Inverses

In each of the two preceding sections we gave an example; these two examples bring out the two nasty properties that the multiplication of linear transformations has, namely, non-commutativity and the existence of divisors of zero. We turn now to the more pleasant properties that linear transformations sometimes have.

It may happen that the linear transformation has one or both of the following two very special properties.

If , then .
To every vector there corresponds (at least) one vector such that .

If ever has both these properties we shall say that is invertible . If is invertible, we define a linear transformation, called the inverse of and denoted by , as follows. If is any vector, we may (by (ii)) find an for which . This is, moreover, uniquely determined, since implies (by (i)) that . We define to be . To prove that is linear, we evaluate . If and , then the linearity of tells us that so that

As a trivial example of an invertible transformation we mention the identity transformation ; clearly . The transformation is not invertible; it violates both the conditions (i) and (ii) about as strongly as they can be violated.

It is immediate from the definition that for any invertible we have we shall now show that these equations serve to characterize .

Theorem 1. If , , and are linear transformations such that then is invertible and .

Proof. If , then , so that (since ) ; in other words, the first condition of the definition of invertibility is satisfied. The second condition is also satisfied, for if is any vector and , then . Multiplying on the left, and on the right, by , we see that . ◻

To show that neither nor is, by itself, sufficient to ensure the invertibility of , we call attention to the differentiation and integration transformations and , defined in Section: Linear transformations , (4) and (5). Although , neither nor is invertible; violates (i), and violates (ii).

In finite-dimensional spaces the situation is much simpler.

Theorem 2. A linear transformation on a finite-dimensional vector space is invertible if and only if implies that , or, alternatively, if and only if every in can be written in the form .

Proof. If is invertible, both conditions are satisfied; this much is trivial. Suppose now that implies that . Then , that is, , implies that , that is, that ; this proves (i). To prove (ii), let be a basis in ; we assert that is also a basis. According to Section: Dimension , Theorem 2, we need only prove linear independence. But means , and, by hypothesis, this implies that ; the linear independence of the now tells us that . It follows, of course, that every vector may be written in the form

Let us assume next that every is an , and let be any basis in . Corresponding to each we may find a (not necessarily unique) for which ; we assert that is also a basis. For implies so that . Consequently every may be written in the form , and implies, as in the argument just given, that . ◻

Theorem 3. If and are invertible, then is invertible and . If is invertible and , then is invertible and . If is invertible, then is invertible and .

Proof. According to Theorem 1, it is sufficient to prove (for the first statement) that the product of with , in both orders, is the identity; this verification we leave to the reader. The proofs of both the remaining statements are identical in principle with this proof of the first statement; the last statement, for example, follows from the fact that the equations are completely symmetric in and . ◻

We conclude our discussion of inverses with the following comment. In the spirit of the preceding section we may, if we like, define rational functions of , whenever possible, by using . We shall not find it useful to do this, except in one case: if is invertible, then we know that is also invertible, ; we shall write for , so that .

EXERCISES

Exercise 1. Which of the linear transformations described in Section: Transformations as vectors , Ex. 1 are invertible?

Exercise 2. A linear transformation is defined on by where , , , and are fixed scalars. Prove that is invertible if and only if .

Exercise 3. If and are linear transformations (on the same vector space), then a necessary and sufficient condition that both and be invertible is that both and be invertible.

Exercise 4. If and are linear transformations on a finite-dimensional vector space, and if , then both and are invertible.

Exercise 5.

If , , , and are linear transformations (all on the same vector space), and if both and are invertible, then there exist linear transformations and such that and
To what extent are the invertibility assumptions in (a) necessary?

Exercise 6.

A linear transformation on a finite-dimensional vector space is invertible if and only if it preserves linear independence. To say that preserves linear independence means that whenever is a linearly independent set in the space on which acts, then is also a linearly independent set in . (The symbol denotes, of course, the set of all vectors of the form , with in .)
Is the assumption of finite-dimensionality needed for the validity of (a)?

Exercise 7. Show that if is a linear transformation such that , then is invertible.

Exercise 8. If and are linear transformations (on the same vector space) and if , then is called a left inverse of and is called a right inverse of . Prove that if has exactly one right inverse, say , then is invertible. (Hint: consider .)

Exercise 9. If is an invertible linear transformation on a finite-dimensional vector space , then there exists a polynomial such that . (Hint: find a non-zero polynomial of least degree such that and prove that its constant term cannot be .)

Exercise 10. Devise a sensible definition of invertibility for linear transformations from one vector space to another. Using that definition, decide which (if any) of the linear transformations described in Section: Transformations as vectors , Ex. 3 are invertible.

EXERCISES

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