Let $V$ be a finite complex vector space. Let $T$ and $N$ be commutative operators in $V$ s.t $T$ is invertible and $N$ is nilpotent. Show that $N$ and $NT$ has the same Jordan's form.
My attempt was that how N is nilpotent, its jordan's form is equal to the rational form. $N$ and $T$ commutes, then $NT$ is also nilpotent, so $NT$ in its jordan's form is also equal to the rational form.
Suppose that $\dim(V)=n$ and $k\in \mathbb{N}$ be the smallest number such that $N^k=0$, we know that both operators, $N$ and $NT$ has $p=x^n$ as the characteristical polynomial and $p_m=x^k$ as the minimal polynomial. Hence, the biggest jordan block has size $k$ for both but then idk how to proceed to guarantee that these Jordan's form has exactly the same settings.