Let $D$ be a diagonal matrix in $M_{2n}(\mathbb{R})$ such that $D^2=I$ and Trace$(D)$=0
Suppose that $e_k$s are the standard vectors in $\mathbb{R}^{2n}$, that is $$e_k=(0,\cdots 0,1,0,\cdots,0)^t$$ where $1$ is in the $k^{th}$ position. Let us consider the following vectors for $k=1,\cdots,n$.
$$v_k=e_{2k}~~,~~~ v_{n+k}=e_{2k-1}$$
Let us consider the matrix $W$ whose $k^{th}$-coulmn $W_k$ is given by $v_k$
Q. How can we find the eigenvalues of $WD$?
I noticed that the $n\times n$ matrix $M=WD$ is "periodic", $M^p=\pm M$ for some even integer $p\leq n$. This identifies the eigenvalues of $M$ as $p$-th roots of $\pm 1$.
For some $n$ I find $p=n$ and the eigenvalues are all distinct: $e^{2k\pi i/n}$ if $M^n=M$ or $e^{(2k+1)\pi i/n}$ if $M^n=-M$, with $k=0,1,2,\ldots n-1$. This is the case for $n=2,4,6,10,12$.
But for other $n$ I find $p<n$, and some eigenvalues have multiplicity greater than 1. This happens for $n=8$ (with $p=6$) and for $n=14$ (with $p=8$). I have not been able to find the rule that governs these cases.
