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 \(\mathcal{X}^{\prime \prime}=\mathcal{X}\) .