Let $Z$ and $A$ be $m \times m$ symmetric matrices such that $Z \succeq 0$, where $\succeq$ means "positive semi-definite". Show that there is a $\mu\geq 0$ such that

$$\begin{equation}\mbox{Tr}(ZA)=\mu\sup_{y\in\mathbb{R}^{m}}\left\{y^{T}Ay\right\}\end{equation}\mbox{ }\mbox{ }\mbox{ }(*)$$ where $\mbox{Tr}$ means "trace".

Also show the converse, that is, given $\mu\geq 0$ and $A$ symmetric matrix of dimension $m \times m$, then there is $Z$ be a symmetric matrices of $m\times m$ such that $Z\succeq 0$ and we have $(*)$.

Remark: I've tried everything but my attempts do not lead to a good destination, I appreciate any suggestion or solution, or if applicable, any reference to a book that addresses this problem.

  • $\begingroup$ What is the sum over inside the sup? $\endgroup$ – Tom Sep 1 '16 at 1:13
  • $\begingroup$ @Tom It was a typo, this corrected already. $\endgroup$ – Diego Fonseca Sep 1 '16 at 1:20
  • $\begingroup$ there is no finite sup with $y$ not restricted. Probably the rayleigh qoutient is meant, sup, actually max, when $|y|=1,$ meaning the sum of $y_j^2$ is one. $\endgroup$ – Will Jagy Sep 1 '16 at 1:23

This is obviously false: take $m=2$, $$ A=\begin{pmatrix}0&0\\0&-1\end{pmatrix},\qquad Z=\begin{pmatrix}1&0\\0&1\end{pmatrix}. $$ Then Tr $ZA=-1$, and $\sup(\dots)=0$.


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