## A fact about nilpotent matrices

Let $N$ be an $n \times n$ matrix over a field $F$. Suppose $N$ is nilpotent; say $N^k = 0$ for some natural number $k$. Prove that $N$ is similar to a matrix whose first superdiagonal consists of 0 and 1 and whose remaining entries are 0.

Suppose $\lambda$ is an eigenvalue of $N$. By this previous exercise, $\lambda^k$ is an eigenvalue of $N^k = 0$, and thus $\lambda^k = 0$, so that $\lambda = 0$. That is, all of the eigenvalues of $N$ are 0.

Thus, if $J$ is the Jordan canonical form of $N$, then the Jordan blocks of $J$ have 1 on the first superdiagonal and 0 elsewhere.

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