Linear Algebra, Theory And Applications, 2012a

10.8. UNIQUENESS 269
Thus the matrix of the transformation is the above. The is the companion matrix of
φ k (λ) r k
= η (λ). In other words, C = C (φ k (λ) r k
)andsoM k has the form claimed in
the theorem.
10.8 Uniqueness
Given A ∈L(V,V )whereV is a vector space having field of scalars F, the above shows
there exists a rational canonical form for A. CouldA have more than one rational canonical
form? Recall the definition of an A cyclic set. For convenience, here it is again.
Definition 10.8.1 Letting x ≠0denote by β x the vectors { x, Ax, A 2 x, ··· ,A m−1 x } where
m is the smallest such that A m x ∈ span ( x, ··· ,A m−1 x ) . This is called an A cyclic set,
denoted by β x .
The following proposition ties these A cyclic sets to polynomials. It is just a review of
ideas used above to prove existence.
Proposition 10.8.2 Let x ≠0and consider { x, Ax, A 2 x, ··· ,A m−1 x } . Then this is an
A cyclic set if and only if there exists a monic polynomial η (λ) such that η (A) x =0and
among 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 irreducible, then for some positive integer p ≤ m, η (λ) =
φ (λ) p .
Lemma 10.8.3 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 { }
β v1
, ··· ,β vs
where β vk
is A cyclic as described above and the rational canonical form for A is the matrix
taken with respect to this basis. Then letting ∣ ∣
∣β vk denote the number of vectors in β vk
, it
follows there is only one possible set of numbers ∣ ∣ βvk∣.
Proof: Say β vj
is associated with the polynomial φ (λ) p j
. Thus, as described above
∣
∣ ∣∣
∣β vj equals pj d. Consider the following table which comes from the A cyclic set
{
vj ,Av j , ··· ,A d−1 v j , ··· ,A pjd−1 }
v j
α j 0 α j 1 α j 2 ··· α j d−1
v j Av j A 2 v j ··· A d−1 v j
φ (A) v j φ (A) Av j φ (A) A 2 v j ··· φ (A) A d−1 v j
.
.
.
.
.
.
.
.
φ (A) pj−1 v j φ (A) pj−1 Av j φ (A) pj−1 A 2 v j ··· φ (A) pj−1 A d−1 v j
In the above, α j k 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 λ d−1 is dp j − 1, which is
less than p j d and the smallest degree of a nonzero polynomial sending v j to 0 is p j d.Also,
each of these vectors is in the span of β vj
and there are dp j of them, just as there are dp j
vectors in β vj
.
Claim: The vectors
Proof of claim: Suppose
{
}
α j 0 , ··· ,αj d−1
are linearly independent.
∑d−1
p
∑ j−1
i=0 k=0
c ik φ (A) k A i v j =0