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We Can prove that $trace(A(P+Q)^{−1}A^T)$ is a jointly convex function of positive variables $[q_1,q_i,...,q_N]$, where $Q=diag(q_1,...,q_N)$, $q_i>0$,∀i, and $P$ is a positive definite matrix. Convexity of a trace of matrices with respect to diagonal elements

Now I am trying to maximize $trace(A(P+Q)^{−1}A^T)$ and find optimal $q_i$'s, with known $A$ and $P$, and a constraint like $\prod q_i=a$. We also know that $trace(A(P+Q)^{−1}A^T)$ is a decreasing function of $q_i$'s. I there a standard way that I can find the optimal solution? I know that it is maximizing a convex function.

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Alireza
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  • When you say that the function is a decreasing function of the $q_i$'s, what order do you put on the vectors of $q$'s? – uniquesolution Sep 21 '15 at 23:26
  • what do you mean by order? I think derivative of function with respect to $q_i$ would be negative ?! – Alireza Sep 21 '15 at 23:29
  • So what you are saying is that if all the $q$'s but one, say, $q_1$, are held fixed, then the function, viewed as a function of $q_1$, is decreasing. Now I understand, thank you. I would have written it otherwise, but it doesn't matter. As for your question. If the constraints happen to define a compact and convex set, then you know that a maximum will be obtained on some extreme point. But unfortunately the constraint $\Pi p_i=a$ does not define a convex set. Can you replace it by some constraint on the arithmetic mean instead of the geometric mean? – uniquesolution Sep 21 '15 at 23:32
  • @uniquesolution the actuall problem that I have is that $q_i=\frac{1}{4^{L_i}}$ and $\sum L_i=b$, so I tried to write the constraint for $q_i$'s, does it help? – Alireza Sep 21 '15 at 23:46
  • I don't see an immediate escape, as being jointly convex in $q_i$ does not seem to imply being jointly convex in $L_i$. I guess you'll have to work hard with lagrange's multipliers. – uniquesolution Sep 21 '15 at 23:51
  • what's the point of working by Lagrange multipliers for this problem? I mean it's maximization on a convex function – Alireza Sep 21 '15 at 23:55
  • True, but it's a convex function on a highly non-convex domain, so the convexity doesn't help you much, does it? Convexity of a function is useful only if you can form convex combinations of it's arguments that still obey your constraints. This is not the case here. – uniquesolution Sep 22 '15 at 00:10
  • @user251257 that's the question – Alireza Sep 22 '15 at 14:31

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