Page 195 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
184
Proof: Suppose N is nilpotent. Then N is not injective and thus
dim range N< dim V (see 3.21). By induction on the dimension of V,
we can assume that the lemma holds on all vector spaces of smaller
dimension. Using range N in place of V and N| range N in place of N,we
thus have vectors u 1 ,...,u j ∈ range N such that
(i) (u 1 ,Nu 1 ,...,N m(u 1 ) u 1 ,...,u j ,Nu j ,...,N m(u j ) u j ) is a basis of
range N;
(ii) (N m(u 1 ) u 1 ,...,N m(u j ) u j ) is a basis of null N ∩ range N.

Because each u r ∈ range N, we can choose v 1 ,...,v j ∈ V such that
Nv r = u r for each r. Note that m(v r ) = m(u r ) + 1 for each r.
The existence of a Let W be a subspace of null N such that
subspace W with this
property follows from 8.41 null N = (null N ∩ range N) ⊕ W
2.13.
and choose a basis of W, which we will label (v j+1 ,...,v k ). Because
v j+1 ,...,v k ∈ null N, we have m(v j+1 ) =· · ·= m(v k ) = 0.
Having constructed v 1 ,...,v k , we now need to show that (a) and
(b) hold. We begin by showing that the alleged basis in (a) is linearly
independent. To do this, suppose

k m(v r )

s
8.42 0 = a r,s N (v r ),
r=1 s=0
where each a r,s ∈ F. Applying N to both sides of the equation above,
we get
k m(v r )

0 = a r,s N s+1 (v r )
r=1 s=0
j m(u r )

s
= a r,s N (u r ).
r=1 s=0

The last equation, along with (i), implies that a r,s = 0 for 1 ≤ r ≤ j,
0 ≤ s ≤ m(v r ) − 1. Thus 8.42 reduces to the equation

0 =a 1,m(v 1 ) N m(v 1 ) v 1 +· · ·+ a j,m(v j ) N m(v j ) v j

+ a j+1,0 v j+1 +· · ·+ a k,0 v k .