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I have the following question, which has stumped me for a long time.

${\max}_{\bf Q \succeq 0}~~~~{\rm det}\left({\bf I}+{\bf AQA^HB^{-1}}\right)$

s.t.$~~~{\rm Tr}\left({\bf Q}\right) \leq 1$.

where $\mathbf{B}=\operatorname{diag}\left[\mu_{1}, \cdots, \mu_{N}\right]$ is a diagonal matrix, and ${\bf A}$ is an arbitrary matrix.

Is there any possibility to find the optimal solution ${\bf Q}$ that has an explicit relationship with the diagonal elements of ${\bf B}$? (because $\left\{\mu_i\right\}$ are also variables.)

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At first glance I'd say that it looks difficult to find a closed form solution that depends directly on $B$. Using simple algebra, one can reshape the objective as:

${\rm det}\left({\bf I}+{\bf AQA^HB^{-1}}\right) = {\rm det}\left({\bf B}+{\bf AQA^H}\right) {\rm det}\left(\bf B^{-1}\right) = {\rm det}\left(\bf B^{-1/2}\right){\rm det}\left({\bf B}+{\bf AQA^H}\right) {\rm det}\left(\bf B^{-1/2}\right) = {\rm det}\left({\bf I}+{\bf B^{-1/2}AQA^HB^{-1/2}}\right)$

From here, write $C = B^{-1/2}A$, and write its SVD as $C = U_C \Sigma_C V_C^T$ and the SVD of $Q$ as $Q = U_Q \Sigma_Q U^T_Q$ (as Q is symmetric), then we have that the objective is:

${\rm det}\left( I + U_C \Sigma_C V_C^T U_Q \Sigma_Q U_Q^T V_C \Sigma_C U_C^T\right)$

We can take the orthonormal matrices $U_C$ out of the objective (as $I = U_C U_C^T$) and (and here I'd need a more formal proof) I'd expect the optimizer to have $U_Q = V_C$, as intuitively it makes sense that the determinant will be maximized when $Q$ is maximally aligned with $C$ in terms of singular vectors (with a similar logic to the Von Neumann trace inequality). Then the objective would look like:

${\rm det}\left( I + \Sigma_C \Sigma_Q \Sigma_C \right) = \prod_i^N \left(1+\sigma_i(C)^2 \cdot \sigma_i(Q)\right)$,

and I guess you could easily find a closed form solution for that under the constraint of $trace(Q) == 1$. But as you can see, the optimizer will depend on the singular values of $C=B^{-1/2}A$, so it won't be straightforward to relate the values of $B$ to the singular values of $C$, at least without assuming anything about $A$.

Is there anything else you can assume about $A$?

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  • $\begingroup$ Is there possible to find a ${\bf Q}$ that achieves an upper bound of ${\rm det}\left({\bf I}+{\bf A}{\bf Q}{\bf A}^H{\bf B}^{-1}\right)$, then ${\bf Q}$ can be expressed in terms of ${\bf B}$ or the diagonal elements of ${\bf B}$? $\endgroup$
    – zhang haiyang
    May 6, 2019 at 1:48
  • $\begingroup$ I try to approximate ${\bf A}$ as a diagonal matrix. But the difficulty is how to ensure the approximated ${\bf A}$ can achieve an upper bound of the objective function. $\endgroup$
    – zhang haiyang
    May 6, 2019 at 1:55
  • $\begingroup$ Shouldn't it be $C = B^{-1/2} A$ or am I missing something? $\endgroup$
    – The Pheromone Kid
    May 6, 2019 at 7:26
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    $\begingroup$ A comment regarding the reformulated objective function: After taking the logarithm, it can be solved by using the water filling algorithm, see, for example, core.ac.uk/download/pdf/41756751.pdf. $\endgroup$
    – The Pheromone Kid
    May 6, 2019 at 7:37
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    $\begingroup$ @zhanghaiyang: Unless you somehow use the structure of $A$, I don't think there's a direct way to relate the entries of $B$ (or of $B^{-1/2}$) to the singular values of $C$. If your $A$ were to admit, for example, a factorization of the form $DQ$ by construction, with $D$ diagonal and $Q$ orthonormal, then you'd be able to get a closed form. But otherwise, for arbitrary $A$, I wouldn't hold out hope. $\endgroup$
    – Biel Roig-Solvas
    May 6, 2019 at 12:07

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