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In the real field. Given a diagonal matrix $D$ and a symmetric matrix $A$. For every skew symmetric matrix $S$, is there always a symmetric matrix $H$ such that $-\operatorname{trace}(DSAS)=\operatorname{trace}(DHAH)$?

If $A$ is also diagonal, this can be easily seen true.

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Why do you ask? – Igor Rivin Aug 14 '11 at 21:54
1  
It is a very specific problem, it does not have interesting background. – Russel Aug 15 '11 at 18:11
up vote 4 down vote accepted

Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then
$$f(\alpha H)=\alpha^2f(H)$$
for every real $\alpha$. Obviously $f(0)=0$. So the problem boils down to find an $H$ such that $f(H)>0$ and another one such that $f(H)<0$.

Now, denoting by $h_k$ the $k-$th column of $H$,
$$(HAH)_{ij}=(h_i)^tAh_j$$
Therefore $f(H)=\sum d_i (h_i)^tAh_i$, if $D=\mathrm{diag}(d_1,\ldots, d_n)$.

If, wlog, $d_1>0$ and $d_2<0$, then we are obviously done.
If every $d_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done.

If every $d_i$ is wlog positive (or null) and $A$ is semi-definite, then $f(H)\geq0$ for every $H$ symmetric and $f(S)\leq 0$ for every $S$ skew-symmetric, so, given $H$ such that $f(H)\neq0$, we can always find $\alpha$ such that $f(\alpha H)=-f(S)$.

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