Maximizing $\mbox{tr} \left( {\bf X}^{-1} {\bf A} {\bf X} {\bf B} + 2 {\bf X} {\bf C} \right)$ subject to ${\bf X} {\bf X}^\top = \gamma {\bf I}$

Question

Can the following optimization statement be converted into a simpler form?

$$\begin{array}{ll} \underset{{\bf X}}{\text{maximize}} & \mbox{tr} \left( {\bf X}^{-1} {\bf A} {\bf X} {\bf B} + 2 {\bf X} {\bf C} \right)\\ \text{subject to} & {\bf X} {\bf X}^\top = \gamma {\bf I}\end{array}$$

where $\bf A$ and $\bf B$ are symmetric matrices, and $\gamma > 0$ is a constant. I tried using Lagrange multipliers and ended up with the equation: ${\bf A} {\bf X} {\bf B} + {\bf C}^\top = {\bf X} {\bf \Lambda}$, where ${\bf \Lambda}$ is the matrix with Lagrange multipliers along its diagonal. I'm not sure how to go about it after that. Please help.

Answer 1

The idempotent Cayley transform $${\rm cay}(S)=(I+S)^{-1}(I-S)\quad\implies\quad {\rm cay}({\rm cay}(S))=S$$ applied to a skew matrix yields an orthogonal matrix.

So starting with an unconstrained matrix $U$, construct a skew matrix, then an orthogonal matrix, and scale that to construct your $X$ matrix, i.e. $$\eqalign{ X &= \gamma^{+1/2}\:{\rm cay}(U-U^T) \\ X^{-1} &= \gamma^{-1/2}\:{\rm cay}(U^T-U) \\ }$$ Rewrite the problem as an unconstrained problem in terms of the $U$ matrix, set the gradient to zero, and solve for the optimal $U=U_{opt}$.

It will be very difficult to find a closed-form solution to this problem, so you will probably use a gradient descent method. Once you've calculated $U_{opt}$ it is easy to construct the corresponding $X_{opt}$

Update

If you visit MatrixCalculus.org and type in

tr(inv(eye+U'-U)*(eye+U-U')*A*inv(eye+U-U')*(eye+U'-U)*B + 2*g*inv(eye+U-U')*(eye+U'-U)*C)

It will calculate the gradient wrt $U$ for you.

Don't forget to specify that $(A,B)$ are symmetric. You'll still need to do some manual simplifications because it doesn't recognize that certain combinations of its temporary variables can be reduced to $X$ or $X^T$.