**Unformatted text preview: **⎛
⎞
⎛
000
010
0
N1 = ⎝ 0 0 0 ⎠ , N2 = ⎝ 0 0 0 ⎠ , N3 = ⎝ 0
000
000
0 T
H
1
0
0 ⎞
0
1⎠.
0 IG
R Example 7.7.2 For a nilpotent matrix L, the theoretical development relies on a complicated
basis for N (L) to derive the structure of the Jordan form N as well as the
Jordan chains that constitute a nonsingular matrix P such that P−1 LP = N.
But, after the dust settled, we saw that a basis for N (L) is not needed to
construct N because N is completely determined simply by ranks of powers of
L. A basis for N (L) is only required to construct the Jordan chains in P. Y
P Question: For the purpose of constructing Jordan chains in P, can we use an
arbitrary basis for N (L) instead of the complicated basis built from the Mi ’s? O
C Answer: No! Consider the nilpotent matrix
⎛
⎞
20
1
L = ⎝ −4 0 −2 ⎠ and its Jordan form
−4 0 −2 ⎛ 01
00
N=⎝
0 0 ⎞
0
0⎠.
0 If P−1 LP = N, where P = [ x1 | x2 | x3 ], then LP = PN implies that
Lx1 = 0, Lx2 = x1 , and Lx3 = 0. In other words, B = {x1 , x3 } must
be a basis for N (L), and Jx1 = {x1 , x2 } must be a Jordan chain built on top
of x1 . If we try to construct such vectors by starting with the naive basis
⎛
⎞
⎛⎞
1
0
x1 = ⎝ 0 ⎠
and x3 = ⎝ 1 ⎠
(7.7.7)
−2
0 Copyright c 2000 SIAM Buy online from SIAM
http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com
582
Chapter 7
Eigenvalues and Eigenvectors
http://www.amazon.com/exec/obidos/ASIN/0898714540
for N (L) obtained by solving Lx = 0 with straightforward Gaussian elimination, we immediately hit a brick wall because x1 ∈ R (L) means Lx2 = x1
is an inconsistent system, so x2 cannot be determined. Similarly, x3 ∈ R (L)
insures that the same diﬃculty occurs if x3 is used in place of x1 . In other
words, even though the vectors in (7.7.7) constitute an otherwise perfectly good
basis for N (L), they can’t be used to build P. It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] Example 7.7.3 Example 7.7.4 Problem: Let Ln...

View
Full Document