Jordan chevaley decomposition and cyclic vectors

Question

Suppose that $g \in GL_{n}(F)$ has jordan chevalley decoposition $ g=su$ where $s$ and $u$ denote the semisimple and unipotent parts respectively. Is it true that if $g$ is cyclic (on the natrual action on $F^{n}$) then either $s$ or $u$ must also be cyclic?

I believe this to be true but I am stuck finding a proof.

**By cyclic I mean that there exists a non-zero vector $v \in F^{n}$ such that the vectors $v,g.v,\dots g^{n-1}.v$ span $F^{n}$.

Thanks

Answer 1

It's false.

Take $g=diag(I_2+J,-I_2+J)$ where $J$ is the nilpotent Jordan block of dimension $2$. $g$ is cyclic.

Then $g=su=diag(I_2,-I_2)diag(I_2+J,I_2-J)$. $s$ and $u$ are not cyclic.