Linear Algebra, 2020a

Section IV. Jordan Form 445
We refer to that result by saying that a matrix or map satisfies its characteristic
polynomial.
1.10 Lemma Any polynomial that is satisfied by T is divisible by T’s minimal
polynomial. That is, for a polynomial f(x), iff(T) is the zero matrix then f(x)
is divisible by the minimal polynomial of T.
Proof Let m(x) be minimal for T. The Division Theorem for Polynomials gives
f(x) =q(x) · m(x)+r(x) where the degree of r is strictly less than the degree of
m. Because T satisfies both f and m, plugging T into that equation gives that
r(T) is the zero matrix. That contradicts the minimality of m unless r is the
zero polynomial.
QED
Combining the prior two lemmas shows that the minimal polynomial divides
the characteristic polynomial. Thus any root of the minimal polynomial is also
a root of the characteristic polynomial.
Thus so far we have that if the minimal polynomial factors as m(x) =
(x − λ 1 ) q1 ···(x − λ i ) q i
then the characteristic polynomial also has the roots λ 1 ,
..., λ i . But as far as what we have established to this point, the characteristic
polynomial might have additional roots: c(x) =(x − λ 1 ) p1 ···(x − λ i ) p i
(x −
λ i+1 ) p i+1
···(x − λ z ) p z
, where 1 q j p j for 1 j i. We finish the proof of
the Cayley-Hamilton Theorem by showing that the characteristic polynomial
has no additional roots so that there are no λ i+1 , λ i+2 , etc.
1.11 Lemma Each linear factor of the characteristic polynomial of a square matrix
is also a linear factor of the minimal polynomial.
Proof Let T be a square matrix with minimal polynomial m(x) of degree n and
assume that x − λ is a factor of the characteristic polynomial of T, so that λ is
an eigenvalue of T. We will show that m(λ) =0, i.e., that x − λ is a factor of m.
Assume that λ is associated with the eigenvector ⃗v and consider the powers
T 2 (⃗v), ..., T n (⃗v). WehaveT 2 (⃗v) =T · λ⃗v = λ · T⃗v = λ 2 ⃗v. The same happens for
all of the powers: T i ⃗v = λ i ⃗v for 1 i n. Thus for any polynomial function
p(x), application of the matrix p(T) to ⃗v equals the result of multiplying ⃗v by
the scalar p(λ).
p(T)⃗v =(c k T k + ···+ c 1 T + c 0 I)⃗v = c k T k ⃗v + ···+ c 1 T⃗v + c 0 ⃗v
= c k λ k ⃗v + ···+ c 1 λ⃗v + c 0 ⃗v = p(λ) · ⃗v
Since m(T) is the zero matrix, ⃗0 = m(T)(⃗v) =m(λ) · ⃗v and hence m(λ) =0, as
⃗v ≠ ⃗0 because it is an eigenvector.
QED
That concludes the proof of the Cayley-Hamilton Theorem.