Page 58 - Linear Algebra Done Right

Chapter 3. Linear Maps
44
3.3
Proposition: If T ∈L(V, W), then range T is a subspace of W.
Proof: Suppose T ∈L(V, W). Then T(0) = 0 (by 3.1), which im-
plies that 0 ∈ range T.
If w 1 ,w 2 ∈ range T, then there exist v 1 ,v 2 ∈ V such that Tv 1 = w 1
and Tv 2 = w 2 . Thus
T(v 1 + v 2 ) = Tv 1 + Tv 2 = w 1 + w 2 ,
and hence w 1 +w 2 ∈ range T. Thus range T is closed under addition.
If w ∈ range T and a ∈ F, then there exists v ∈ V such that Tv = w.
Thus
T(av) = aTv = aw,
and hence aw ∈ range T. Thus range T is closed under scalar multipli-
cation.
We have shown that range T contains 0 and is closed under addition
and scalar multiplication. Thus range T is a subspace of W.

Many mathematicians A linear map T : V → W is called surjective if its range equals W.
use the term onto, For example, the differentiation map T ∈L(P(R), P(R)) defined by
which means the same Tp = p is surjective because its range equals P(R). As another exam-

2
as surjective. ple, the linear map T ∈L(P(R), P(R)) defined by (Tp)(x) = x p(x) is
not surjective because its range does not equal P(R). As a final exam-
ple, you should verify that the backward shift T ∈L(F , F ) defined
∞
∞
by
T(x 1 ,x 2 ,x 3 ,...) = (x 2 ,x 3 ,...)
is surjective.
Whether a linear map is surjective can depend upon what we are
thinking of as the target space. For example, fix a positive integer m.
The differentiation map T ∈L(P m (R), P m (R)) defined by Tp = p
is not surjective because the polynomial x m is not in the range of T.
However, the differentiation map T ∈L(P m (R), P m−1 (R)) defined by
Tp = p is surjective because its range equals P m−1 (R), which is now

the target space.
The next theorem, which is the key result in this chapter, states that
the dimension of the null space plus the dimension of the range of a
linear map on a finite-dimensional vector space equals the dimension
of the domain.