Page 400 - Matrix Analysis & Applied Linear Algebra

396 Chapter 5 Norms, Inner Products, and Orthogonality
Example 5.10.1
2 0 0

Problem: Determine the index of A = 0 1 1 .
0 −1 −1
Solution: A is singular (because rank (A)=2), so index(A) > 0. Since
   
400 800
3
2
A =  000  and A =  000  ,
000 000

3
we see that rank (A) >rank A 2 = rank A , so index(A)=2. Alternately,
       
2 0 4 8
 
2 3
R (A)= span    1  ,R A = span   ,R A = span   ,
0
,
0
0
0 −1 0 0
 

so R (A) ⊃ R A 2 = R A 3 implies index(A)=2.
Nilpotent Matrices
k
• N n×n is said to be nilpotent whenever N = 0 for some positive
integer k.
k
• k = index(N)is the smallest positive integer such that N = 0.
(Some authors refer to index(N)as the index of nilpotency.)
Proof. To prove that k = index(N)is the smallest positive integer such that
k
p
N = 0, suppose p is a positive integer such that N = 0, but N p−1 = 0.

We know from (5.10.3) that R N 0 ⊃ R (N) ⊃ ··· ⊃ R N k = R N k+1 =

R N k+2 = ··· , and this makes it clear that it’s impossible to have p<k or
p>k, so p = k is the only choice.
Example 5.10.2
Problem: Verify that
 
010
N =  001 
000
is a nilpotent matrix, and determine its index.

Solution: Computing the powers
   
001 000
2
3
N =  000  and N =  000  ,
000 000
reveals that N is indeed nilpotent, and it shows that index(N)= 3because
3
2
N = 0, but N = 0.