256 CHAPTER 9. CANONICAL FORMS

Proof: By Theorem 9.2.5 the matrix of A with respect to {B1, · · · , Bq} is of the formgiven in 9.13. Now by Theorem 9.3.5 the basis Bk may be chosen in the form

{βv1 , · · · , βvs

}where each βvk

is an A cyclic set of vectors and also it can be assumed the lengths of theseβvk

are decreasing. Thus

Vk = span(βv1

)⊕ · · · ⊕ span

(βvs

)and it only remains to consider the matrix of A restricted to span

(βvk

). Then you can

apply Theorem 9.2.5 to get the result in 9.14. Say

βvk= vk, Avk, · · · , Ad−1vk

where η (A) vk = 0 and the degree of η (λ) is d, the smallest degree such that this is so, η beinga monic polynomial. Then η (λ) must divide ϕk (λ)

mk . By Corollary 7.3.11, η (λ) = ϕk (λ)rk

where rk ≤ mk. It remains to consider the matrix of A restricted to span(βvk

). Say

η (λ) = ϕk (λ)rk = a0 + a1λ+ · · ·+ ad−1λ

d−1 + λd

Thus, since η (A) vk = 0,

Advk = −a0vk − a1Avk − · · · − ad−1Ad−1vk

Recall the formalism for finding the matrix of A restricted to this invariant subspace.(Avk A2vk A3vk · · · −a0vk − a1Avk − · · · − ad−1A

d−1vk

)=

(vk Avk A2vk · · · Ad−1vk

)

0 0 0 · · · −a01 0 −a1

0 1. . .

.... . .

. . . 0 −ad−2

0 0 1 −ad−1

Thus the matrix of the transformation is the above. This is the companion matrix ofϕk (λ)

rk = η (λ). In other words, C = C (ϕk (λ)rk) and so Mk has the form claimed in the

theorem. ■

9.8 Uniqueness

Given A ∈ L (V, V ) where V is a vector space having field of scalars F, the above showsthere exists a rational canonical form for A. Could A have more than one rational canonicalform? Recall the definition of an A cyclic set. For convenience, here it is again.

Definition 9.8.1 Letting x ̸= 0 denote by βx the vectors{x,Ax,A2x, · · · , Am−1x

}where

m is the smallest such that Amx ∈ span(x, · · · , Am−1x

).

The following proposition ties these A cyclic sets to polynomials. It is just a review ofideas used above to prove existence.

Proposition 9.8.2 Let x ̸= 0 and consider{x,Ax,A2x, · · · , Am−1x

}. Then this is an A

cyclic set if and only if there exists a monic polynomial η (λ) such that η (A)x = 0 andamong all such polynomials ψ (λ) satisfying ψ (A)x = 0, η (λ) has the smallest degree.If V = ker (ϕ (λ)

m) where ϕ (λ) is monic and irreducible, then for some positive integer

p ≤ m, η (λ) = ϕ (λ)p.

256 CHAPTER 9. CANONICAL FORMSProof: By Theorem 9.2.5 the matrix of A with respect to {B,,--- ,B,} is of the formgiven in 9.13. Now by Theorem 9.3.5 the basis By, may be chosen in the form {f,,,,-++ ,8,, }where each 6, is an A cyclic set of vectors and also it can be assumed the lengths of theseB,, are decreasing. ThusV, = span (3,,) ®--+ P span (3,,)and it only remains to consider the matrix of A restricted to span (,,) . Then you canapply Theorem 9.2.5 to get the result in 9.14. Sayd-1By, = Uk, AUK, A Ukwhere 7) (A) vz, = 0 and the degree of 7) (A) is d, the smallest degree such that this is so, 7 beinga monic polynomial. Then 7 (A) must divide ¢;, (A)""* . By Corollary 7.3.11, 7 (A) = @, (A)"®where rz, < mx. It remains to consider the matrix of A restricted to span (3,,,) . Say7 (A) = dy (\)"* = ay +aj,\ +++» + aq_1dA1 + AfThus, since 7 (A) vz, = 0,Atv, = —aguk — a, Av; eee ag At ‘upRecall the formalism for finding the matrix of A restricted to this invariant subspace.( Avr; A’ up A® uz ses) —a9UR — G1 AUR + — ag—1 At! vp ) =0 O QO «:- —do1 O ay( vy, AvpE A? vu, tee At ly, ) 0 10 —ag—20 0 1 -@q-1Thus the matrix of the transformation is the above. This is the companion matrix ofob, (A)"® = 7 (A). In other words, C = C (¢; (A)"*) and so M* has the form claimed in thetheorem.9.8 UniquenessGiven A € £L(V,V) where V is a vector space having field of scalars F, the above showsthere exists a rational canonical form for A. Could A have more than one rational canonicalform? Recall the definition of an A cyclic set. For convenience, here it is again.Definition 9.8.1 Letting x #4 0 denote by £., the vectors {x, Ax, A*x, nee ,A™ta} wherem is the smallest such that A™x € span (x, ee ,A™tx) .The following proposition ties these A cyclic sets to polynomials. It is just a review ofideas used above to prove existence.Proposition 9.8.2 Let x #4 0 and consider {x, Ax, Ax, tee Am ta}. Then this is an Acyclic set if and only if there exists a monic polynomial n(A) such that n(A)a = 0 andamong all such polynomials w (A) satisfying p(A)a = 0, (A) has the smallest degree.If V = ker(¢(A)"”) where @(A) is monic and irreducible, then for some positive integerp<m,n(a) = eA)’.