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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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  • $\begingroup$ btw: ^H means hermitian transpose $\endgroup$
    – user34079
    Commented May 17, 2013 at 6:56
  • $\begingroup$ since $Det A = (Det V^H V) \prod_n D_n$, this is trivial $\endgroup$
    – Carlo Beenakker
    Commented May 17, 2013 at 7:28
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    $\begingroup$ Not sure about that: $V^HV$ is a rank-$M$ $2M\times 2M$ matrix, so its determinant would be zero. $\endgroup$
    – Federico Poloni
    Commented May 17, 2013 at 7:49
  • $\begingroup$ 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. $\endgroup$
    – Carlo Beenakker
    Commented May 17, 2013 at 10:44

1 Answer 1

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Here is one way to solve your problem numerically using CVX under Matlab:

% assuming we have defined v and m above
cvx_begin sdp
    variable d(2*m,2*m) diagonal
    maximize det_rootn(v*d*v')
    subject to
         d >= 0;
         trace(d)==1;
cvx_end
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