128 CHAPTER 6. DIRECT SUMS AND BLOCK DIAGONAL MATRICES

and each ker(

φ i (L)ki)

is invariant with respect to L. Letting L j be the restriction of L to

ker(

φ j (L)k j),

it follows that the minimum polynomial of L j equals φ j (λ )k j . Also p≤ n.

Proof: By Theorem 6.1.6, the minimum polynomial p(λ ) is of the form a∏pi=1 φ i (λ )

ki

where φ i (λ ) is monic and irreducible with φ i (λ ) ̸= φ j (λ ) if i ̸= j. Since p(λ ) is monic,it follows that a = 1. Since L commutes with itself, all of these φ i (L)

ki commute. Also

φ i (L) : ker(

φ j (L)k j)→ ker

(φ j (L)

k j)

because all of these operators commute.Now consider φ i (L) . Is it one to one on ker

(φ j (L)

k j)

? Suppose not. Suppose that for

some j ̸= i,φ i (L) is not one to one on ker(

φ j (L)k j). We know that φ i (λ ) ,φ j (λ )

k j arerelatively prime meaning the monic polynomial of greatest degree which divides them bothis 1. Why is this? If some polynomial divided both, then it would need to be φ i (λ ) or 1because φ i (λ ) is irreducible. But φ i (λ ) cannot divide φ j (λ )

k j unless it equals φ j (λ ) , thisby Corollary 6.1.7 and they are assumed unequal. Hence there are polynomials l (λ ) ,m(λ )

such that 1 = l (λ )φ i (λ )+m(λ )φ j (λ )k j . By what we mean by equality of polynomials,

that coefficients of equal powers of λ are equal, it follows that for I the identity transfor-mation,

I = l (L)φ i (L)+m(L)φ j (L)k j

Say v ∈ ker(

φ j (L)k j)

and v ̸= 0 while φ i (L)v = 0. Then from the above equation,

v = l (L)φ i (L)v+m(L)φ j (L)k j v = 0+0 = 0

a contradiction. Thus φ i (L) and hence φ i (L)ki is one to one on ker

(φ j (L)

k j). (Re-

call that, since these commute, φ i (L) maps ker(

φ i (L)ki)

to ker(

φ i (L)ki)

.) On Vj ≡

ker(

φ j (L)k j),φ i (L) actually has an inverse. In fact, the above equation says that for

v ∈Vj,v = l (L)φ i (L)v. hence an inverse for φ i (L)m is l (L)m.

Thus, from Lemma 6.1.5,

V = ker

(p

∏i=1

φ i (L)ki

)= ker

(φ 1 (L)

k1)⊕·· ·⊕ker

(φ p (L)

kp)

Next consider the claim about the minimum polynomial of L j. Denote this minimum

polynomial as p j (λ ). Then since φ j (L)k j = φ j (L j)

k j = 0 on ker(

φ j (L)k j), it must be the

case that p j (λ ) must divide φ j (λ )k j and so by Corollary 6.1.7 this means p j (λ ) = φ j (λ )

r j

where r j ≤ k j. If r j < k j, consider the polynomial

p

∏i=1,i̸= j

φ i (λ )ki φ j (λ )

r j ≡ r (λ )