Page 177 - DMTH502_LINEAR_ALGEBRA

Unit 13: Annihilating Polynomials

In Example 1 in unit 12 we discussed the linear operator T on R which is represented in the Notes
2
standard basis by the matrix.
0 1
A = 1 0 .

2
The characteristic polynomial is x + 1, which has no real roots. To determine the minimal
polynomial, forget about T and concentrate on A. As a complex 2  2 matrix, A has the characteristic
values i and –i. Both roots must appear in the minimal polynomial. Thus the minimal polynomial
2
2
2
is divisible by x + 1. It is trivial to verify that A + I = 0. So the minimal polynomial is x + 1.
Theorem 2 (Cayley-Hamilton): Let T be a linear operator on a finite dimensional vector space V.
If f is the characteristic polynomial for T, then f(T) = 0; in other words, the minimal polynomial
divides the characteristic polynomial for T.
Proof: The proof, although short, may be difficult to understand. Aside from brevity, it has the
virtue of providing an illuminating and far from trivial application of the general theory of
determinants.
Let K be the commutative ring with identity consisting of all polynomials in T. Of course, K is
actually a commutative algebra with identity over the scalar field. Choose an ordered basis
{ , , } for V, and let A be the matrix which represents T in the given basis. Then
1 n
n
T i A ji j , 1 j n
i j

These equations may be written in the equivalent form
n
( T A ) I 0, 1 i . n
ij ji j
j 1
Let B denote the element of K nn with entries

B T A . I
ij ij ji
When n = 2
T A I A I
B 11 21
A I T A I
12 22
and

det B = ( –T A I )(T A I )– A A I
12
22
11
21
= T 2 (A 11 A 22 )T (A A 12 A A 21 )I
12
11
= f(T)
where f is the characteristic polynomial:
f x 2 (trace A )x det .
A
For the case n > 2, it is also clear that
det B = f(T)

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