Linear Algebra, Theory And Applications, 2012a

252 LINEAR TRANSFORMATIONS CANONICAL FORMS
Lemma 10.3.2 Let W be an A invariant (AW ⊆ W ) subspace of ker (φ (A) m ) for m a
positive integer where φ (λ) is an irreducible monic polynomial of degree d. Thenifη (λ) is
a monic polynomial of smallest degree such that for
then
x ∈ ker (φ (A) m ) \{0} ,
η (A) x =0,
η (λ) =φ (λ) k
for some positive integer k. Thus if r is the degree of η, then r = kd. Also, for a cyclic set,
β x ≡ { x, Ax, ··· ,A r−1 x }
is linearly independent. Recall that r is the smallest such that A r x is a linear combination
of { x, Ax, ··· ,A r−1 x } .
Now let U be an A invariant subspace of ker (φ (A)) .
If {v 1 , ··· ,v s } is a basis for W then if x ∈ U \ W,
{v 1 , ··· ,v s ,β x }
is linearly independent.
There exist vectors x 1 , ··· ,x p each in U such that
{v 1 , ··· ,v s ,β x1
, ··· ,β xp
}
is a basis for
U + W.
Proof: Consider the first claim. If η (A) x =0, then writing
φ (λ) m = η (λ) g (λ)+r (λ)
where either r (λ) = 0 or the degree of r (λ) is less than that of η (λ) , the latter possibility
cannot occur because if it did, r (A) x = 0 and this would contradict the definition of η (λ).
Therefore r (λ) =0andsoη (λ) divides φ (λ) m . From Corollary 8.3.11,
η (λ) =φ (λ) k
for some integer, k ≤ m. Since x ≠0, it follows k>0. In particular, the degree of η (λ)
equals kd.
Now consider x ≠0,x ∈ ker (φ (A) m ) and the vectors β x . Do these vectors yield a
linearly independent set? The vectors are { x, Ax, A 2 x, ··· ,A r−1 x } where A r x is in
span ( x, Ax, A 2 x, ··· ,A r−1 x )
and r is as small as possible for this to happen. Suppose then that there are scalars d j ,not
all zero such that
∑r−1
d j A j x =0,x≠0. (10.3)
j=0
Suppose m is the largest nonzero scalar in the above linear combination. d m ≠0,m≤ r − 1.
Then A m x is a linear combination of the preceeding vectors in the list, which contradicts
the definition of r. Thus from the first part, r = kd for some positive integer k.