# proof of cyclic vector theorem

First, let’s assume $f$ has a cyclic vector $v$. Then $B=\{v,f(v),...,f^{n-1}(v)\}$ is a basis for $V$. Suppose $g$ is a linear transformation which commutes with $f$. Consider the coordinates $(\alpha_{0},...,\alpha_{n-1})$ of $g(v)$ in B, that is

 $g(v)=\sum_{i=0}^{n-1}\alpha_{i}f^{i}(v).$

Let

 $P=\sum_{i=0}^{n-1}\alpha_{i}X^{i}\in k[X].$

We show that $g=P(f)$. For $w\in V$, write

 $w=\sum_{j=0}^{n-1}\beta_{j}f^{j}(v),$

then

 $\displaystyle g(w)$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{n-1}\beta_{j}g(f^{j}(v))=\sum_{j=0}^{n-1}\beta_{j}f^{% j}(g(v))$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{n-1}\beta_{j}f^{j}(\sum_{i=0}^{n-1}\alpha_{i}f^{i}(v)% )=\sum_{j=0}^{n-1}\beta_{j}\sum_{i=0}^{n-1}\alpha_{i}f^{j+i}(v)=\sum_{j=0}^{n-% 1}\sum_{i=0}^{n-1}\beta_{j}\alpha_{i}f^{j+i}(v)$ $\displaystyle=$ $\displaystyle\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\beta_{j}\alpha_{i}f^{j+i}(v)=% \sum_{i=0}^{n-1}\alpha_{i}f^{i}(\sum_{j=0}^{n-1}\beta_{j}f^{j}(v))=\sum_{i=0}^% {n-1}\alpha_{i}f^{i}(w)$

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

 $B=\{v_{1},f(v_{1}),...,f^{j_{1}}(v_{1}),v_{2},f(v_{2}),...,f^{j_{2}}(v_{2}),..% .,v_{r},f(v_{r}),...,f^{j_{r}}(v_{r})\}.$

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

 $g(f^{k}(v_{1}))=\left\{\begin{array}[]{ll}0&\textrm{for every }k=0,\ldots,j_{1% }\\ f^{k_{i}}(v_{i})&\textrm{for every }i=2,\ldots,r\textrm{ and }k_{i}=0,\ldots,j% _{i}.\end{array}\right.$

The fact that $f$ and $g$ commute is a consequence of $g$ being defined as zero on one $f$-invariant subspace and as the identity on its complementary $f$-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 $k=0,...,j_{1}-1$, then

 $(gf)(f^{k}(v_{1}))=g(f^{k+1}(v_{1}))=0\quad\mbox{ and }\quad(fg)(f^{k}(v_{1}))% =f(g(f^{k}(v_{1}))=f(0)=0.$

If $k=j_{1}$, we know there are $\lambda_{0},...,\lambda_{j_{1}}$ such that

 $f^{j_{1}+1}(v_{1})=\sum_{k=0}^{j_{1}}\lambda_{k}f^{k}(v_{1}),$

so

 $(gf)(f^{j_{1}}(v_{1}))=\sum_{k=0}^{j_{1}}\lambda_{k}g(f^{k}(v_{1}))=0\quad% \mbox{ and }\quad(fg)(f^{j_{1}}(v_{1}))=f(0)=0.$

Now, let $i=2,...,r$ and $k_{i}=0,...,j_{i}-1$, then

 $(gf)(f^{k_{i}}(v_{i}))=g(f^{k_{i}+1}(v_{i}))=f^{k_{i}+1}(v_{i})\quad\mbox{ and% }\quad(fg)(f^{k_{i}}(v_{i}))=f(g(f^{k_{i}}(v_{i}))=f^{k_{i}+1}(v_{i}).$

In the case $k_{i}=j_{i}$, we know there are $\lambda_{0,i},...,\lambda_{j_{i},i}$ such that

 $f^{j_{i}+1}(v_{i})=\sum_{k=0}^{j_{i}}\lambda_{k,i}f^{k}(v_{i})$

then

 $(gf)(f^{j_{i}}(v_{i}))=g(f^{j_{i}+1}(v_{i}))=\sum_{k=0}^{j_{i}}\lambda_{k,i}g(% f^{k}(v_{i}))=\sum_{k=0}^{j_{i}}\lambda_{k,i}f^{k}(v_{i})=f^{j_{i}+1}(v_{i}),$

and

 $(fg)(f^{j_{i}}(v_{i}))=f(g(f^{j_{i}}(v_{i}))=f(f^{j_{i}}(v_{i}))=f^{j_{i}+1}(v% _{i}).$

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

 $P=\sum_{k=0}^{h}c_{k}X^{k}\in K[X]$

such that $g=P(f)$. Then, $0=g(v_{1})=P(f)(v_{1})$, and so the annihilator polynomial $m_{v_{1}}$ of $v_{1}$ divides $P$. But then, as the annihilator $m_{v_{2}}$ of $v_{2}$ divides $m_{v_{1}}$ (see the cyclic decomposition theorem), we have that $m_{v_{2}}$ divides $P$, and then $0=P(f)(v_{2})=g(v_{2})=v_{2}$ which is absurd because $v_{2}$ is a vector of the basis $B$. This finishes the proof.

Title proof of cyclic vector theorem ProofOfCyclicVectorTheorem 2013-03-22 14:14:42 2013-03-22 14:14:42 CWoo (3771) CWoo (3771) 12 CWoo (3771) Proof msc 15A04