I have a block matrix
$$M=\begin{bmatrix}
I_0& I_1& \cdots& I_1\\
I_2& I_0& \ddots& \vdots\\
\vdots& \ddots& \ddots& I_1\\
I_2& \cdots& I_2& I_0\\
\end{bmatrix}_{n \times n}$$
with
$$I_0=\begin{bmatrix}
0& 1\\
1& 0\\
\end{bmatrix}, \qquad
I_1=\begin{bmatrix}
0& 1\\
-1& 0\\
\end{bmatrix},\qquad
I_2=\begin{bmatrix}
0& -1\\
1& 0\\
\end{bmatrix}.$$
I want to find all its eigenvalues $\{\lambda_1,\lambda_2,\ldots,\lambda_{2n}\}$, where $\lambda_1 < \lambda_2 < \cdots < \lambda_{2n}$.
Due to the chiral symmetry, we can find $\lambda_i=-\lambda_{2n+1-i}$ for all $i$.
Best Answer
For the signed circulant matrix $$U:=\left[\begin{matrix} & 1 & & & \\ & & \ddots & & \\ & & & 1 &\\ -1 & & & & \end{matrix}\right] \mbox{ in } M_n(\mathbb{C}),$$ one has $$M= 1 \otimes I_0 + (U+U^2+ \cdots + U^{n-1}) \otimes I_1 \mbox{ in } M_n(\mathbb{C})\otimes M_2(\mathbb{C}).$$ For $\omega:=\exp\frac{i\pi}{n}$, the unitary matrix $U$ has eigenvalues $\{ \omega^k : k=1,3,5,\ldots,2n-1\}$ and eigenvectors $v_k:=[\begin{smallmatrix} 1 & \omega^k & \omega^{2k} & \cdots &\omega^{(n-1)k}\end{smallmatrix}]^{\mathrm{T}}/\sqrt{n}$ in $\ell_2^n$. Accordingly, the matrix $M$ is decomposed into the direct sum of $$I_0+ (\omega^k+\omega^{2k}+ \cdots + \omega^{(n-1)k})I_1 =\left[\begin{matrix} 0 & \lambda_k\\ \overline{\lambda_k} & 0 \end{matrix}\right] \mbox{ acting on } \mathbb{C}v_k \otimes \ell_2^2,$$ where $$\lambda_k=\frac{2}{1-\omega^k}.$$ The eigenvalues of $M$ are $\pm|\lambda_k|$, $k=1,3,\ldots,2(n-1)$ with eigenvectors $ v_k \otimes [\begin{smallmatrix} 1 & \pm \mathrm{sgn}(\overline{\lambda_k}) \end{smallmatrix}]^{\mathrm{T}}/\sqrt{2}$ in $\ell_2^n \otimes \ell_2^n$. Note that since $|\lambda_k|=|\lambda_{2n-k}|$, all eigenvalues except for $1$ (corresponding to the case when $n$ is odd and $\lambda_n=-1$) have multiplicity $2$.