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in two dimensions there is a simple trick to study the spectrum of operators of the form

$$\textbf{A}:=\left( \begin{matrix}0 && A^* \\ A && 0 \end{matrix}\right)$$

The trick is to square the matrix to get $$\textbf{A}^2=\left( \begin{matrix}A^*A && 0 \\ 0 && AA^* \end{matrix}\right).$$

This matrix is now in very simple form, we may study the operators on the diagonal and recover the spectrum of $\textbf{A}$ by taking $\pm$ the square root of the spectrum of the operators on the diagonal (spectral mapping theorem).

This is a very common method that is successfully applied for example to Dirac operators.

I ask: Is there a way to generalize this method to dimension $3$ when studying operators of the form

$$\textbf{D}:=\left( \begin{matrix}0 && A^* &&B^* \\ A && 0 && C^* \\ B && C && 0\end{matrix}\right)?$$

Would it help if $A,B,C \in \mathbb{C}$?

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    $\begingroup$ the special feature of $A$ that allows for your "trick" is not that it has vanishing diagonal blocks, but that it has a chiral symmetry: there exists a unitary $U$ such that $AU=-UA$ (just take $U={{1\, 0}\choose{0\, -1}}$); if your $D$ does not have a chiral symmetry, which it will not have in general, the "trick" you mention does not apply. $\endgroup$
    – Carlo Beenakker
    Oct 30, 2017 at 22:24
  • $\begingroup$ @CarloBeenakker I do not understand your argument fully, I am not saying that squaring could be sufficient here, but it could be that one first has to plug $\textbf{D}$ into some polynomial $F(X)=a_1X+a_2X^2+a_3X^3$ for example. Should your comment rule out such a possibility, too? $\endgroup$
    – Zehner
    Oct 30, 2017 at 22:30

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