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Let $H$ be an $n\times n$ Hermitian matrix and $A$ an $n\times n$ anti-Hermitian matrix, i.e. $H^\dagger = H$, $A^\dagger = -A$. Consider their sum $S= H+A$. Let $\{\sigma_i(S)\}_{i=1,\dots,r}$ denote the singular values of $S$. By convention, they are decreasingly ordered. Knowing the spectrum of $H$ and $A$, what is the maximum achievable value for $\sigma_1(S)$?

In other words, if $\sigma_1(S)$ denotes the greatest singular value of $S$, the question is to compute the maximum \begin{equation} \underset{U, V}{\text{max}}\;\sigma_1(U H U^\dagger + V A V^\dagger)\;, \end{equation} where $U$ and $V$ are arbitrary $n\times n$ unitary matrices.

I believe a similar question is completely settled for the sum of two Hermitian matrices. I would like to know if similar results are known in this more general scenario.

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    $\begingroup$ you could just choose $U$ and $V$ such that $H$ and $A$ are diagonal, so that $\sigma_1^2=h_{\rm max}^2+a_{\rm max}^2$, with $h_{\rm max}$ and $ia_{\rm max}$ the largest eigenvalues of $H$ and $A$. $\endgroup$
    – Carlo Beenakker
    Commented Aug 10, 2017 at 18:34
  • $\begingroup$ The $\max$ in the displayed equation is obviously equal to $\|H\|+\|A\|$, but it's not the same as $\sigma_1(S)$ (what this is depends on the details of the situation). $\endgroup$
    – Christian Remling
    Commented Aug 10, 2017 at 23:13

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