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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