Natural isomorphisms

There is now only one more possible doubt that the reader might (or, at any rate, should) have. Many of our preceding results were consequences of such reflexivity relations as ; do these remain valid after the brackets-to-parentheses revolution? More to the point is the following way of asking the question. Everything we say about a unitary space must also be true about the unitary space ; in particular it is also in a natural conjugate isomorphic relation with its dual space . If now to every vector in we make correspond a vector in , by first applying the natural conjugate isomorphism from to and then going the same way from to , then this mapping is a rival for the title of natural mapping from to , a title already awarded in Chapter I to a seemingly different correspondence. What is the relation between the two natural correspondences? Our statements about the coincidence, except for trivial modifications, of the parenthesis and bracket theories, are really justified by the fact, which we shall n ow prove, that the two mappings are the same. (It should not be surprising, since , that after two applications the bothersome conjugation disappears.) The proof is shorter than the introduction to it.

Let be any element of ; to it there corresponds the linear functional in , defined by , and to , in turn, there corresponds the linear functional in , defined by . Both these correspondences are given by the mapping introduced in this chapter. Earlier (see Section: Reflexivity ) the correspondent in of in was defined by for all in ; we must show that , as we here defined it, satisfies this identity. Let be any linear functional on (that is, any element of ); we have (The middle equality comes from the definition of inner product in .) This settles all our problems.

EXERCISES

Exercise 1. If and are subspaces of a finite-dimensional inner product space, then and

Exercise 2. If for each in , find a vector in such that .

Exercise 3. If is a vector in an inner product space, if is a linear transformation on that space, and if for every vector , then is a linear functional; find a vector such that for every .

Exercise 4.

If is a linear transformation on a finite-dimensional inner product space, then ; a necessary and sufficient condition that is that . (Hint: look at matrices.) This property of traces can often be used to obtain otherwise elusive algebraic facts about products of transformations and their adjoints.
Prove by a trace argument, and also directly, that if are linear transformations on a finite-dimensional inner product space and if , then .
If , then .
If commutes with and if commutes with , then commutes with B. (Hint: if and , then

Exercise 5.

Suppose that is a unitary space, and form the set of all ordered pairs with and in (that is, the direct sum of with itself). Prove that the equation defines an inner product in the direct sum .
If is defined by , then .
The graph of a linear transformation on is the set of all those elements of for which . Prove that the graph of every linear transformation on is a subspace of .
If is a linear transformation on with graph , then the graph of is the orthogonal complement (in ) of the image under (see (b)) of the graph of .

Exercise 6.

If for every linear transformation on a finite-dimensional inner product space , then is a norm (on the space of all linear transformations).
Is the norm induced by an inner product?

Exercise 7.

Two linear transformations and on an inner product space are called congruent if there exists an invertible linear transformation such that . (The concept is frequently defined for the "quadratic forms" associated with linear transformations and not for the linear transformations themselves; this is largely a matter of taste. Note that if and , then implies that .) Prove that congruence is an equivalence relation.
If and are congruent, then so also are and .
Does there exist a linear transformation such that is congruent to a scalar , but ?
Do there exist linear transformations and such that and are congruent, but and are not?
If two invertible transformations are congruent, then so are their inverses.

EXERCISES

Quick Links

Social Media