Which polynomial kernels are cyclic, and how to find a cyclic generator Let $T$ be a $\Bbbk$-linear operator on a vector space $V$. Given $f\in \Bbbk[x]$ consider the operator $f(T)$ on $V$. Write $K(f)=\operatorname{Ker}f(T)$ and call such $T$-invariant subspaces polynomial kernels. Let $T_{K(f)}$ be the induced operator on the $T$-invariant subspace $K(f)\leq V$.
Questions.


*

*Precisely for which $f\in \Bbbk[x]$ is $T_{K(f)}$ a cyclic operator?

*What's an example of a $T$-cyclic subspace of $V$ which is not a polynomial kernel?

*What's the algorithm for finding a cyclic generator in the first case? 


(Suppose $T$ has a minimal polynomial and that $f\in \Bbbk[x]$ is a monic divisor of it.)
 A: Your second question is very simple: for $T$ a scalar operator, the only polynomial kernels are the whole space and the zero subspace, so every nontrivial subspace is not a polynomial kernel. The same remains true for any cyclic subspace when $T$ is for instance nilpotent with a kernel of dimension${}>1$ (since in the $T$ nilpotent case, every nonzero polynomial kernel contains $\ker(T)$).
Now for the first question.
First of all one can reduce the question to individual irreducible factors of $f$, since $K$ is canonically a direct sum of kernels of appropriate powers of such irreducible factors (indeed, the projections to the summands can be written as polynomials in $f$). This means that $K$ is cyclic is and only if all the summands in this decomposition are.
Now assuming $f=p^m$ for some irreducible $p\in k[X]$, on has that $K$ is cyclic if and only if $\ker(p[T])$ is cyclic. One can see that this is so because $\ker(p[T|_W])$ is cyclic for any cyclic submodule $W$ of $K$ (so the stated condition is necessary), and $K$ can always (non canonically) be written as a direct sum of such submodules (whence the condition is sufficient, since a submodule $M$ annihilated by $p$ is cyclic if and only if $\dim(M)\in\{0,\deg(p)\}$).
For your third question, you need to choose a vector$~v$ such that no strict divisor$~d$ of $f$ has $d[T](v)=0$, and this can be done separately for each irreducible factor of$~f$.
You can find more information on this question by looking at the several questions on this site around the question when the minimal polynomial of an operator is equal to its characteristic polynomial.
