Page 71 - Linear Algebra Done Right

Invertibility
Proposition: If V and W are finite dimensional, then L(V, W)
3.20
is finite dimensional and
dim L(V, W) = (dim V)(dim W). 57
Proof: This follows from the equation dim Mat(m, n, F) = mn,
3.18, and 3.19.
A linear map from a vector space to itself is called an operator.If The deepest and most
we want to specify the vector space, we say that a linear map T : V → V important parts of
is an operator on V. Because we are so often interested in linear maps linear algebra, as well
from a vector space into itself, we use the notation L(V) to denote the as most of the rest of
set of all operators on V. In other words, L(V) =L(V, V). this book, deal with
Recall from 3.17 that a linear map is invertible if it is injective and operators.
surjective. For a linear map of a vector space into itself, you might
wonder whether injectivity alone, or surjectivity alone, is enough to
imply invertibility. On infinite-dimensional vector spaces neither con-
dition alone implies invertibility. We can see this from some examples
2
we have already considered. The multiplication by x operator (from
P(R) to itself) is injective but not surjective. The backward shift (from
F ∞ to itself) is surjective but not injective. In view of these examples,
the next theorem is remarkable—it states that for maps from a finite-
dimensional vector space to itself, either injectivity or surjectivity alone
implies the other condition.

3.21 Theorem: Suppose V is finite dimensional. If T ∈L(V), then
the following are equivalent:
(a) T is invertible;
(b) T is injective;
(c) T is surjective.

Proof: Suppose T ∈L(V). Clearly (a) implies (b).
Now suppose (b) holds, so that T is injective. Thus null T ={0}
(by 3.2). From 3.4 we have

dim range T = dim V − dim null T
= dim V,

which implies that range T equals V (see Exercise 11 in Chapter 2). Thus
T is surjective. Hence (b) implies (c).