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I have a matrix

$$ A= \begin{pmatrix} 0 & a & d & c\\ \bar a & 0 & b & d \\ \bar d & \bar b & 0 & a \\ \bar c & \bar d & \bar a & 0 \end{pmatrix} $$

As you can see, the matrix is always self-adjoint for any $a, b, c, d \in \mathbb C$.

But it has a funny property (that I found by playing with some numbers):

If $a,b,c$ are arbitrary real numbers and also $d$ is real, then the spectrum of $A$ is in general not symmetric with respect to zero. To illustrate this, we take $d := 2$, $a := 5$, $b := 3$, $c := 4$ then the eigenvalues are

$$\sigma(A):=\{10.5178, -6.54138, -3.51783, -0.458619\}$$

But once I take $d \in i \mathbb R$, the spectrum becomes immediately symmetric. In fact, $d := 2 i$, $a := 5$, $b := 3$, $c := 4$ leads to eigenvalues

$$\sigma(A)=\{-9.05607, 9.05607, -0.993809, 0.993809\}$$

Is there any particular symmetry that only exists for $d \in i\mathbb R$ that implies this nice inflection symmetry?

I am less interested in a brute-force computation of the spectrum than of an explanation of what symmetry causes the inflection symmetry.

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  • $\begingroup$ Is it symmetric for all choices of $a$, $b$ and $c$ (real and or complex)? $\endgroup$ – M. Winter Aug 5 at 10:01
  • $\begingroup$ @M.Winter good question, now I am tempted to say that it actually only works for $a,b,c$ real. $\endgroup$ – Sascha Aug 5 at 10:09
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For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity: $$X=\left( \begin{array}{cccc} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ i & 0 & 0 & 0 \\ \end{array} \right),\;\;XA+AX=0,\;\;X^2=I.$$ Hence the spectrum of $A$ has $\pm$ symmetry: $$\det (\lambda-A)=\det(\lambda X^2-XAX)=\det(\lambda+X^2A)=\det(\lambda+A),$$ so if $\lambda$ is an eigenvalue then also $-\lambda$.

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  • $\begingroup$ how did you get $\det(\lambda I-A)=\det(\lambda X^2-XAX)$? $\endgroup$ – vidyarthi Aug 5 at 11:50
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    $\begingroup$ Since $X^2=1$, you can multiply $\det(\lambda-A)$ with $\det X^2=(\det X)^2$; and since the product of determinants is the determinant of the matrix product, you have $\det(\lambda-A)=\det[X(\lambda-A)X]=\det(\lambda X^2-XAX)$. $\endgroup$ – Carlo Beenakker Aug 5 at 11:53
  • $\begingroup$ great! a useful trick in evaluating some determinants $\endgroup$ – vidyarthi Aug 5 at 11:54
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An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.

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