How to maximize the determinant of a matrix of the form VDV^H

Hi,

I have a matrix of the form $A=VDV^H$, where $V$ is a $M \times 2M$ complex matrix, $D$ is a $2M \times 2M$ diagonal real matrix, thus the dimension of $A$ is $M \times M$.

My problem is how to maximize the determinant of $A$ by choosing the diagonal $D$ (suppose the sum of diagonals of $D$ equals to $1$)?

Since the dimension of $V$ is $M \times 2M$, I don't know how to solve this problem. Thanks.

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btw: ^H means hermitian transpose – user34079 May 17 '13 at 6:56
since $Det A = (Det V^H V) \prod_n D_n$, this is trivial – Carlo Beenakker May 17 '13 at 7:28
Not sure about that: $V^HV$ is a rank-$M$ $2M\times 2M$ matrix, so its determinant would be zero. – Federico Poloni May 17 '13 at 7:49
thank you, Federico, I stand corrected; the correct formula is $${\rm Det}A=\sum_{S}|{\rm Det}V_{S}|^2\prod_{k\in S}D_{kk}$$ where $S$ is a subset of $M$ indices out of $1,2,...2M$ and $V_S$ is an $M\times M$ matrix constructed from $V$ by deleting the $M$ columns that are not in $S$. it would seem that to maximize this is in general not trivial. – Carlo Beenakker May 17 '13 at 10:44

% assuming we have defined v and m above