Let $F$ be a field and $V$ be a finite dimensional vector space. Suppose $T,S:V \longrightarrow V$ are two linear maps that are conjugate. Now consider some $f(x) \in F[x]$. I want to show that the rank of the maps $f(T),f(S): V \longrightarrow V$ are the same. First, we note that by the Rank-Nullity theorem,
$dim(Ker(f(T))) + dim(Im(f(T))) = dim(Ker(f(S))) + dim(Im(f(S)))$
and thus, $rnk(f(T)) = rnk(f(S))$ if and only if $null(f(T)) = null(f(S))$. Therefore, I am trying to show that the kernel of $f(T)$ is equal to the kernel of $f(S)$.
I begin with the fact that $Ker(f(T)) = \{v \in V | f(T)(v) = 0\}$. Similarly for $Ker(f(S))$. Not sure how to proceed.