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 $A$ has one or both of the following two very special properties.

If $x_{1} \neq x_{2}$ , then $A x_{1} \neq A x_{2}$ .
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 $A^{- 1} (α_{1} y_{1} + α_{2} y_{2})$ . If $A x_{1} = y_{1}$ and $A x_{2} = y_{2}$ , then the linearity of $A$ tells us that $A (α_{1} x_{1} + α_{2} x_{2}) = α_{1} y_{1} + α_{2} y_{2},$ so that $A^{- 1} (α_{1} y_{1} + α_{2} y_{2}) = α_{1} x_{1} + α_{2} x_{2} = α_{1} A^{- 1} y_{1} + α_{2} A^{- 1} y_{2} .$

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 $𝒱$ is invertible if and only if $A x = 0$ implies that $x = 0$ , or, alternatively, if and only if every $y$ in $𝒱$ 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 ${x_{1}, \dots, x_{n}}$ be a basis in $𝒱$ ; we assert that ${A x_{1}, \dots, A x_{n}}$ is also a basis. According to Section: Dimension , Theorem 2, we need only prove linear independence. But $\sum_{i} α_{i} A x_{i} = 0$ means $A (\sum_{i} α_{i} x_{i}) = 0$ , and, by hypothesis, this implies that $\sum_{i} α_{i} x_{i} = 0$ ; the linear independence of the $x_{i}$ now tells us that $α_{1} = \dots = α_{n} = 0$ . It follows, of course, that every vector $y$ may be written in the form $y = \sum_{i} α_{i} A x_{i} = A (\sum_{i} α_{i} x_{i}) .$

Let us assume next that every $y$ is an $A x$ , and let ${y_{1}, \dots, y_{n}}$ be any basis in $𝒱$ . Corresponding to each $y_{i}$ we may find a (not necessarily unique) $x_{i}$ for which $y_{i} = A x_{i}$ ; we assert that ${x_{1}, \dots, x_{n}}$ is also a basis. For $\sum_{i} α_{i} x_{i} = 0$ implies $\sum_{i} α_{i} A x_{i} = \sum_{i} α_{i} y_{i} = 0,$ so that $α_{1} = \dots = α_{n} = 0$ . Consequently every $x$ may be written in the form $x = \sum_{i} α_{i} x_{i}$ , 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 $α \neq 0$ , then $α A$ is invertible and $(α A)^{- 1} = \frac{1}{α} A^{- 1}$ . 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, $n = 1, 2, \dots$ ; 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 $ℂ^{2}$ by $A (ξ_{1}, ξ_{2}) = (α ξ_{1} + β ξ_{2}, γ ξ_{1} + δ ξ_{2})$ where $α$ , $β$ , $γ$ , and $δ$ are fixed scalars. Prove that $A$ is invertible if and only if $α δ - β γ \neq 0$ .

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.

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 $B X + A Y = D .$
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 $A$ preserves linear independence means that whenever $𝒳$ is a linearly independent set in the space $𝒱$ on which $A$ acts, then $A 𝒳$ is also a linearly independent set in $𝒱$ . (The symbol $A 𝒳$ denotes, of course, the set of all vectors of the form $A x$ , with $x$ in $𝒳$ .)
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 $𝒱$ , 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.

EXERCISES

Quick Links

Social Media