# How to find the optimal ${\bf Q}$ that maximizes $\left|{\bf I}+{\bf AQA^HB^{-1}}\right|$

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.)

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 $$A$$?

• 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 ${\bf B}$? – zhang haiyang May 6 at 1:48
• 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. – zhang haiyang May 6 at 1:55
• Shouldn't it be $C = B^{-1/2} A$ or am I missing something? – The Pheromone Kid May 6 at 7:26
• 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. – The Pheromone Kid May 6 at 7:37
• @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. – Biel Roig-Solvas May 6 at 12:07