If $N$ is a $k \times k$ elementary nilpotent matrix, i.e. $N^k = 0$ but $N^{k-1} \ne 0$, then show that $N^\top$ is similar to $N$. Now use the Jordan form to prove that every complex $n \times n$ matrix is similar to its transpose.
I have figured out the second part, and am struggling with connecting nilpotency to being similar to the transpose.
Best Answer
To show that a complex $k \times k$ elementary nilpotent matrix is similar to its transpose, it suffices to show their Jordan Canonical Forms are equal.
But a complex nilpotent matrix $N$ has only zero complex eigenvalues, hence its Jordan Canonical Form is zero along the diagonal. Were the maximum size of its Jordan Blocks to be $k'\times k'$ for some $k' < k$, then $N^{k'} = 0$, contradicting the elementary nilpotency of $N$. Hence $N$ has exactly one Jordon block (of size $k \times k$) in its Jordan Canonical Form, with zeros along the diagonal.
By the same argument, $N^t$ being elementary nilpotent, its Jordan Canonical Form also has exactly one Jordon Block with zeros along the diagonal. Thus $N$ and $N^t$ are similar, for their Jordon Canonical Forms of $N$ and $N^t$ are the same.