Page 57 - Linear Algebra Done Right

Null Spaces and Ranges
A linear map T : V → W is called injective if whenever u, v ∈ V
use the term
and Tu = Tv, we have u = v. The next proposition says that we
can check whether a linear map is injective by checking whether 0 is Many mathematicians 43
one-to-one, which
the only vector that gets mapped to 0. As a simple application of this means the same as
proposition, we see that of the three linear maps whose null spaces we injective.
2
computed earlier in this section (differentiation, multiplication by x ,
2
and backward shift), only multiplication by x is injective.
3.2 Proposition: Let T ∈L(V, W). Then T is injective if and only
if null T ={0}.
Proof: First suppose that T is injective. We want to prove that
null T ={0}. We already know that {0}⊂ null T (by 3.1). To prove the
inclusion in the other direction, suppose v ∈ null T. Then

T(v) = 0 = T(0).
Because T is injective, the equation above implies that v = 0. Thus
null T ={0}, as desired.
To prove the implication in the other direction, now suppose that
null T ={0}. We want to prove that T is injective. To do this, suppose
u, v ∈ V and Tu = Tv. Then

0 = Tu − Tv = T(u − v).
Thus u − v is in null T, which equals {0}. Hence u − v = 0, which
implies that u = v. Hence T is injective, as desired.

For T ∈L(V, W), the range of T, denoted range T, is the subset of Some mathematicians
W consisting of those vectors that are of the form Tv for some v ∈ V: use the word image,
which means the same
range T ={Tv : v ∈ V}.
as range.
For example, if T ∈L(P(R), P(R)) is the differentiation map defined by
Tp = p , then range T =P(R) because for every polynomial q ∈P(R)

there exists a polynomial p ∈P(R) such that p = q.

As another example, if T ∈L(P(R), P(R)) is the linear map of
2
2
multiplication by x defined by (Tp)(x) = x p(x), then the range
m
2
of T is the set of polynomials of the form a 2 x +· · ·+ a m x , where
a 2 ,...,a m ∈ R.
The next proposition shows that the range of any linear map is a
subspace of the target space.