Example Problems and Solutions 703A–9–9. If an n*nmatrixAhasndistinct eigenvalues, then the minimal polynomial of Ais identical to
the characteristic polynomial. Also, if the multiple eigenvalues of Aare linked in a Jordan chain,
the minimal polynomial and the characteristic polynomial are identical. If, however, the multiple
eigenvalues of Aare not linked in a Jordan chain, the minimal polynomial is of lower degree than
the characteristic polynomial.
Using the following matrices AandBas examples, verify the foregoing statements about the
minimal polynomial when multiple eigenvalues are involved:
Solution.First, consider the matrix A. The characteristic polynomial is given byThus, the eigenvalues of Aare 2, 2, and 1. It can be shown that the Jordan canonical form of Aisand the multiple eigenvalues are linked in the Jordan chain as shown.
To determine the minimal polynomial, let us first obtain adj(lI-A). It is given byNotice that there is no common divisor of all the elements of Hence,d(l)=1.
Thus, the minimal polynomial f(l)is identical to the characteristic polynomial, orA simple calculation proves that= C
0
0
0
0
0
0
0
0
0
S = 0
= C
8
0
0
72
8
21
28
0
1
S - 5 C
4
0
0
16
4
9
12
0
1
S + 8 C
2
0
0
1
2
3
4
0
1
S- 4 C
1
0
0
0
1
0
0
0
1
S
A^3 - 5 A^2 + 8 A- 4 I
=l^3 - 5 l^2 + 8 l- 4f(l)=∑l I-A∑=(l-2)^2 (l-1)adj(l I-A).adj(l I-A)= C
(l- 2 )(l- 1 )
0
0(l+ 11 )
(l- 2 )(l- 1 )
3 (l- 2 )4 (l- 2 )
0
(l- 2 )^2S
C
2
0
0
1
2
0
0
0
1
S
∑l I-A∑= 3
l- 2
0
0- 1
l- 2- 3
- 4
0
l- 13 =(l-2)^2 (l-1)
A= C
2
0
0
1
2
3
4
0
1
S, B= C
2
0
0
0
2
3
0
0
1
S