Cyclic subspaces, cyclic vectors, and vector annihilators

Let $T:V\to V$ be an endomorphism and $v\in V$. The $T$-cyclic subspace generated by $v$ is $Z(v;T):=\{p(T)v:p\in F[x]\}=\mathrm{span}\{v,Tv,T^{2}v,\dots \}\mathrm{.}$ It is the smallest $T$-invariant subspace containing $v$. The vector $v$ is a cyclic vector for $T$ when $Z(v;T)=V$, and $T$ is cyclic when it has a cyclic vector.

The annihilator ideal of $v$ is ${\mathrm{Ann}}_T(v):=\{p\in F[x]:p(T)v=0\}\mathrm{.}$ When $V$ is finite-dimensional this ideal is nonzero and has a unique monic generator, whose existence is proved in The vector annihilator is the unique monic generator of ${\mathrm{Ann}}_T(v)$ and divides the minimal polynomial ↗; that generator is the vector annihilator $m_{T,v}$. For $v=0$, the ideal is all of $F[x]$ and $m_{T,0}=1$.

Finite dimensionality is not a convenience: it is what forces the ideal to be nonzero, and without it the ideal can vanish. On $V={⨁}_{n\ge 0}F{e}_n$ with $T(e_n)=e_{n+1}$ and $v=e_0$, every nonzero $p={\sum }_{k\le m}{a}_k{x}^k$ gives $p(T)v={\sum }_{k\le m}{a}_k{e}_k\mathrm{\ne }0$, so ${\mathrm{Ann}}_T(v)=(0)$ and there is no monic generator. It is not necessary, however: for $T=0$ on any $V$ and any $v\mathrm{\ne }0$, $p(T)v=p(0)v$, so ${\mathrm{Ann}}_T(v)=(x)$ and $m_{T,v}=x$.