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 \(A\) has one or both of the following two very special properties.

  1. If \(x_{1} \neq x_{2}\) , then \(A x_{1} \neq A x_{2}\) .
  2. To every vector \(y\) there corresponds (at least) one vector \(x\) such that \(A x=y\) .

If ever \(A\) has both these properties we shall say that \(A\) is invertible . If \(A\) is invertible, we define a linear transformation, called the inverse of \(A\) and denoted by \(A^{-1}\) , as follows. If \(y_{0}\) is any vector, we may (by (ii)) find an \(x_{0}\) for which \(A x_{0}=y_{0}\) . This \(x_{0}\) is, moreover, uniquely determined, since \(x_{0} \neq x_{1}\) implies (by (i)) that \(y_{0}=A x_{0} \neq A x_{1}\) . We define \(A^{-1} y_{0}\) to be \(x_{0}\) . To prove that \(A^{-1}\) is linear, we evaluate \(A^{-1}(\alpha_{1} y_{1}+\alpha_{2} y_{2})\) . If \(A x_{1}=y_{1}\) and \(A x_{2}=y_{2}\) , then the linearity of \(A\) tells us that \[A(\alpha_{1} x_{1}+\alpha_{2} x_{2})=\alpha_{1} y_{1}+\alpha_{2} y_{2},\] so that \[A^{-1}(\alpha_{1} y_{1}+\alpha_{2} y_{2})=\alpha_{1} x_{1}+\alpha_{2} x_{2}=\alpha_{1} A^{-1} y_{1}+\alpha_{2} A^{-1} y_{2}.\] 

As a trivial example of an invertible transformation we mention the identity transformation \(1\) ; clearly \(1^{-1}=1\) . The transformation \(0\) 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 \(A\) we have \[A A^{-1}=A^{-1} A=1\] we shall now show that these equations serve to characterize \(A^{-1}\) .

Theorem 1. If \(A\) , \(B\) , and \(C\) are linear transformations such that \[A B=C A=1\] then \(A\) is invertible and \(A^{-1}=B=C\) .

Proof. If \(A x_{1}=A x_{2}\) , then \(C A x_{1}=C A x_{2}\) , so that (since \(C A=1\) ) \(x_{1}=x_{2}\) ; in other words, the first condition of the definition of invertibility is satisfied. The second condition is also satisfied, for if \(y\) is any vector and \(x=B y\) , then \(y=A B y=A x\) . Multiplying \(A B=1\) on the left, and \(C A=1\) on the right, by \(A^{-1}\) , we see that \(A^{-1}=B=C\) . ◻

To show that neither \(A B=1\) nor \(C A=1\) is, by itself, sufficient to ensure the invertibility of \(A\) , we call attention to the differentiation and integration transformations \(D\) and \(S\) , defined in Section: Linear transformations , (4) and (5). Although \(D S=1\) , neither \(D\) nor \(S\) is invertible; \(D\) violates (i), and \(S\) violates (ii).

In finite-dimensional spaces the situation is much simpler.

Theorem 2. A linear transformation \(A\) on a finite-dimensional vector space \(\mathcal{V}\) is invertible if and only if \(A x=0\) implies that \(x=0\) , or, alternatively, if and only if every \(y\) in \(\mathcal{V}\) can be written in the form \(y=A x\) .

Proof. If \(A\) is invertible, both conditions are satisfied; this much is trivial. Suppose now that \(A x=0\) implies that \(x=0\) . Then \(u \neq v\) , that is, \(u-v \neq 0\) , implies that \(A(u-v) \neq 0\) , that is, that \(A u \neq A v\) ; this proves (i). To prove (ii), let \(\{x_{1}, \ldots, x_{n}\}\) be a basis in \(\mathcal{V}\) ; we assert that \(\{A x_{1}, \ldots, A x_{n}\}\) is also a basis. According to Section: Dimension , Theorem 2, we need only prove linear independence. But \(\sum_{i} \alpha_{i} A x_{i}=0\) means \(A\big(\sum_{i} \alpha_{i} x_{i}\big)=0\) , and, by hypothesis, this implies that \(\sum_{i} \alpha_{i} x_{i}=0\) ; the linear independence of the \(x_{i}\) now tells us that \(\alpha_{1}=\cdots=\alpha_{n}=0\) . It follows, of course, that every vector \(y\) may be written in the form \[y=\sum_{i} \alpha_{i} A x_{i}=A\Big(\sum_{i} \alpha_{i} x_{i}\Big).\] 

