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Am looking for a proof of non-convexity of the quotient of two matrix trace functions as given by $\frac{\operatorname{Tr}X^TAX}{\operatorname{Tr}X^TBX}$, when $TrX^TBX>0$ for two different positive semi-definite matrices , $A$ and $B$ such that $A \neq \alpha B$ for any scalar $\alpha$.

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Thanks for making the point. I have made the necessary changes. – user23600 Apr 18 '12 at 20:42
In fact not equal up to a constant ($A=\alpha B$ where $\alpha$ is a scalar). – Davide Giraudo Apr 18 '12 at 20:45
That's right again, taking the inverse into consideration. – user23600 Apr 18 '12 at 20:47
You also have a definition issue, because it is perfectly possible that for some $X$ your denominator is zero (for example if $B$ is not invertible and $X$ is the projection onto the kernel of $B$). – Martin Argerami Apr 18 '12 at 22:34
Updated and awaiting the proof. – user23600 Apr 19 '12 at 0:52

I'm not exactly sure what you are looking for, but a specific counterexample demonstrates 'non-convexity':


$$ A = \left( \begin{array}{cc} 1 & 0 \\ 0 & 10 \end{array} \right), B = \left( \begin{array}{cc} 10 & 0 \\ 0 & 1 \end{array} \right), X(t) = \left( \begin{array}{cc} 1 & 0 \\ 0 & t \end{array} \right). $$ Some manipulation yields: $$\phi(t) = \frac{\mathbb{tr}(X^T A X)}{\mathbb{tr}(X^T B X)} = \frac{10 t^2+1}{t^2+10}.$$ The function $\phi$ is not convex (for example, $\phi(5) > \frac{1}{2}(\phi(0)+\phi(10))$.

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