# Eigenvalues of a specific matrix

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 $$i$$.

• I don't know about the chiral symmetry. But $I_0^2=I$, $I_1^2=-I$, and $I_0I_1+I_1I_0=0$ (where $I$ is the $2\times2$ identity matrix) imply $M^2=-(n-2)E$, where $E_{i,j}=I$ for all $i,j$. Hence $M^2/(-n(n-2))$ is an orthogonal projection of rank $2$. This means that the nonzero eigenvalues of $M^2$ is only $-n(n-2)$ with multiplicity $2$. The nonzero eigenvalues of $M$ are simple and $\pm i\sqrt{n(n-2)}$ (because the trace is zero). Mar 2, 2023 at 9:10
• @NarutakaOZAWA I am so sorry. I ignore an important property, i.e., M=M^T. Mar 2, 2023 at 11:35
• It can be shown that finding the eigenvalues of $M$ is equivalent to finding those of $A$ or $A^2$ where $A$ is the matrix with a zero diagonal and entries $a_{i,j}=1$ for $j>i$; $a_{i,j}=-1$ for $i>j$. If you try with $n$ you see the eigenvalues (for $A$ and $M$) gets complicated with radicals in terms of (large) $n$. Mar 2, 2023 at 15:19

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 $$2$$.