Page 194 - Linear Algebra Done Right

Jordan Form
Jordan Form
We know that if V is a complex vector space, then for every T ∈L(V) 183
there is a basis of V with respect to which T has a nice upper-triangular
matrix (see 8.28). In this section we will see that we can do even better—
there is a basis of V with respect to which the matrix of T contains zeros
everywhere except possibly on the diagonal and the line directly above
the diagonal.
We begin by describing the nilpotent operators. Consider, for ex-
n
ample, the nilpotent operator N ∈L(F ) defined by
N(z 1 ,...,z n ) = (0,z 1 ,...,z n−1 ).

n
If v = (1, 0,..., 0), then clearly (v,Nv,...,N n−1 v) is a basis of F and
(N n−1 v) is a basis of null N, which has dimension 1.
5
As another example, consider the nilpotent operator N ∈L(F ) de-
fined by

8.39 N(z 1 ,z 2 ,z 3 ,z 4 ,z 5 ) = (0,z 1 ,z 2 , 0,z 4 ).

Unlike the nilpotent operator discussed in the previous paragraph, for
5
this nilpotent operator there does not exist a vector v ∈ F such that
3
2
5
4
(v, Nv, N v, N v, N v) is a basis of F . However, if v 1 = (1, 0, 0, 0, 0)
2
and v 2 = (0, 0, 0, 1, 0), then (v 1 ,Nv 1 ,N v 1 ,v 2 ,Nv 2 ) is a basis of F 5
2
and (N v 1 ,Nv 2 ) is a basis of null N, which has dimension 2.
Suppose N ∈L(V) is nilpotent. For each nonzero vector v ∈ V, let
m(v) denote the largest nonnegative integer such that N m(v) v = 0. For Obviously m(v)
5
example, if N ∈L(F ) is defined by 8.39, then m(1, 0, 0, 0, 0) = 2. depends on N as well
The lemma below shows that every nilpotent operator N ∈L(V) as on v, but the choice
behaves similarly to the example defined by 8.39, in the sense that there of N will be clear from
is a finite collection of vectors v 1 ,...,v k ∈ V such that the nonzero the context.
j
vectors of the form N v r form a basis of V; here r varies from 1 to k
and j varies from 0 to m(v r ).
8.40 Lemma: If N ∈L(V) is nilpotent, then there exist vectors
v 1 ,...,v k ∈ V such that

(a) (v 1 ,Nv 1 ,...,N m(v 1 ) v 1 ,...,v k ,Nv k ,...,N m(v k ) v k ) is a basis of V;
(b) (N m(v 1 ) v 1 ,...,N m(v k ) v k ) is a basis of null N.