Let $V$ be a finite-dimensional vector space, and let $S,T \in L(V)$ which commute, i.e. $ST = TS$. Suppose $T$ is nilpotent with $\dim\operatorname{Nul}T = 2$. Prove that $S$ has at most $2$ distinct eigenvalues.
I tried to directly prove this one, but I failed. Then I tried to prove if $S$ has $3$ different eigenvalues, then I want to show $\dim\operatorname{Nul}T \neq 2$, but I also failed.
I doubt that "at most $2$ distinct eigenvalues" is trying to say that the eigenspace of $S$ is at most $2$. Even though this is not totally right since if $S$ has $2$ distinct eigenvalues, it can also have $2$ to more dimensional eigenspace, but if there are $3$ distinct eigenvalues for $S$, then the eigenspace of it at least has dimension $3$. Thus probably there is something that restricts the dimension of the eigenspace of operator $S$.
Then since $\dim\operatorname{Nul}T=2$, this implies that if the null space of $T$ continues to enlarge, when $T^{n-1}$ is will be the maximum. I don't know whether this is useful. I also don't know how to use $ST=TS$ this given information.
Any help on this? Thanks
Hints
\begin{align} x\in\ker{T}&\Rightarrow TSx=STx=0\\ &\Rightarrow Sx\in\ker{T}\ . \end{align} Therefore $\ S(\ker{T})\subseteq\ker{T}\ $. Let $\ \lambda\ $ be an eigenvalue of $\ S\ $ with corresponding eigenvector $\ v\ $. Since $\ T^0v=v\ne0\ $, and $\ T\ $ is nilpotent, there's a smallest natural number $\ k\ $ such that $\ T^kv\ne0\ $. Since $$ ST^kv=T^kSv=\lambda T^kv\ , $$ $\ T^kv\ $ is also an eigenvector of $\ S\ $ corresponding to the eigenvalue $\ \lambda\ $. But $\ T^kv\in\ker T\ $ by the definition of $\ k\ $. Thus every eigenvalue of $\ S\ $ is also an eigenvalue of $\ S_{/\ker T}\in L(\ker T)\ $. Can you finish it off from here?