proof of cyclic vector theorem

First, let’s assume $f$ has a cyclic vector $v$. Then $B=\{v,f(v),\mathrm{\dots },f^{n-1}(v)\}$ is a basis for $V$. Suppose $g$ is a linear transformation which commutes with $f$. Consider the coordinates $({\alpha }_0,\mathrm{\dots },{\alpha }_{n-1})$ 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

$g(w)$$=$${\displaystyle \sum _{j=0}^{n-1}}{\beta }_{j}g(f^j(v))={\displaystyle \sum _{j=0}^{n-1}}{\beta }_j{f}^j(g(v))$

$=$${\displaystyle \sum _{j=0}^{n-1}}{\beta }_j{f}^j({\displaystyle \sum _{i=0}^{n-1}}{\alpha }_i{f}^i(v))={\displaystyle \sum _{j=0}^{n-1}}{\beta }_j{\displaystyle \sum _{i=0}^{n-1}}{\alpha }_i{f}^{j+i}(v)={\displaystyle \sum _{j=0}^{n-1}}{\displaystyle \sum _{i=0}^{n-1}}{\beta }_j{\alpha }_i{f}^{j+i}(v)$

$=$${\displaystyle \sum _{i=0}^{n-1}}{\displaystyle \sum _{j=0}^{n-1}}{\beta }_j{\alpha }_i{f}^{j+i}(v)={\displaystyle \sum _{i=0}^{n-1}}{\alpha }_i{f}^i({\displaystyle \sum _{j=0}^{n-1}}{\beta }_j{f}^j(v))={\displaystyle \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

$$B=\{v_1,f(v_1),\mathrm{\dots },f^{j_1}(v_1),v_2,f(v_2),\mathrm{\dots },f^{j_2}(v_2),\mathrm{\dots },v_r,f(v_r),\mathrm{\dots },f^{j_r}(v_r)\}.$$

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

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 $k=0,\mathrm{\dots },j_1-1$, then

$$(gf)(f^k(v_1))=g(f^{k+1}(v_1))=0\mathit{\hspace{1em}}\text{ and }\mathit{\hspace{1em}}(fg)(f^k(v_1))=f(g(f^k(v_1))=f(0)=0.$$

If $k=j_1$, we know there are ${\lambda }_0,\mathrm{\dots },{\lambda }_{j_1}$ such that

$$f^{j_1+1}(v_1)=\sum _{k=0}^{j_1}{\lambda }_k{f}^k(v_1),$$

so

$$(gf)(f^{j_1}(v_1))=\sum _{k=0}^{j_1}{\lambda }_{k}g(f^k(v_1))=0\mathit{\hspace{1em}}\text{ and }\mathit{\hspace{1em}}(fg)(f^{j_1}(v_1))=f(0)=0.$$

Now, let $i=2,\mathrm{\dots },r$ and $k_i=0,\mathrm{\dots },j_i-1$, then

$$(gf)(f^{k_i}(v_i))=g(f^{k_i+1}(v_i))=f^{k_i+1}(v_i)\mathit{\hspace{1em}}\text{ and }\mathit{\hspace{1em}}(fg)(f^{k_i}(v_i))=f(g(f^{k_i}(v_i))=f^{k_i+1}(v_i).$$

In the case $k_i=j_i$, we know there are ${\lambda }_{0,i},\mathrm{\dots },{\lambda }_{j_i,i}$ 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.