Editing Dual space (section)

=== Infinite-dimensional case ===

If <math>V</math>  is not finite-dimensional but has a [[basis (linear algebra)|basis]]<ref group=nb name="choice">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 <math>\R^\N</math> has a basis.
It is also needed to show that the dual of an infinite-dimensional vector space <math>V</math> is nonzero, and hence that the natural map from <math>V</math> to its double dual is injective.</ref> <math>\mathbf{e}_\alpha</math> indexed by an infinite set <math>A</math>, then the same construction as in the finite-dimensional case yields [[linearly independent]] elements <math>\mathbf{e}^\alpha</math> (<math>\alpha\in A</math>) of the dual space, but they will not form a basis.

For instance, consider the space <math>\R^\infty</math>, whose elements are those [[sequence]]s of real numbers that contain only finitely many non-zero entries, which has a basis indexed by the natural numbers <math>\N</math>. For <math>i \in \N</math>, <math>\mathbf{e}_i</math> is the sequence consisting of all zeroes except in the <math>i</math>-th position, which is 1.
The dual space of <math>\R^\infty</math> is (isomorphic to) <math>\R^\N</math>, the space of ''all'' sequences of real numbers: each real sequence <math>(a_n)</math> defines a function where the element <math>(x_n)</math> of <math>\R^\infty</math> is sent to the number

:<math>\sum_n a_nx_n,</math>

which is a finite sum because there are only finitely many nonzero <math>x_n</math>. The [[dimension (vector space)|dimension]] of <math>\R^\infty</math> is [[countably infinite]], whereas <math>\R^\N</math> does not have a countable basis.

This observation generalizes to any<ref group=nb name="choice"/> infinite-dimensional vector space <math>V</math> over any field <math>F</math>: a choice of basis <math>\{\mathbf{e}_\alpha:\alpha\in A\}</math> identifies <math>V</math> with the space <math>(F^A)_0</math> of functions <math>f:A\to F</math> such that <math>f_\alpha=f(\alpha)</math> is nonzero for only finitely many <math>\alpha\in A</math>, where such a function <math>f</math> is identified with the vector

:<math>\sum_{\alpha\in A} f_\alpha\mathbf{e}_\alpha</math>

in <math>V</math> (the sum is finite by the assumption on <math>f</math>, and any <math>v\in V</math> may be written uniquely in this way by the definition of the basis).

The dual space of <math>V</math> may then be identified with the space <math>F^A</math> of ''all'' functions from <math>A</math> to <math>F</math>: a linear functional <math>T</math> on <math>V</math> is uniquely determined by the values <math>\theta_\alpha=T(\mathbf{e}_\alpha)</math> it takes on the basis of <math>V</math>, and any function <math>\theta:A\to F</math> (with <math>\theta(\alpha)=\theta_\alpha</math>) defines a linear functional <math>T</math> on <math>V</math> by

:<math>T\left (\sum_{\alpha\in A} f_\alpha \mathbf{e}_\alpha\right) = \sum_{\alpha \in A} f_\alpha T(e_\alpha) = \sum_{\alpha\in A} f_\alpha \theta_\alpha.</math>

Again, the sum is finite because <math>f_\alpha</math> is nonzero for only finitely many <math>\alpha</math>.

The set <math>(F^A)_0</math> may be identified (essentially by definition) with the [[Direct sum of modules|direct sum]] of infinitely many copies of <math>F</math> (viewed as a 1-dimensional vector space over itself) indexed by <math>A</math>, i.e. there are linear isomorphisms
:<math> V\cong (F^A)_0\cong\bigoplus_{\alpha\in A} F.</math>

On the other hand, <math>F^A</math> is (again by definition), the [[direct product]] of infinitely many copies of <math>F</math> indexed by <math>A</math>, and so the identification
:<math>V^* \cong \left (\bigoplus_{\alpha\in A}F\right )^* \cong \prod_{\alpha\in A}F^* \cong \prod_{\alpha\in A}F \cong F^A</math>
is a special case of a [[Direct sum of modules#Properties|general result]] relating direct sums (of [[module (mathematics)|module]]s) 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 <math>V</math> is an infinite-dimensional <math>F</math>-vector space, the arithmetical properties of [[cardinal numbers]] implies that 
:<math>\mathrm{dim}(V)=|A|<|F|^{|A|}=|V^\ast|=\mathrm{max}(|\mathrm{dim}(V^\ast)|, |F|),</math>
where cardinalities are denoted as [[absolute value]]s. For proving that <math>\mathrm{dim}(V)< \mathrm{dim}(V^*),</math> it suffices to prove that <math>|F|\le |\mathrm{dim}(V^\ast)|,</math> which can be done with an argument similar to [[Cantor's diagonal argument]].<ref>{{cite book|title=Elements of mathematics: Algebra I, Chapters 1 - 3|author=Nicolas Bourbaki|page=400|editor=Hermann|isbn=0201006391|year=1974|publisher=Addison-Wesley Publishing Company |language=en}}</ref> The exact dimension of the dual is given by the [[Erdős–Kaplansky theorem]].