Consider $N \times N$ matrices
$$A = \begin{bmatrix}
0 & 0 & \cdots & 0 & 1 \\
1 & 0 & 0 & & 0 \\
\vdots & 1 & 0 & \ddots & \vdots \\
0 & & \ddots & \ddots & 0 \\
0 & 0 & \cdots &1 & 0 \\
\end{bmatrix}$$
and
$$B=\operatorname{diag}( \cos(2\pi\cdot 0/N),…,\cos(2\pi\cdot (N-1)/N)).$$
Does anybody know why the eigenvalues of $i(A+A^T)+2B$ are invariant under 90° rotations?- Numerics seem to imply this. What I mean by this is that if $\lambda$ is an eigenvalue, then also $e^{i \frac{\pi}{2}} \lambda.$
Best Answer
Assume $N$ is even (this is false when N is odd). Let $X=2B, Y=A+A^T$. Let $$P = \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & \zeta & \zeta^2 & \cdots & \zeta^{N-1} \\ 1 & \zeta^2 & \zeta^4 & \cdots & \zeta^{2(N-1)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & \zeta^{N-1} & \zeta^{2(N-1)} & \cdots & \zeta^{(N-1)^2} \\ \end{bmatrix}$$ $$Q = \text{diag}(1, -1, 1, -1, \ldots, 1, -1)$$ where $\zeta=e^{\frac{2i\pi}{N}}$.
One can easily check that $$PXP^{-1}=Y$$ $$PYP^{-1}=X$$ $$QXQ^{-1}=X$$ $$QYQ^{-1}=-Y$$ from which it follows that $$P(X+iY)P^{-1}=Y+iX=i(X-iY)$$ $$Q(X+iY)Q^{-1}=X-iY$$ This shows that $X+iY$ is conjugate to $i(X+iY)$, so its eigenvalues are invariant under multiplication by $i$.