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## Homework Statement

If ##A## is an ##n \times n## nilpotent matrix, then the characteristic polynomial of ##A## is ##x^n##

## Homework Equations

## The Attempt at a Solution

Suppose that ##A## has an eigenvalue with corresponding eigenvector such that ##A v = \lambda v##. Then ##A^k v = \lambda^k v = 0##, and since ##v \ne \vec{0}##, ##\lambda^k = 0 \implies \lambda = 0##. Since ##0## is the only eigenvalue, and since the characteristic polynomial must be of degree n, the characteristic polynomial must be ##x^n##.