4
$\begingroup$

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.)

$\endgroup$
3
$\begingroup$

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 $A$?

$\endgroup$
  • $\begingroup$ 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 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 at 1:55
  • $\begingroup$ Shouldn't it be $C = B^{-1/2} A$ or am I missing something? $\endgroup$ – The Pheromone Kid May 6 at 7:26
  • 1
    $\begingroup$ 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 at 7:37
  • 1
    $\begingroup$ @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 at 12:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.