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.



Spectral symmetry of a certain structured matrix
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 $-\lambda$.