Let us assume next that every \(y\) is an \(A x\) , and let \(\{y_{1}, \ldots, y_{n}\}\) be any basis in \(\mathcal{V}\) . Corresponding to each \(y_{i}\) we may find a (not necessarily unique) \(x_{i}\) for which \(y_{i}=A x_{i}\) ; we assert that \(\{x_{1}, \ldots, x_{n}\}\) is also a basis. For \(\sum_{i} \alpha_{i} x_{i}=0\) implies \[\sum_{i} \alpha_{i} A x_{i}=\sum_{i} \alpha_{i} y_{i}=0,\] so that \(\alpha_{1}=\cdots=\alpha_{n}=0\) . Consequently every \(x\) may be written in the form \(x=\sum_{i} \alpha_{i} x_{i}\) , and \(A x=0\) implies, as in the argument just given, that \(x=0\) . ◻

Theorem 3. If \(A\) and \(B\) are invertible, then \(A B\) is invertible and \((A B)^{-1}=B^{-1} A^{-1}\) . If \(A\) is invertible and \(\alpha \neq 0\) , then \(\alpha A\) is invertible and \((\alpha A)^{-1}=\frac{1}{\alpha} A^{-1}\) . If \(A\) is invertible, then \(A^{-1}\) is invertible and \((A^{-1})^{-1}=A\) .

Proof. According to Theorem 1, it is sufficient to prove (for the first statement) that the product of \(A B\) with \(B^{-1} A^{-1}\) , 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 \(A A^{-1}=A^{-1} A=1\) are completely symmetric in \(A\) and \(A^{-1}\) . ◻

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 \(A\) , whenever possible, by using \(A^{-1}\) . We shall not find it useful to do this, except in one case: if \(A\) is invertible, then we know that \(A^{n}\) is also invertible, \(n=1,2, \ldots\) ; we shall write \(A^{-n}\) for \((A^{n})^{-1}\) , so that \(A^{-n}=(A^{-1})^{n}\) .

EXERCISES

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

Exercise 2. A linear transformation \(A\) is defined on \(\mathbb{C}^{2}\) by \[A(\xi_{1}, \xi_{2})=(\alpha \xi_{1}+\beta \xi_{2}, \gamma \xi_{1}+\delta \xi_{2})\] where \(\alpha\) , \(\beta\) , \(\gamma\) , and \(\delta\) are fixed scalars. Prove that \(A\) is invertible if and only if \(\alpha \delta - \beta \gamma \neq 0\) .

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

Exercise 4. If \(A\) and \(B\) are linear transformations on a finite-dimensional vector space, and if \(A B=1\) , then both \(A\) and \(B\) are invertible.

Exercise 5. 

  1. If \(A\) , \(B\) , \(C\) , and \(D\) are linear transformations (all on the same vector space), and if both \(A+B\) and \(A-B\) are invertible, then there exist linear transformations \(X\) and \(Y\) such that \[A X+B Y=C\] and \[B X+A Y=D.\] 
  2. To what extent are the invertibility assumptions in (a) necessary?

Exercise 6. 

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

Exercise 7. Show that if \(A\) is a linear transformation such that \(A^{2}-A+1=0\) , then \(A\) is invertible.

Exercise 8. If \(A\) and \(B\) are linear transformations (on the same vector space) and if \(A B=1\) , then \(A\) is called a left inverse of \(B\) and \(B\) is called a right inverse of \(A\) . Prove that if \(A\) has exactly one right inverse, say \(B\) , then \(A\) is invertible. (Hint: consider \(B A+B-1\) .)

Exercise 9. If \(A\) is an invertible linear transformation on a finite-dimensional vector space \(\mathcal{V}\) , then there exists a polynomial \(p\) such that \(A^{-1}=p(A)\) . (Hint: find a non-zero polynomial \(q\) of least degree such that \(q(A)=0\) and prove that its constant term cannot be \(0\) .)

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.