Cyclic space and commuting linear transforms Suppose that $V$ is a $n$ dimensional vector space, $\sigma$, $\tau$ are two linear transformation on it. Suppose further that $V$ is a cyclic space of $\sigma$, that is, there exists a $\alpha\in V$ such that $V=span\{\sigma^i(\alpha); i=0,1,\cdots\}$, $\sigma\tau=\tau\sigma$. Show that there exists a polynomial $f$ such that $\tau=f(\sigma)$.
I have done some problems like this, what I need is to find the $f$ by solving a linear system  . But I have no idea now.
 A: Say the base field of $V$ is $F$.
Do you know some basic module theory? It’s easier to do this in the language of modules. (It better organizes the arguments.)
Let $R = F[X]$, then define an action of $R$ on $V$ via $f.v = f(σ)(v)$.
By this, $V$ becomes a cyclic $R$-module (because $V$ is a cyclic space of $σ$) and $τ\colon V → V$ is $R$-linear (because $στ = τσ$). Now it’s two simple steps:


*

*As $V$ is cyclic, say $V = 〈α〉$ as an $R$-module, $τ(α) = f.α$ for some $f ∈ R$, that is $τ(α) = f(σ)(α)$.

*As both $τ$ and $f(σ)$ are $R$-linear and $V$ is generated by $α$, from $τ(α) = f(σ)(α)$ you can already deduce that $τ = f(σ)$ everywhere.
A: Choose any $b=(\tau\alpha) \in V$ then $b= c_1\alpha+c_2\sigma(\alpha)+c_3\sigma^2(\alpha)+ \cdots + +c_{n+1}\sigma^n(\alpha)$. Now apply $\sigma$ on both side what we have is $\sigma(b)= (\sigma(\tau\alpha))=(\tau(\sigma(\alpha)))=c_1\sigma\alpha+c_2\sigma^2\alpha+c_3\sigma^3\alpha+ \cdots + +c_{n+1}\sigma^{n+1}\alpha =(c_1+c_2\sigma+c_3\sigma^2+ \cdots + +c_{n+1}\sigma^n)(\sigma\alpha) $. Similarly we'll get $(\tau(\sigma^k(\alpha))) =(c_1+c_2\sigma+c_3\sigma^2+ \cdots + +c_{n+1}\sigma^n)(\sigma^k \alpha)$. We know ${\sigma^i \alpha}$ forms a basis of V & $\tau$ agrees to be a polynomial on that so $\exists f(x)=(c_1+c_2x+c_3x^2+ \cdots + +c_{n+1}x^n)$ s.t $\tau=f(\sigma)$.
