If N is an nxn nilpotent matrix over a field, then Nⁿ = 0

Let N be an n \times n matrix over a field F, and suppose N^k = 0. Prove that N^n = 0.


We saw in this previous exercise that N is similar to a matrix J which has no nonzero entries on or below the main diagonal. (Specifically, J is the Jordan canonical form of N.) As we showed in this previous exercise, J^n = 0. Since N = P^{-1}JP, we have N^n = 0.

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