Page 178 - Linear Algebra Done Right

Generalized Eigenvectors
167
The Latin word nil
An operator is called nilpotent if some power of it equals 0. For
4
example, the operator N ∈L(F ) defined by
means nothing or zero;
the Latin word potent
N(z 1 ,z 2 ,z 3 ,z 4 ) = (z 3 ,z 4 , 0, 0)
means power. Thus
2
is nilpotent because N = 0. As another example, the operator of dif- nilpotent literally
ferentiation on P m (R) is nilpotent because the (m + 1) st derivative of means zero power.
any polynomial of degree at most m equals 0. Note that on this space of
dimension m + 1, we need to raise the nilpotent operator to the power
m + 1 to get 0. The next corollary shows that we never need to use a
power higher than the dimension of the space.
8.8 Corollary: Suppose N ∈L(V) is nilpotent. Then N dim V = 0.
Proof: Because N is nilpotent, every vector in V is a generalized
eigenvector corresponding to the eigenvalue 0. Thus from 8.7 we see
that null N dim V = V, as desired.

Having dealt with null spaces of powers of operators, we now turn
our attention to ranges. Suppose T ∈L(V) and k is a nonnegative
integer. If w ∈ range T k+1 , then there exists v ∈ V with

k
k
w = T k+1 v = T (Tv) ∈ range T .
k
Thus range T k+1 ⊂ range T . In other words, we have
1
0
k
V = range T ⊃ range T ⊃· · ·h range T ⊃ range T k+1 ⊃· · · . These inclusions go in
the opposite direction
The proposition below shows that the inclusions above become equal- from the corresponding
ities once the power reaches the dimension of V. inclusions for null
spaces (8.4).
8.9 Proposition: If T ∈L(V), then
range T dim V = range T dim V+1 = range T dim V+2 = ··· .

Proof: We could prove this from scratch, but instead let’s make use
of the corresponding result already proved for null spaces. Suppose
m> dim V. Then
dim range T m = dim V − dim null T m
= dim V − dim null T dim V

= dim range T dim V ,