9.8. UNIQUENESS 257

The following is the main consideration for proving uniqueness. It will depend on whatwas already shown for the Jordan canonical form. This will apply to the nilpotent matrixϕ (A).

Lemma 9.8.3 Let V be a vector space and A ∈ L (V, V ) has minimal polynomial ϕ (λ)m

where ϕ (λ) is irreducible and has degree d. Let the basis for V consist of{βv1 , · · · , βvs

}where βvk

is A cyclic as described above and the rational canonical form for A is the matrix

taken with respect to this basis. Then letting∣∣βvk

∣∣ denote the number of vectors in βvk, it

follows there is only one possible set of numbers∣∣βvk

∣∣.Proof: Say βvj is associated with the polynomial ϕ (λ)

pj . Thus, as described above∣∣∣βvj

∣∣∣ equals pjd. Consider the following table which comes from the A cyclic set{vj , Avj , · · · , Ad−1vj , · · · , Apjd−1vj

}αj0 αj

1 αj2 · · · αj

d−1

vj Avj A2vj · · · Ad−1vj

ϕ (A) vj ϕ (A)Avj ϕ (A)A2vj · · · ϕ (A)Ad−1vj...

......

...

ϕ (A)pj−1

vj ϕ (A)pj−1

Avj ϕ (A)pj−1

A2vj · · · ϕ (A)pj−1

Ad−1vj

In the above, αjk signifies the vectors below it in the kth column. None of these vectors

below the top row are equal to 0 because the degree of ϕ (λ)pj−1

λd−1 is dpj − 1, which isless than pjd and the smallest degree of a nonzero polynomial sending vj to 0 is pjd. Also,each of these vectors is in the span of βvj and there are dpj of them, just as there are dpjvectors in βvj .

Claim: The vectors{αj0, · · · , α

jd−1

}are linearly independent.

Proof of claim: Suppose

d−1∑i=0

pj−1∑k=0

cikϕ (A)kAivj = 0

Then multiplying both sides by ϕ (A)pj−1

this yields

d−1∑i=0

ci0ϕ (A)pj−1

Aivj = 0

this is because if k ≥ 1, you have a typical term of the form

cikϕ (A)pj−1

ϕ (A)kAivj = Aiϕ (A)

k−1cikϕ (A)

pj vj = 0

Now if any of the ci0 is nonzero this would imply there exists a polynomial having degreesmaller than pjd which sends vj to 0. In fact, the polynomial would have degree d−1+pj−1.Since this does not happen, it follows each ci0 = 0. Thus

d−1∑i=0

pj−1∑k=1

cikϕ (A)kAivj = 0