240 CHAPTER 9. CANONICAL FORMS

Proof: Suppose that there are scalars ak, not all zero such that

m−1∑k=0

akAkx = 0

Then letting ar be the last nonzero scalar in the sum, you can divide by ar and solve forArx as a linear combination of the Ajx for j < r ≤ m − 1 contrary to the definition of m.■

Now here is a nice lemma which has been pretty much discussed earlier.

Lemma 9.3.3 Suppose W is a subspace of V where V is a finite dimensional vector spaceand L ∈ L (V, V ) and suppose LW = LV. Then V =W + ker (L).

Proof: Let a basis for LV = LW be {Lw1, · · · , Lwm} , wi ∈ W . Then let y ∈ V. ThusLy =

∑mi=1 ciLwi and so

L

=z︷ ︸︸ ︷

y −m∑i=1

ciwi

 ≡ Lz = 0

It follows that z ∈ ker (L) and so y =∑m

i=1 ciwi + z ∈W + ker (L). ■For more on the next lemma and the following theorem, see Hofman and Kunze [15]. I

am following the presentation in Friedberg Insel and Spence [10]. See also Herstein [14] fora different approach to canonical forms. To help organize the ideas in the lemma, here is adiagram.

ker(ϕ(A)m)

Wv1, ..., vs

U ⊆ ker(ϕ(A))

βx1, βx2

, ..., βxp

Lemma 9.3.4 Let W be an A invariant (AW ⊆W ) subspace of ker (ϕ (A)m) for m a pos-

itive integer where ϕ (λ) is an irreducible monic polynomial of degree d. Let U be an Ainvariant subspace of ker (ϕ (A)) .

If {v1, · · · , vs} is a basis for W then if x ∈ U \W,

{v1, · · · , vs, βx}

is linearly independent.There exist vectors x1, · · · , xp each in U such that{

v1, · · · , vs, βx1, · · · , βxp

}is a basis for

U +W.

Also, if x ∈ ker (ϕ (A)m) , |βx| = kd where k ≤ m. Here |βx| is the length of βx, the degree of

the monic polynomial η (λ) satisfying η (A)x = 0 with η (λ) having smallest possible degree.