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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 $i$.

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    $\begingroup$ 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). $\endgroup$ Mar 2, 2023 at 9:10
  • $\begingroup$ @NarutakaOZAWA I am so sorry. I ignore an important property, i.e., M=M^T. $\endgroup$
    – Young Q
    Mar 2, 2023 at 11:35
  • $\begingroup$ 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) $n$. $\endgroup$
    – Toni Mhax
    Mar 2, 2023 at 15:19

1 Answer 1

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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 $2$.

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  • $\begingroup$ I have two questions. 1. Can we use the method to solve the eigenvalues of any Toeplitz matrices? 2. Can we solve its eigenstates? $\endgroup$
    – Young Q
    Mar 3, 2023 at 9:03
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    $\begingroup$ I revised my answer to include eigenvectors. The answer to your question 1 is negative. There is no general methods of calculating eigenvalues of a general (hermitian) Toeplitz matrices (and there is a lot of research on this topics). $\endgroup$ Mar 3, 2023 at 14:19
  • $\begingroup$ I understand. If the diagonal terms (i.e., I_0) are not identical, is the method is ineffective? Here, I_0, I_1, and I_2 can be regarded as the numbers. $\endgroup$
    – Young Q
    Mar 5, 2023 at 16:28

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