TITLE: If $N$ is nilpotent, then $N \sim N^2 \Longleftrightarrow N=0$
QUESTION [7 upvotes]: Given that $N$ is nilpotent, then $N$ is similar to $N^2$ if and only if $N=0$ ($N$ is the zero matrix).  The "$\Longleftarrow$" direction is easy enough, because $N = 0 = N^2$ so they're equal and trivially similar.
But showing $N \text{ nilpotent and } N \sim N^2\Longrightarrow N=0$ is turning out to be more difficult.  I'm comfortable using well-known facts about $n \times n$ nilpotent matrices such as:
\begin{align}
N \text{ nilpotent } &\Longleftrightarrow \text {all eigenvalues }=0\\
&\Longleftrightarrow p_N = \lambda^n\\
&\Longleftrightarrow m_N =\lambda^k\,,\, \text{ for some k, }1\leq k \leq n\\
&\Longleftrightarrow {\rm tr}N^q = 0\,,\, \text{ where }q\in \mathbb{N}\text{, minimal, s.t. } N^q=0
\end{align}
Note that $p_N$ above is the characteristic and $m_N$ is the minimal polynomial.  My hunch is to use that trace equals 0 for some power $q$ and the fact that each entry $c_{ij}$ of $N^2$ is
\begin{align}
c_{ij} = \sum_{k=1}^n n_{ik} n_{kj}
\end{align}
to get some cancellations but I can't quite see it yet.  Any thoughts?

REPLY [2 votes]: Increasing powers of nilpotent matrices are of decreasing ranks (until $N^k=0$ )
It can be easily seen from Jordan forms of nilpotent matrices. See more   detailed description here.
Two non-zero matrices of different ranks can't be similar.
Conclusion: the given in the question nilpotent matrix is zero matrix.