Page 180 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes p, x(x + 2), x(x – 2), x – 4. The three quadratic polynomials can be eliminated because it is obvious
2
A
A
I
at a glance that A 2 2 , A 2 2 , A 2 4 . Therefore p is the minimal polynomial for A. In
particular 0, 2, and – 2 are the characteristic values of A. One of the factors x, x – 2, x + 2 must be
repeated twice in the characteristic polynomial. Evidently, rank (A) = 2. Consequently there is a
two-dimensional space of characteristic vectors associated with the characteristic value 0. From
2
2
Theorem 2, it should now be clear that the characteristic polynomial is x (x – 4) and that A is
similar over the field of rational numbers to the matrix
0 0 0 0
0 0 0 0
0 0 2 0 .
0 0 0 2
Example 3: Verify Cayley-Hamilton’s theorem for the linear transformation T represented
by the matrix A.
0 0 1
A = 3 1 0
2 1 4

Solution: The characteristic polynomial is given by

0 x 0 1
A x I = 3 1 x 0
2 1 4 x

= – x 1 x 4 x 3 2 2x

= x 4 5x x 2 5 2x
= x 3 5x 2 6x 5 0

or f(x) = x 3 5x 2 6x 5 0
Now

0 0 1 0 0 1 2 1 4
2
A = 3 1 0 3 1 0 3 1 3
2 1 4 2 1 4 5 5 14

2 1 4 0 0 1 5 5 14
3 1 3 3 1 0 3 4 15
A =
3
5 5 14 2 1 4 13 19 51
So
2
3
f(A) = A – 5A + 6A – 5I
5 5 14 10 5 20 0 0 6 5 0 0
= 3 4 15 15 5 15 18 6 0 0 5 0
13 19 51 25 25 14 12 6 24 0 0 5

0 0 0
= 0 0 0 0
0 0 0

where 0 being null matrix. So f(A) = 0

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