Reflexivity

It is natural to think that if the dual space of a vector space , and the relations between a space and its dual, are of any interest at all for , then they are of just as much interest for . In other words, we propose now to form the dual space of ; for simplicity of notation we shall denote it by . The verbal description of an element of 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 appears; by means of it, it is easy to discuss and its relation to .

If we consider the symbol for some fixed , we obtain nothing new: is merely another way of writing the value of the function at the vector . If, however, we consider the symbol for some fixed , then we observe that the function of the vectors in , whose value at is , is a scalar-valued function that happens to be linear (see Section: Brackets , (2)); in other words, defines a linear functional on , and, consequently, an element of .

By this method we have exhibited some linear functionals on ; have we exhibited them all? For the finite-dimensional case the following theorem furnishes the affirmative answer.

Theorem 1. If is a finite-dimensional vector space, then corresponding to every linear functional on there is a vector in such that for every in ; the correspondence between and is an isomorphism.

The correspondence described in this statement is called the natural correspondence between and .

Proof. Let us view the correspondence from the standpoint of going from to ; in other words, to every in we make correspond a vector in defined by for every in . Since depends linearly on , the transformation is linear.

We shall show that this transformation is one-to-one, as far as it goes. We assert, in other words, that if and are in , and if and are the corresponding vectors in (so that and for all in ), and if , then . To say that means that for every in ; the desired conclusion follows from Section: Dual bases , Theorem 3.

The last two paragraphs together show that the set of those linear functionals on (that is, elements of ) that do have the desired form (that is, is identically equal to for a suitable in ) is a subspace of which is isomorphic to and which is, therefore, -dimensional. But the -dimensionality of implies that of , which in turn implies that is -dimensional. It follows that must coincide with the -dimensional subspace just described, and the proof of the theorem is complete. ◻

It is important to observe that the theorem shows not only that and 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 : for finite-dimensional vector spaces we shall identify with (by the natural isomorphism), and we shall say that the element of is the same as the element of whenever for all in . In this language it is very easy to express the relation between a basis , in , and the dual basis of its dual basis, in ; the symmetry of the relation shows that .

Quick Links

Social Media