite dimensional case yields linearly independent elements e^α (α∈A) 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 i∈N, e_i is the sequence which is zero apart from the ith term, which is one. The dual space of R^∞ is R^N, the space of all sequences of real numbers: such a sequence (a_n) is applied to an element (x_n) of R^∞ to give the number ∑_na_nx_n, which is a finite sum because there are only finitely many nonzero x_n. The dimension of R^∞ is countably infinite, whereas R^N does not have a countable basis.

This observation generalizes to any^[2] infinite dimensional vector space V over any field F: a choice of basis {e_α:α∈A} identifies V with the space (F^A)₀ of functions f:A→F such that f_α=f(α) is nonzero for only finitely many α∈A, 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 v∈V may be written in this way by the definition of a basis).

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

$T\biggl(\sum_{\alpha\in A} f_\alpha e_\alpha\biggr) = \sum_{\alpha\in A} \theta_\alpha f_\alpha.$

Again the sum is finite because f_α is nonzero for only finitely many α.

Note that (F^A)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

$V\cong (\mathbb F^A)_0\cong\bigoplus_{\alpha\in A} \mathbb{F}.$

On the other hand F^A is (again by definition), the direct product of infinitely many copies of F indexed by A, and so the identification

$V^*\cong \biggl(\bigoplus_{\alpha\in A}\mathbb{F}\biggr)^*\cong \prod_{\alpha\in A}\mathbb{F}^*\cong\prod_{\alpha\in A}\mathbb{F}\cong \mathbb{F}^A$

is a special case of a general result relating direct sums (of modules) to direct products.

Thus if the basis is infinite, then there are always more vectors in the dual space 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.

[edit] 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 (MacLane & Birkhoff 1999, §VI.4). 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 w ∈ V 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 assumed to be 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

$\Phi_{\langle\cdot,\cdot\rangle} : V\to \bar{V}^*.$

The conjugate space $\scriptstyle{\bar{V}^*}$ can be identified with the set of all additive complex-valued functionals ƒ : V → C such that

$f(\alpha v) = \bar{\alpha}f(v).$

[edit] 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^[2]; it is an isomorphism if and only if V is finite-dimensional. (Infinite-dimensional Hilbert spaces are not a counterexample to this, as they are isomorphic to their continuous duals, not to their algebraic duals.)

[edit] Transpose of a linear map

If ƒ : V → W is a linear map, then the transpose (or dual) ƒ* : W* → V* is defined by

$f^* (\varphi) = \varphi \circ f \,$		(1)

for every φ ∈ W*. The resulting functional ƒ*(φ) is in V*, and is called as the pullback of φ along ƒ.

The following identity holds for all φ ∈ W* and v ∈ V:

$[f^*(\varphi), v] = [\varphi, f(v)] \,$

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 (Halmos 1974, §44), and is formally similar to the definition of an adjoint.

The assignment $f \mapsto \, f^*$ 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 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 ^tA 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 by 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.

[edit] Quotient spaces and annihilators

Let S be a subset of V then the annihilator of S in V^*, denoted here S^o is the collection of linear functionals ƒ ∈ V^* such that [ƒ,s] = 0 for all s ∈ S. That is, S^o consists of all linear functionals ƒ : V → F such that the restriction to S vanishes: ƒ|_S = 0.

The annihilator of a subset is itself a vector space. In particular, ∅^o = V^* is all of V^* (vacuously), whereas V^o = 0 is the zero subspace. Furthermore, the assignment of an annihilator to a subset of V reverses inclusions, so that if S ⊂ T ⊂ V 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 A_i 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 o o = W

after identifying W with its image in the second dual space under the double duality isomorphism V ≈ V′′. 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 ƒ : V → F factors through V/W if and only if W is in the kernel of ƒ. There is thus an isomorphism

$\left(V/W\right)' \cong W^o.$

As a particular consequence, if V is a direct sum of two subspaces A and B, then V′ is a direct sum of A^o and B^o.

[edit] 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. This gives rise to the notion of the "continuous dual space" which is a linear subspace of the algebraic dual space V*, denoted 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. In topological contexts sometimes V* may also be used for just the continuous dual space and 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. A norm ||φ|| of a continuous linear functional on V is defined by

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

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, this 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.

[edit] Examples

Let 1 < p < ∞ be a real number and consider the Banach space ℓ^p of all sequences a = (a_n) 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 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^∞ (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 ℓ¹.

[edit] Further properties

In analogy with the case of the algebraic double dual, there is always a naturally defined injective^[3] continuous linear operator Ψ : V → V ′′ from V into its continuous double dual V ′′. In case V is normed, 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.

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.

[edit] See also

Duality (mathematics)
Duality (projective geometry)
Reciprocal lattice - dual space basis, in crystallography

[edit] Notes

^ In many areas, such as quantum mechanics, $\langle\cdot,\cdot\rangle$ is reserved for a sesquilinear form defined on V × V.
^ ^a ^b ^c Several 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 R^N has a basis. It is also needed to show that the dual of an infinite dimensional vector space V is nonempty, and hence that the natural map from V to its double dual is injective.
^ Injectivity holds if and only if V is Hausdorff. Otherwise, the kernel is the smallest closed subspace containing {0}.

[edit] References

Bourbaki, Nicolas (1989), Elements of mathematics, Algebra I, Springer-Verlag, ISBN 3-540-64243-9
Halmos, Paul (1974), Finite dimensional vector spaces, Springer, ISBN 0387900934
Lang, Serge (2002), Algebra, Springer-Verlag, ISBN 0-387-95385-X .
MacLane, Saunders; Birkhoff, Garrett (1999), Algebra (3rd ed.), AMS Chelsea Publishing, ISBN 0-8218-1646-2 .
Misner, Charles W.; Thorne, Kip S.; Wheeler, John A. (1973), Gravitation, W. H. Freeman, ISBN 0-7167-0344-0
Rudin, Walter (1991), Functional analysis, McGraw-Hill Science, ISBN 978-0070542365

[0] In many areas, such as quantum mechanics, $\langle\cdot,\cdot\rangle$ is reserved for a sesquilinear form defined on V × V.

[choice-1] Several 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 R^N has a basis. It is also needed to show that the dual of an infinite dimensional vector space V is nonempty, and hence that the natural map from V to its double dual is injective.

[2] Injectivity holds if and only if V is Hausdorff. Otherwise, the kernel is the smallest closed subspace containing {0}.

[1]

[2]

[3]

Dual space

From Wikipedia, the free encyclopedia

Contents

[edit] Algebraic dual space

[edit] The finite dimensional case

[edit] The infinite dimensional case

[edit] Bilinear products and dual spaces

[edit] Injection into the double-dual

[edit] Transpose of a linear map

[edit] Quotient spaces and annihilators

[edit] Continuous dual space

[edit] Examples

[edit] Further properties

[edit] See also

[edit] Notes

[edit] References

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