I am trying to find matrices whose $n^{th}$ power is the identify, i.e., $M^n=1$, but where $M$ is Hermitian. I appreciate that this question is less strict than the title suggests, but it is certainly related, and a full representation of the group in terms of Hermitian matrices would be ideal!
The case $n=2$ is fine. In particular, the Weyl-Brauer matrices have all the relevant properties.
So far I have not been able to find such a matrix for $n=3$. Does anyone know if this is possible. If not, do you have a proof?
This question is physically motivated by trying find a quantum mechanical system which returns to an initial state before it can explore the full Hilbert space (a very particular type of ergodicity breaking).
A slightly relaxed problem: Find $M^n = D$, where $D$ is diagonal and $M$ is Hermitian.
Say that $M$ is a Hermitian matrix such that $M^n =1$. Now $M$ is Hermitian, so all of its eigenvalues are real. Since $M^n = 1$, each eigenvalue $\lambda$ of $M$ is a real number such that $\lambda^n = 1$. This implies $\lambda = \pm1$.
Now $M$ is diagonalizable since it is Hermitian, so $M$ is similar to a diagonal matrix with all entries $\pm 1$. Thus $M^2 = 1$.