Reflexivity

It is natural to think that if the dual space \mathcal{V}^\prime of a vector space $𝒱$ , and the relations between a space and its dual, are of any interest at all for $𝒱$ , 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 $𝒱$ 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 $𝒱$ 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 $𝒱$ such that $z_{0} (y)$ $= [x_{0}, y] = y (x_{0})$ for every $y$ in \mathcal{V}^{\prime} ; the correspondence $z_{0} ⇄ x_{0}$ between \mathcal{V}^{\prime\prime} and $𝒱$ is an isomorphism.

The correspondence described in this statement is called the natural correspondence between \mathcal{V}^{\prime \prime} and $𝒱$ .

Proof. Let us view the correspondence from the standpoint of going from $𝒱$ to \mathcal{V}^{\prime \prime} ; in other words, to every $x_{0}$ in $𝒱$ 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 $x_{0} \to z_{0}$ 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 $𝒱$ , 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 $𝒱$ ) is a subspace of \mathcal{V}^{\prime \prime} which is isomorphic to $𝒱$ and which is, therefore, $n$ -dimensional. But the $n$ -dimensionality of $𝒱$ 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 $𝒱$ 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 $𝒱$ (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 $𝒱$ 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 $𝒳$ , in $𝒱$ , and the dual basis of its dual basis, in \mathcal{V}^{\prime \prime} ; the symmetry of the relation $[x_{i}, y_{j}] = δ_{i j}$ shows that \mathcal{X}^{\prime \prime}=\mathcal{X} .

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