Page 179 - DMTH502_LINEAR_ALGEBRA

Unit 13: Annihilating Polynomials


Now B B = (det B)I, so that Notes
n

B
B B ij kj det .
ki
i 1
Therefore
n
det B
0 = ki j
j 1
= (det )B , k 1 k . n
The Cayley-Hamilton theorem is useful to us at this point primarily because it narrows down
the search for the minimal polynomials of various operators. If we know the matrix A which
represents T in some ordered basis, then we can compute the characteristic polynomial f. We
know that the minimal polynomial p divides f and that the two polynomials have the same
roots. There is no method for computing precisely the roots of a polynomial (unless its degree
is small); however, if f factors
1
dk
d
f (x c )  (x c ) ,C  ,C , distinct, d 1 .... (4)
1 k 11 k i
then
rk
p (x c ) r 1 (x c ) , 1 r d .... (5)
1 k j j
That is all we can say in general. If f is the polynomial (4) and has degree n, then for every
polynomial p as in (5) we can find an n n matrix which has f as its characteristic polynomial and
p as its minimal polynomial. We shall not prove this now. But, we want to emphasize the fact
that the knowledge that the characteristic polynomial has the form (4) tells us that the minimal
polynomial has the form (5), and it tells us nothing else about p.

Example 2: Let A be the 4 4 (rational) matrix

0 1 0 1
1 0 1 0
A =
0 1 0 1
1 0 1 0

The powers of A are easy to compute

2 0 2 0
0 2 0 2
2
A =
2 0 2 0
0 2 0 2

0 4 0 4
4 0 4 0
A =
3
0 4 0 4
4 0 4 0
3
3
Thus A = 4A, i.e., if p = x – 4x = x(x + 2) (x – 2), then p(A) = 0. The minimal polynomial for A must
divide p. That minimal polynomial is obviously not of degree 1, since that would mean that
A was a scalar multiple of the identity. Hence, the candidates for the minimal polynomial are:

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