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I have two real functions $f(X),g(X)$ where the argument $X$ is a real matrix. The solution $X^*$ for the problem of minimizing $f$ is ending up maximizing $g$ as well. I am looking for a way to prove or at least understand the condition/reasoning behind why this is happening for these two functions.

$f(X)=Tr(X^TAX)-Tr(X^TBX)$ and $g(X)=\frac{Tr(X^TBX)}{Tr(X^Th(X)X)}$ and

$h(X)=diag(XX^T\mathbb{1})-XX^T$. $A$ and $B$ are p.s.d matrices.

$B$ is a matrix that does not depend on $X$.

Coming to $A$, this is what I do, I first set $A=h(X)$ for some chosen $X$ and then I keep that matrix constant, in the optimization of $f(.)$, that is it is only optimized on the $X$ terms other than $A$. Let me know if you need more detail/clarification on how I use $A$ in this optimization.

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You have given very little information. If $g=f$ then it is obvious why this will occur. If you want a meaningful answer, you need to elaborate. –  copper.hat Jun 19 '13 at 5:38
    
@copper.hat Added more details on the functions –  qlinck Jun 19 '13 at 15:56
    
A brief glance would suggest that this is coincidental, since $f$ depends on $A,B$, whereas $g$ depends on $B$ only. Both $f,g$ are homogeneous of degree 2, so either the maximum is zero, or it is unbounded above. –  copper.hat Jun 19 '13 at 17:47
    
@copper.hat I made a gross error while typing the question. I apologize. The observation is that the minimizer (not maximizer) of $f(.)$ is ending up maximizing $g(.)$. Can you look at this version instead? I have corrected it and I also inverted the previous definition of $g(.)$. I can see that minimizing the second term in $f(.)$ with the negative sign can maximize the numerator in $g(.)$. Also $f(.)$ is a sum of a convex and a concave function. –  qlinck Jun 19 '13 at 20:55
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Same basic comment. $f(\lambda X) = \lambda^2 f(X)$ and $g(\lambda X) = \frac{1}{\lambda^2} g(X)$, so the $\sup, \inf$ are either $0$ or $\pm \infty$. –  copper.hat Jun 19 '13 at 22:49

1 Answer 1

Edit: Answered before new details were added. I'll revisit this.

This will be true generically when $g$ is a monotonic transformation of $f$ (transformation by a strictly increasing function). When this is the case: $g(x)\geq g(\tilde{x})\Leftrightarrow f(x)\geq f(\tilde{x}) $

Since $f(x^*)\geq f(x)$ for all $x$ in the domain, $g(x^*)\geq g(x)$ for all $x$ in the domain as well.

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How is this relevant to the OP's question? It is not at all clear to me that $g$ is a monotonic transformation of $f$? –  copper.hat Jun 20 '13 at 0:40

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