Spectral symmetry of a certain structured matrix

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.

• Is it symmetric for all choices of $a$, $b$ and $c$ (real and or complex)? – M. Winter Aug 5 at 10:01
• @M.Winter good question, now I am tempted to say that it actually only works for $a,b,c$ real. – Sascha Aug 5 at 10:09

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 $$-\lambda$$.
• how did you get $\det(\lambda I-A)=\det(\lambda X^2-XAX)$? – vidyarthi Aug 5 at 11:50
• 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 $\det(\lambda-A)=\det[X(\lambda-A)X]=\det(\lambda X^2-XAX)$. – Carlo Beenakker Aug 5 at 11:53
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 $$\pm\lambda$$.