Page 172 - A Course in Linear Algebra with Applications

156 Chapter Six: Linear Transformations

Example 6.1.5
Consider the functions F and G with domain and codomain
l
R which are defined by F(x) = 2x — 1 and G(x) = (x + )/2.
Then G is an inverse of F since F o G and G o F are both
equal to lp^. Indeed

F o G(x) = F(G(x)) = F((x + l)/2) = 2((x + l)/2) - 1 = 2;,

with a similar computation for 6* o F(x).

Not every function has an inverse; in fact a basic theorem
asserts that only the bijective ones do.
Theorem 6.1.3
A function F : X —> Y has an inverse if and only if it is
bijective.
Proof
Suppose first that F has an inverse function G : Y —> X.
If F(xi) = F(x 2), then, on applying G to both sides, we
obtain G o F(xi) = G o F(x 2). But G o F is the identity
function on X, so xi = x 2. Hence F is injective. Next let y
be any element of Y; then, since FoG is the identity function,
y = F o G(y) = F(G(y)), which shows that / belongs to the
j
image of F and F is surjective. Therefore F is bijective.
Conversely, assume that F is a bijective function. We
need to find an inverse function G : Y —> X for F. To this end
let y belong to F; then, since F is surjective, y = F{x) for
some a; in X; moreover x is uniquely determined by y since
F is injective. This allows us to define G(y) to be x. Then
G(F(a;)) = G(y) = x and F(G(y)) = F(ar) = j/. Here it is
necessary to observe that every element of X is of the form
G(y) for some y in Y, so that G(F(x)) equals x for all elements
x of X. Therefore G is an inverse function for F.

The next observation is that when inverse functions do
exist, they are unique.