Page 291 - 35Linear Algebra

16.2 Image 291

Theorem 16.2.1. A function f : S → T has an inverse function g: T → S
if and only if f is bijective.

Proof. This is an “if and only if” statement so the proof has two parts.

1. (Existence of an inverse ⇒ bijective.)

Suppose that f has an inverse function g. We need to show f is bijec-
tive, which we break down into injective and surjective.

0
• The function f is injective: Suppose that we have s, s ∈ S such
0
that f(s) = f(s ). We must have that g(f(s)) = s for any s ∈ S, so
0
0
0
in particular g(f(s)) = s and g(f(s )) = s . But since f(s) = f(s ),
0
0
we have g(f(s)) = g(f(s )) so s = s . Therefore, f is injective.
• The function f is surjective: Let t be any element of T. We must
have that f(g(t)) = t. Thus, g(t) is an element of S which maps
to t. So f is surjective.
2. (Bijectivity ⇒ existence of an inverse.) Suppose that f is bijective.
Hence f is surjective, so every element t ∈ T has at least one pre-
image. Being bijective, f is also injective, so every t has no more than
one pre-image. Therefore, to construct an inverse function g, we simply
define g(t) to be the unique pre-image f −1 (t) of t.

Now let us specialize to functions f that are linear maps between two
vector spaces. Everything we said above for arbitrary functions is exactly
the same for linear functions. However, the structure of vector spaces lets
us say much more about one-to-one and onto functions whose domains are
vector spaces than we can say about functions on general sets. For example,
we know that a linear function always sends 0 V to 0 W , i.e.,

f(0 V ) = 0 W

In Review Exercise 2, you will show that a linear transformation is one-to-one
if and only if 0 V is the only vector that is sent to 0 W . Linear functions are
unlike arbitrary functions between sets in that, by looking at just one (very
special) vector, we can figure out whether f is one-to-one!

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