Page 175 - DMTH502_LINEAR_ALGEBRA

Unit 13: Annihilating Polynomials

From the last remark about operators and matrices it follows that similar matrices have the Notes
same minimal polynomial. That fact is also clear from the definitions because

–1
–1
A
( f P AP ) = P f ( )P
for every polynomial f.
There is another basic remark which we should make about minimal polynomials of matrices.
Suppose that A is an n n matrix with entries in the field F. Suppose that F is a field which
1
contains F as a subfield. (For example, A might be a matrix with rational entries, while F is the
1
field of real numbers. Or, A might be a matrix with real entries, while F is the field of complex
1
numbers.) We may regard A either as an n n matrix over F or as an n n matrix over F . On
1
the surface, it might appear that we obtain two different minimal polynomials for A. Fortunately
that is not the case; and we must see why. What is the definition of the minimal polynomial for
A, regarded as an n n matrix over the field F? We consider all monic polynomials with
coefficients in F which annihilate A, and we choose the one of least degree. If f is a monic
polynomial over F:

k 1
f = x k a x i … (1)
j
j 0
then f(A) = 0 merely says that we have a linear relation between the powers of A:
A k a A k 1  a A a I 0 … (2)
1
k
1
0
The degree of the minimal polynomial is the least positive integer k such that there is a linear
k
relation of the form (2) between the powers I, A , ,A Furthermore, by the uniqueness of the
.
minimal polynomial, there is for that k one and only one relation of the form (2); i.e., once the
minimal k is determined, there are unique scalars a 0 , , a k 1 in F such that (2) holds. They are the
coefficients of minimal polynomial.
2
.
Now (for each k) we have in (2) a system of n linear equations for the ‘unknowns’ a 0 , , a k 1 Since
the entries of A lie in F, the coefficients of the system of equations (2) are in F. Therefore, if the
system has a solution with a 0 , , a k 1 in F it has a solution with a 0 , , a k 1 in F. It should now be
1
clear that the two minimal polynomials are the same.
What do we know thus far about the minimal polynomial for a linear operator on an n-dimensional
space? Only that its degree does not exceed n . That turns out to be a rather poor estimate, since
2
the degree cannot exceed n. We shall prove shortly that the operator is annihilated by its
characteristic polynomial. First, let us observe a more elementary fact.
Theorem 1: Let T be a linear operator on an n-dimensional vector space V [or, let A be an
n n matrix]. The characteristic and minimal polynomials for T [for A] have the same roots,
except for multiplicities.
Proof. Let p be the minimal polynomial for T. Let c be a scalar. What we want to show is that
p(c) = 0 if and only if c is a characteristic value of T.
First, suppose p(c) = 0. Then
p = (x – c)q
where q is a polynomial. Since deg q < deg p, the definition of the minimal polynomial p tells us
T
that ( ) 0.q T Choose a vector  such that q(T) 0. Let q ( ) . Then
0 = p(T)
= (T – cI)q(T)

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