238 CHAPTER 9. CANONICAL FORMS

polynomial for Ak then it must divide ϕk (λ)mk and so by Corollary 7.3.11 η (λ) = ϕk (λ)

rk

where rk ≤ mk. Could rk < mk? No, this is not possible because then p (λ) would failto be the minimal polynomial for A. You could substitute for the term ϕk (λ)

mk in thefactorization of p (λ) with ϕk (λ)

rk and the resulting polynomial p′ would satisfy p′ (A) = 0.Here is why. From Theorem 9.2.3, a typical x ∈ V is of the form

q∑i=1

vi, vi ∈ Vi

Then since all the factors commute,

p′ (A)

(q∑

i=1

vi

)=

q∏i ̸=k

ϕi (A)mi ϕk (A)

rk

(q∑

i=1

vi

)

For j ̸= k

q∏i ̸=k

ϕi (A)mi ϕk (A)

rk vj =

q∏i ̸=k,j

ϕi (A)mi ϕk (A)

rk ϕj (A)mj vj = 0

If j = k,q∏

i ̸=k

ϕi (A)mi ϕk (A)

rk vk = 0

which shows p′ (λ) is a monic polynomial having smaller degree than p (λ) such that p′ (A) =0. Thus the minimal polynomial for Ak is ϕk (λ)

mk as claimed. ■How does Theorem 9.2.3 relate to matrices?

Theorem 9.2.5 Suppose V is a vector space with field of scalars F and A ∈ L (V, V ).Suppose also

V = V1 ⊕ · · · ⊕ Vq

where each Vk is A invariant. (AVk ⊆ Vk) Also let βk be an ordered basis for Vk and let Ak

denote the restriction of A to Vk. Letting Mk denote the matrix of Ak with respect to thisbasis, it follows the matrix of A with respect to the basis

{β1, · · · , βq

}is

M1 0. . .

0 Mq

Proof: Let β denote the ordered basis

{β1, · · · , βq

}, |βk| being the number of vectors

in βk. Let qk : F|βk| → Vk be the usual map such that the following diagram commutes.

Ak

Vk → Vk

qk ↑ ◦ ↑ qkF|βk| → F|βk|

Mk

Thus Akqk = qkMk. Then if q is the map from Fn to V corresponding to the ordered basis

β just described,

q(

0 · · · x · · · 0)T

= qkx,