Page 56 - Linear Algebra Done Right

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only functions whose derivative equals the zero function are the con-
stant functions, so in this case the null space of T equals the set of
constant functions. Chapter 3. Linear Maps
2
In the multiplication by x example, we defined T ∈L(P(R), P(R))
2
2
by (Tp)(x) = x p(x). The only polynomial p such that x p(x) = 0
for all x ∈ R is the 0 polynomial. Thus in this case we have
null T ={0}.
In the backward shift example, we defined T ∈L(F , F ) by
∞
∞
T(x 1 ,x 2 ,x 3 ,...) = (x 2 ,x 3 ,...).
Clearly T(x 1 ,x 2 ,x 3 ,...) equals 0 if and only if x 2 ,x 3 ,... are all 0. Thus
in this case we have

null T ={(a, 0, 0,...) : a ∈ F}.

The next proposition shows that the null space of any linear map is
a subspace of the domain. In particular, 0 is in the null space of every
linear map.

3.1 Proposition: If T ∈L(V, W), then null T is a subspace of V.

Proof: Suppose T ∈L(V, W). By additivity, we have

T(0) = T(0 + 0) = T(0) + T(0),

which implies that T(0) = 0. Thus 0 ∈ null T.
If u, v ∈ null T, then

T(u + v) = Tu + Tv = 0 + 0 = 0,

and hence u + v ∈ null T. Thus null T is closed under addition.
If u ∈ null T and a ∈ F, then

T(au) = aTu = a0 = 0,

and hence au ∈ null T. Thus null T is closed under scalar multiplica-
tion.
We have shown that null T contains 0 and is closed under addition
and scalar multiplication. Thus null T is a subspace of V.