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I have a matrix of the form

$$D = \left( \begin{array}{cc} 0 & C \\ C^{\dagger} & 0 \end{array} \right)$$

where $C$ is not necessarily hermitian. In general, can we say anything about the eigenvalues of $D$ in terms of $C$? In particular, I'm interested in the case where every eigenvalue of $C$ has algebraic multiplicity 2, and what that implies about the multiplicities of the eigenvalues of $D$.

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    $\begingroup$ If you square it, the resulting matrix has nonnegative eigenvalues of $CC*$ with multiplicity multiplied by two (since $CC^*$ and $C^*C$ are unitarily conjugate). The corresponding eigenvalues of $D$ are the square roots. This has relatively little to do with the eigenvalues of $C$ itself, rather the singular values. I'm sure you can figure out the details---or was this part of an exercise? Most of the time, when $\dagger$ for conjugacy transpose, it's an exercise. $\endgroup$
    – David Handelman
    Commented Jul 25, 2017 at 3:16
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    $\begingroup$ As @David said, it is known that the eigenvectors of $D$ are precisely the singular pairs of $C$. $\endgroup$
    – Surb
    Commented Jul 25, 2017 at 12:59

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