proof of cyclic vector theorem


First, let’s assume f has a cyclic vector v. Then B={v,f(v),,fn-1(v)} is a basis for V. Suppose g is a linear transformation which commutes with f. Consider the coordinatesMathworldPlanetmathPlanetmath (α0,,αn-1) of g(v) in B, that is

g(v)=i=0n-1αifi(v).

Let

P=i=0n-1αiXik[X].

We show that g=P(f). For wV, write

w=j=0n-1βjfj(v),

then

g(w) = j=0n-1βjg(fj(v))=j=0n-1βjfj(g(v))
= j=0n-1βjfj(i=0n-1αifi(v))=j=0n-1βji=0n-1αifj+i(v)=j=0n-1i=0n-1βjαifj+i(v)
= i=0n-1j=0n-1βjαifj+i(v)=i=0n-1αifi(j=0n-1βjfj(v))=i=0n-1αifi(w)

Now, to finish the proof, suppose f doesn’t have a cyclic vector (we want to see that there is a linear transformation g which commutes with f but is not a polynomial evaluated in f). As f doesn’t have a cyclic vector, then due to the cyclic decomposition theorem V has a basis of the form

B={v1,f(v1),,fj1(v1),v2,f(v2),,fj2(v2),,vr,f(vr),,fjr(vr)}.

Let g be the linear transformation defined in B as follows:

g(fk(v1))={0for every k=0,,j1fki(vi)for every i=2,,r and ki=0,,ji.

The fact that f and g commute is a consequence of g being defined as zero on one f-invariant subspacePlanetmathPlanetmath and as the identity on its complementaryPlanetmathPlanetmath f-invariant subspace. Observe that it’s enough to see that g and f commute in the basis B (this fact is trivial). We see that, if k=0,,j1-1, then

(gf)(fk(v1))=g(fk+1(v1))=0 and (fg)(fk(v1))=f(g(fk(v1))=f(0)=0.

If k=j1, we know there are λ0,,λj1 such that

fj1+1(v1)=k=0j1λkfk(v1),

so

(gf)(fj1(v1))=k=0j1λkg(fk(v1))=0 and (fg)(fj1(v1))=f(0)=0.

Now, let i=2,,r and ki=0,,ji-1, then

(gf)(fki(vi))=g(fki+1(vi))=fki+1(vi) and (fg)(fki(vi))=f(g(fki(vi))=fki+1(vi).

In the case ki=ji, we know there are λ0,i,,λji,i such that

fji+1(vi)=k=0jiλk,ifk(vi)

then

(gf)(fji(vi))=g(fji+1(vi))=k=0jiλk,ig(fk(vi))=k=0jiλk,ifk(vi)=fji+1(vi),

and

(fg)(fji(vi))=f(g(fji(vi))=f(fji(vi))=fji+1(vi).

This proves that g and f commute in B. Suppose now that g is a polynomial evaluated in f. So there is a

P=k=0hckXkK[X]

such that g=P(f). Then, 0=g(v1)=P(f)(v1), and so the annihilator polynomial mv1 of v1 divides P. But then, as the annihilatorMathworldPlanetmathPlanetmath mv2 of v2 divides mv1 (see the cyclic decomposition theorem), we have that mv2 divides P, and then 0=P(f)(v2)=g(v2)=v2 which is absurd because v2 is a vector of the basis B. This finishes the proof.

Title proof of cyclic vector theorem
Canonical name ProofOfCyclicVectorTheorem
Date of creation 2013-03-22 14:14:42
Last modified on 2013-03-22 14:14:42
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 12
Author CWoo (3771)
Entry type Proof
Classification msc 15A04