9.3. CYCLIC SETS 241

Proof: Claim: If x ∈ kerϕ (A) , and |βx| denotes the length of βx, then |βx| = d thedegree of the irreducible polynomial ϕ(λ) and so

βx ={x,Ax,A2x, · · · , Ad−1x

}also span (βx) is A invariant, A (span (βx)) ⊆ span (βx).

Proof of the claim: Let m = |βx| . That is, there exists monic η (λ) of degree m andη (A)x = 0 with m is as small as possible for this to happen. Then from the usual process ofdivision of polynomials, there exist l (λ) , r (λ) such that r (λ) = 0 or else has smaller degreethan that of η (λ) such that

ϕ (λ) = η (λ) l (λ) + r (λ)

If deg (r (λ)) < deg (η (λ)) , then the equation implies 0 = ϕ (A)x = r (A)x and so m wasincorrectly chosen. Hence r (λ) = 0 and so if l (λ) ̸= 1, then η (λ) divides ϕ (λ) contraryto the assumption that ϕ (λ) is irreducible. Hence l (λ) = 1 and η (λ) = ϕ (λ) . The claimabout span (βx) is obvious because A

dx ∈ span (βx). This shows the claim.Suppose now x ∈ U \W where U ⊆ ker (ϕ (A)). Consider

{v1, · · · , vs, βx} .

Is this set of vectors independent? Suppose

s∑i=1

aivi +

d∑j=1

djAj−1x = 0.

If z ≡∑d

j=1 djAj−1x, then z ∈W ∩ span

(x,Ax, · · · , Ad−1x

). Then also for each m ≤ d−1,

Amz ∈W ∩ span(x,Ax, · · · , Ad−1x

)because W, span

(x,Ax, · · · , Ad−1x

)are A invariant. Therefore,

span(z,Az, · · · , Ad−1z

)⊆ W ∩ span

(x,Ax, · · · , Ad−1x

)⊆ span

(x,Ax, · · · , Ad−1x

)(9.3)

Suppose z ̸= 0. Then from the Lemma 9.3.2 above,{z,Az, · · · , Ad−1z

}must be linearly

independent. Therefore,

d = dim(span

(z,Az, · · · , Ad−1z

))≤ dim

(W ∩ span

(x,Ax, · · · , Ad−1x

))≤ dim

(span

(x,Ax, · · · , Ad−1x

))= d

ThusW ∩ span

(x,Ax, · · · , Ad−1x

)= span

(x,Ax, · · · , Ad−1x

)which would require x ∈ W but this is assumed not to take place. Hence z = 0 and sothe linear independence of the {v1, · · · , vs} implies each ai = 0. Then the linear indepen-dence of

{x,Ax, · · · , Ad−1x

}, which follows from Lemma 9.3.2, shows each dj = 0. Thus{

v1, · · · , vs, x, Ax, · · · , Ad−1x}is linearly independent as claimed.

Let x ∈ U \W ⊆ ker (ϕ (A)) . Then it was just shown that {v1, · · · , vs, βx} is linearlyindependent. Let W1 be given by

y ∈ span (v1, · · · , vs, βx) ≡W1

9.3. CYCLIC SETS 241Proof: Claim: If x € ker¢(A), and |,,| denotes the length of 6,, then |f,,| = d thedegree of the irreducible polynomial (A) and soB, = {x, Ax, A*a, oo ,At lax}also span (3,,) is A invariant, A (span (G,,)) C span (G,).Proof of the claim: Let m = |f,,|. That is, there exists monic 7 (A) of degree m and1 (A) a = 0 with m is as small as possible for this to happen. Then from the usual process ofdivision of polynomials, there exist 1 (A) ,r (A) such that r (A) = 0 or else has smaller degreethan that of 7 (A) such that6 (0) =n (LO) +r (Q)If deg (r (A)) < deg (7 (A)), then the equation implies 0 = ¢(A)a =r (A) and so m wasincorrectly chosen. Hence r (A) = 0 and so if (A) 4 1, then 7 (A) divides ¢(A) contraryto the assumption that @ (A) is irreducible. Hence /(A) = 1 and 7 (A) = (A). The claimabout span (@,,) is obvious because A@x € span (Z,,). This shows the claim.Suppose now 7 € U \ W where U C ker (¢(A)). Consider{v1,+°: Us; B,}.Is this set of vectors independent? Supposes dSo ain; + So dj AT x =0.i=l j=lIfz= an d;AJ~*x, then z € WMspan (2, Az,--- , A7~*x) . Then also for each m < d—1,A™z € W span (x, Aa,-:: At tr)because W, span (2, Ag,--- , Atl x) are A invariant. Therefore,span (z, Az,--- ,A¢ tz) W Mspan (x, Ax,--- , A?~'2)cC span (a, Ax,+:: ,At tz) (9.3)Suppose z 4 0. Then from the Lemma 9.3.2 above, {z, Az, oo At tz must be linearlyindependent. Therefore,d = dim (span (z, Az,--- ,At12)) < dim (WM span (2, Az, --- At r))< dim (span (a, Aa,+:: ,At'2)) =dThusW span (x, Ax,--- ,A*~!x) = span (2, Az,--- , A‘ a)which would require x € W but this is assumed not to take place. Hence z = 0 and sothe linear independence of the {v1,--- ,us} implies each a; = 0. Then the linear indepen-dence of {x, Ag,--: Atleh, which follows from Lemma 9.3.2, shows each dj = 0. Thus{vr, +++ Us,0,AX,--- Atty} is linearly independent as claimed.Let « € U\ W C ker (¢(A)). Then it was just shown that {v1,--- ,vs,6,,} is linearlyindependent. Let W; be given byYE span (v1, -°- Us, Bx) = Wi