Does the matrix equation $\text{A}^k=\epsilon \text{A}$ have any $\text{A}\neq I_n$ solutions?

Question

The question was recently posed to myself and some peers of whether or not the equation $$\text{A}^k=\epsilon\text{A}\ \Big|\ k, \epsilon \in \mathbb{R}, $$ where A is $n\times n\ \big|\ n>1,$ has any matrix solutions A $\neq I_n.$ My curiosity is killing, and I cannot find any sufficient resources online.

Answer 1

For any positive integer $k$ and any $\epsilon$, the solutions of $A^k = \epsilon A$ include all diagonalizable matrices whose eigenvalues are roots of the polynomial $z^k - \epsilon z$ (thus $0$ and $k-1$'th roots of $\epsilon$). For non-integer $k$, if you specify a branch of $z^k$, you can define $A^k$ by the holomorphic functional calculus using that branch if the branch is analytic in a neighbourhood of the eigenvalues of $A$; $A$ will then satisfy $A^k = \epsilon A$ if $A$ is diagonalizable and its eigenvalues are zeros of $z^k - \epsilon z$ (using that branch of $z^k$).

EDIT: For $\epsilon \ne 0$ and positive integer $k$, the roots of $z^k - \epsilon z$ are all simple, so all solutions $A$ are diagonalizable (any matrix whose minimal polynomial has simple roots is diagonalizable).