Existence of a symmetric matrix. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T21:25:50Z http://mathoverflow.net/feeds/question/72887 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72887/existence-of-a-symmetric-matrix Existence of a symmetric matrix. Zae Kwong 2011-08-14T21:03:32Z 2011-08-15T02:46:13Z <p>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 $-\operatorname{trace}(DSAS)=\operatorname{trace}(DHAH)$? </p> <p>If $A$ is also diagonal, this can be easily seen true. </p> http://mathoverflow.net/questions/72887/existence-of-a-symmetric-matrix/72895#72895 Answer by Samuele for Existence of a symmetric matrix. Samuele 2011-08-14T23:46:51Z 2011-08-15T02:46:13Z <p>Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then<br> $$f(\alpha H)=\alpha^2f(H)$$<br> 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 $f(H)&lt;0$. </p> <p>Now, denoting by $h_k$ the $k-$th column of $H$,<br> $$(HAH)_{ij}=(h_i)^tAh_j$$<br> Therefore $f(H)=\sum d_i (h_i)^tAh_i$, if $D=\mathrm{diag}(d_1,\ldots, d_n)$.</p> <p>If, wlog, $d_1>0$ and $d_2&lt;0$, then we are obviously done.<br> If every $d_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done. </p> <p>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 $f(\alpha H)=-f(S)$.</p>