Let $M$ be a nondiagonalizable matrix with entries in $\mathbb{C}$. Suppose that $M^n$ is diagonalizable for some positive integer $n$. Prove that $\det(M) = 0$.
I'm not quite sure how to approach this problem. It seems to me that it would be most useful to show that $M$ must have 0 as an eigenvalue, but I'm not quite sure how to do that. It seems to me that $M$ must be nilpotent, such that when it annihilates it is trivially diagonal.
Let $\lambda_1,\ldots,\lambda_N$ be the eigenvalues of $M$. By reindexing the eigenvalues if necessary, we may assume that $\lambda_1^n,\ldots,\lambda_k^n$ (with $k\le N$) are the distinct eigenvalues of $M^n$.
Since $M^n$ is diagonalisable, its minimal polynomial is $f(x)=\prod_{i=1}^k(x-\lambda_i^n)$. It follows that $g(x)=\prod_{i=1}^k(x^n-\lambda_i^n)$ is annihilating polynomial of $M$. Let $\omega$ be a primitive $n$-th root of unity. Then $g(x)=\prod_{i=1}^k\prod_{r=0}^{n-1}(x-\lambda_i\omega^r)$. Since $\lambda_1^n,\ldots,\lambda_k^n$ are distinct, if none of them is zero, then $\lambda_i\omega^r\ne\lambda_j\omega^s$ whenever $(i,r)\ne(j,s)$. Hence $g$ is a product of distinct linear factors, and so is the minimal polynomial of $M$. This contradicts the assumption that $M$ is non-diagonalisable. Therefore some eigenvalue of $M$ must be zero and $\det M=0$.