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By coincidence I noticed that the following two matrices yield the same eigenvalues

\begin{pmatrix} A & B \\ B^* & A \end{pmatrix} and \begin{pmatrix} 0& A+b1_{\mathbb C^{2 \times 2}} \\ A+b^* 1_{\mathbb C^{2 \times 2}} & 0 \end{pmatrix} where $A = \begin{pmatrix} 0 & a \\ a^* & 0 \end{pmatrix}$ and $B=\begin{pmatrix} b & 0 \\ 0 & b^* \end{pmatrix}$ for any complex numbers, $a,b \in \mathbb{C}.$ Can one understand this somehow?

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    $\begingroup$ One can replace $a^*$ by $c$ and $b^*$ by $d$, and the eigenvalues of both matrices are the same, by a brute force computation of characteristic polynomials. $\endgroup$
    – Richard Stanley
    Commented Oct 6, 2020 at 17:30
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    $\begingroup$ okay fine, but is there an elegant unitary transform that takes one into the other? $\endgroup$
    – Sascha
    Commented Oct 6, 2020 at 17:58
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    $\begingroup$ @LSpice their traces are the same... $\endgroup$
    – Sascha
    Commented Oct 6, 2020 at 18:35
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    $\begingroup$ @LSpice They both have trace 0. In fact, they have the same characteristic polynomial. $\endgroup$
    – Richard Stanley
    Commented Oct 6, 2020 at 18:35
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    $\begingroup$ Oops, sorry, missed what $A$ was! $\endgroup$
    – LSpice
    Commented Oct 6, 2020 at 19:07

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Define the unitary matrix $$U=\left( \begin{array}{cccc} i e^{i \pi /4} & 0 & 0 & 0 \\ 0 & 0 & 0 & -e^{-i \pi /4} \\ 0 & 0 & i e^{i \pi /4} & 0 \\ 0 & -e^{-i \pi /4} & 0 & 0 \\ \end{array} \right),$$ then $$U\begin{pmatrix} A & B \\ B^* & A \end{pmatrix}U^{-1}=\begin{pmatrix} 0 & A+b\mathbb{1} \\ A+b^\ast\mathbb{1} & 0 \end{pmatrix}$$ has the desired form.

Note also that the eigenvalues $\lambda$ of this matrix come in inverse pairs $\pm \lambda$, because it anticommutes with $\begin{pmatrix} \mathbb{1} & 0 \\ 0& -\mathbb{1} \end{pmatrix}$ (chiral symmetry).

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