Linear Algebra, Theory And Applications, 2012a

10.3. CYCLIC SETS 253
Since β x is independent for each x ≠ 0, it follows that whenever x ≠0,
{
x, Ax, A 2 x, ··· ,A d−1 x }
is linearly independent because, the length of β x is a multiple of d and is therefore, at least
as long as the above list. However, if x ∈ ker (φ (A)) , then β x is equal to the above list.
This is because φ (λ) isofdegreed so β x is no longer than the above list. However, from
the first part β x has length equal to kd for some integer k so it is at least as long.
Suppose now x ∈ U \ W where U ⊆ ker (φ (A)). Consider
{
v1 , ··· ,v s ,x,Ax,A 2 x, ··· ,A d−1 x } .
Is this set of vectors independent? First note that
span ( x, Ax, A 2 x, ··· ,A d−1 x )
is A invariant because, from what was just shown, { x, Ax, A 2 x, ··· ,A d−1 x } = β x and so
A d x is a linear combination of the other vectors in the above list. Suppose now that
s∑
a i v i +
i=1
d∑
d j A j−1 x =0.
j=1
If z ≡ ∑ d
j=1 d jA j−1 x, then z ∈ W ∩span ( x, Ax, ··· ,A d−1 x ) . Then also for each m ≤ d−1,
A m z ∈ W ∩ span ( x, Ax, ··· ,A d−1 x )
because W, span ( x, Ax, ··· ,A d−1 x ) are A invariant, and so
span ( z, Az, ··· ,A d−1 z ) ⊆ W ∩ span ( x, Ax, ··· ,A d−1 x )
⊆ span ( x, Ax, ··· ,A d−1 x ) (10.4)
Suppose z ≠0. Then from the above, { z, Az, ··· ,A d−1 z } must be linearly independent.
Therefore,
d =dim ( span ( z, Az, ··· ,A d−1 z )) ≤ dim ( W ∩ span ( x, Ax, ··· ,A d−1 x ))
≤ dim ( span ( x, Ax, ··· ,A d−1 x )) = d
Thus
span ( z, Az, ··· ,A d−1 z ) ⊆ span ( x, Ax, ··· ,A d−1 x )
and both have the same dimension and so the two sets are equal. It follows from (10.4)
W ∩ span ( x, Ax, ··· ,A d−1 x ) =span ( x, Ax, ··· ,A d−1 x )
which would require x ∈ W but this is assumed not to take place. Hence z =0andsothe
linearly independence of the {v 1 , ··· ,v s } implies each a i =0. Then the linear independence
of { x, Ax, ··· ,A d−1 x } which follows from the first part of the argument shows each d j =0.
Thus { v 1 , ··· ,v s ,x,Ax,··· ,A d−1 x } is linearly independent as claimed.
Let x ∈ U \ W ⊆ ker (φ (A)) . Then it was just shown that {v 1 , ··· ,v s ,β x } is linearly
independent. Recall that β x = { x, Ax, A 2 x, ··· ,A d−1 x } because x ∈ ker (φ (A)). Also, if
y ∈ span ( v 1 , ··· ,v s ,x,Ax,A 2 x, ··· ,A d−1 x ) ≡ W 1