Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 8F / c / Solution

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for some $m>0$, then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A|_E=\lambda I+N$ has $\lambda\ne0$ and
$$
(\lambda I+N)^m=\lambda^mI+m\lambda^{m-1}N+\cdots.
$$
The first nonzero nilpotent term prevents this power from being diagonalizable.


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 Incoming links: Diagonalizability inherited from an invertible power