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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.

Important_man74
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  • The implication seems to be that the Jordan Normal Form switches from being merely block diagonal to diagonal... – Important_man74 Jan 06 '24 at 07:13
  • Following @MarianoSuárez-Álvarez's comment, you may check https://math.stackexchange.com/questions/910635/power-of-a-matrix-given-its-jordan-form – Vezen BU Jan 06 '24 at 08:03
  • $M$ is not necessarily nilpotent. Consider $\pmatrix{0&1&0\ 0&0&0\ 0&0&1}$. – user1551 Jan 06 '24 at 08:12
  • So the idea is, if $J$ is the Jordan Canonical Form of $M$ than raising $J^k$ just raises each block to $J^k$. If $0$ is an eigenvalue, then its Jordan Block is nilpotent, so it will eventually become 0. This consequently implies that the geometric multiplicity of the other eigenvalues was equal to their algebraic multiplicity (else there would still be a Jordan block larger than $1\times 1$ in $J^k$). – Important_man74 Jan 06 '24 at 08:56
  • If $M$ did not have $0$ as an eigenvalue and since $M$ was nondiagonalizable, there must have been at least one Jordan Block which was larger than $1\times 1$ which was not nilpotent, hence it would never be diagonalizable. – Important_man74 Jan 06 '24 at 08:58
  • the argument I gave here https://math.stackexchange.com/questions/4030914/a-in-m-n-mathbbc-invertible-and-a2-is-diagonalizable-prove-a-is-dia/ runs verbatim – user8675309 Jan 06 '24 at 18:30

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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 $\det M=0$.

user1551
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