It is natural to think that if the dual space \mathcal{V}^\prime of a vector space \mathcal{V} , and the relations between a space and its dual, are of any interest at all for \mathcal{V} , then they are of just as much interest for \mathcal{V}^\prime . In other words, we propose now to form the dual space (\mathcal{V}^{\prime})^{\prime} of \mathcal{V}^{\prime} ; for simplicity of notation we shall denote it by \mathcal{V}^{\prime \prime} . The verbal description of an element of \mathcal{V}^{\prime \prime} is clumsy: such an element is a linear functional of linear functionals. It is, however, at this point that the greatest advantage of the notation [x, y] appears; by means of it, it is easy to discuss \mathcal{V} and its relation to \mathcal{V}^{\prime \prime} .
If we consider the symbol [x, y] for some fixed y=y_{0} , we obtain nothing new: [x, y_{0}] is merely another way of writing the value y_{0}(x) of the function y_{0} at the vector x . If, however, we consider the symbol [x, y] for some fixed x=x_{0} , then we observe that the function of the vectors in \mathcal{V}^{\prime} , whose value at y is [x_{0}, y] , is a scalar-valued function that happens to be linear (see Section: Brackets , (2)); in other words, [x_{0}, y] defines a linear functional on \mathcal{V}^{\prime} , and, consequently, an element of \mathcal{V}^{\prime \prime} .
By this method we have exhibited some linear functionals on \mathcal{V}^{\prime} ; have we exhibited them all? For the finite-dimensional case the following theorem furnishes the affirmative answer.
Theorem 1. If \mathcal{V} is a finite-dimensional vector space, then corresponding to every linear functional z_{0} on \mathcal{V}^{\prime} there is a vector x_{0} in \mathcal{V} such that z_{0}(y) =[x_{0}, y]=y(x_{0}) for every y in \mathcal{V}^{\prime} ; the correspondence z_{0} \rightleftarrows x_{0} between \mathcal{V}^{\prime\prime} and \mathcal{V} is an isomorphism.
The correspondence described in this statement is called the natural correspondence between \mathcal{V}^{\prime \prime} and \mathcal{V} .
Proof. Let us view the correspondence from the standpoint of going from \mathcal{V} to \mathcal{V}^{\prime \prime} ; in other words, to every x_{0} in \mathcal{V} we make correspond a vector z_{0} in \mathcal{V}^{\prime \prime} defined by z_{0}(y)=y(x_{0}) for every y in \mathcal{V}^{\prime} . Since [x, y] depends linearly on x , the transformation x_{0} \rightarrow z_{0} is linear.
We shall show that this transformation is one-to-one, as far as it goes. We assert, in other words, that if x_{1} and x_{2} are in \mathcal{V} , and if z_{1} and z_{2} are the corresponding vectors in \mathcal{V}^{\prime \prime} (so that z_{1}(y)=[x_{1}, y] and z_{2}(y)=[x_{2}, y] for all y in \mathcal{V}^{\prime} ), and if z_{1}=z_{2} , then x_{1}=x_{2} . To say that z_{1}=z_{2} means that [x_{1}, y]=[x_{2}, y] for every y in \mathcal{V}^{\prime} ; the desired conclusion follows from Section: Dual bases , Theorem 3.
The last two paragraphs together show that the set of those linear functionals z on \mathcal{V}^{\prime} (that is, elements of \mathcal{V}^{\prime \prime} ) that do have the desired form (that is, z(y) is identically equal to [x, y] for a suitable x in \mathcal{V} ) is a subspace of \mathcal{V}^{\prime \prime} which is isomorphic to \mathcal{V} and which is, therefore, n -dimensional. But the n -dimensionality of \mathcal{V} implies that of \mathcal{V}^{\prime} , which in turn implies that \mathcal{V}^{\prime \prime} is n -dimensional. It follows that \mathcal{V}^{\prime \prime} must coincide with the n -dimensional subspace just described, and the proof of the theorem is complete. ◻
It is important to observe that the theorem shows not only that \mathcal{V} and \mathcal{V}^{\prime \prime} are isomorphic—this much is trivial from the fact that they have the same dimension—but that the natural correspondence is an isomorphism. This property of vector spaces is called reflexivity; every finite-dimensional vector space is reflexive.
It is frequently convenient to be mildly sloppy about \mathcal{V}^{\prime \prime} : for finite-dimensional vector spaces we shall identify \mathcal{V}^{\prime \prime} with \mathcal{V} (by the natural isomorphism), and we shall say that the element z_{0} of \mathcal{V}^{\prime \prime} is the same as the element x_{0} of \mathcal{V} whenever z_{0}(y)=[x_{0}, y] for all y in \mathcal{V}^{\prime} . In this language it is very easy to express the relation between a basis \mathcal{X} , in \mathcal{V} , and the dual basis of its dual basis, in \mathcal{V}^{\prime \prime} ; the symmetry of the relation [x_{i}, y_{j}]=\delta_{i j} shows that \mathcal{X}^{\prime \prime}=\mathcal{X} .
