Page 191 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
180
A polynomial p ∈P(F) is said to divide a polynomial q ∈P(F) if
p divides q if we can take the remainder r in 4.6 to be 0. For exam-
Note that (z − λ) there exists a polynomial s ∈P(F) such that q = sp. In other words,
2
3
2
divides a polynomial q ple, the polynomial (1 + 3z) divides 5 + 32z + 57z + 18z because
3
2
2
if and only if λ is a 5 + 32z + 57z + 18z = (2z + 5)(1 + 3z) . Obviously every nonzero
root of q. This follows constant polynomial divides every polynomial.
immediately from 4.1. The next result completely characterizes the polynomials that when
applied to an operator give the 0 operator.
8.34 Theorem: Let T ∈L(V) and let q ∈P(F). Then q(T) = 0 if
and only if the minimal polynomial of T divides q.
Proof: Let p denote the minimal polynomial of T.
First we prove the easy direction. Suppose that p divides q. Thus
there exists a polynomial s ∈P(F) such that q = sp. We have

q(T) = s(T)p(T) = s(T)0 = 0,

as desired.
To prove the other direction, suppose that q(T) = 0. By the division
algorithm (4.5), there exist polynomials s, r ∈P(F) such that

8.35 q = sp + r

and deg r< deg p. We have

0 = q(T) = s(T)p(T) + r(T) = r(T).

Because p is the minimal polynomial of T and deg r< deg p, the equa-
tion above implies that r = 0. Thus 8.35 becomes the equation q = sp,
and hence p divides q, as desired.

Now we describe the eigenvalues of an operator in terms of its min-
imal polynomial.

8.36 Theorem: Let T ∈L(V). Then the roots of the minimal poly-
nomial of T are precisely the eigenvalues of T.