How to maximize $\langle X, Y \rangle$ over $X$ under constraint s.t $\mbox{Tr} (X^TAX)=1$?

Question

Maximize $\langle X, Y \rangle$ over $X$ subject to the constraint $\mbox{Tr} (X^TAX) = 1$.

Here $A$ is positive semi-definite and also all matrices have real entries and $Y$ is a known matrix given before hand. Is there a closed form solution to this problem? Or a good closed-form approximation only in the case that there is no closed form solution at all. Only if these two cannot be obtained I may be interested in iterative solutions. But those I can obtain myself as well, so you may provide those but that is not quite what am looking for, unless if they have a guarantee of global optima in finite steps! So will upvote them but not consider them an answer unless: 1) It's a closed-form solution (main priority for me) or 2) if not atleast an iterative solution with a guarantee of global optima in finite steps..which I will accept as an answer as well.Thanks.

Answer 1

Given $\mathrm A \in \mathbb R^{m \times m}$ and $\mathrm C \in \mathbb R^{m \times n}$, we have the quadratically constrained linear program (QCLP)

$$\begin{array}{ll} \text{maximize} & \langle \mathrm C , \mathrm X \rangle\\ \text{subject to} & \mbox{tr} \left( \mathrm X^{\top} \mathrm A \mathrm X \right) = 1\end{array}$$

If $\mathrm A$ is symmetric and positive definite, then there exists an invertible matrix $\mathrm Q \in \mathbb R^{m \times m}$ such that $\mathrm A = \mathrm Q^{\top} \mathrm Q$. Hence,

$$\mbox{tr} \left( \mathrm X^{\top} \mathrm A \mathrm X \right) = \mbox{tr} \left( \mathrm X^{\top} \mathrm Q^{\top} \mathrm Q \mathrm X \right) = \| \mathrm Q \mathrm X \|_{\text{F}}^2$$

Let $\mathrm Y := \mathrm Q \mathrm X$. Hence, we have the following QCLP in $\mathrm Y \in \mathbb R^{m \times n}$

$$\begin{array}{ll} \text{maximize} & \langle \mathrm Q^{-\top}\mathrm C , \mathrm Y \rangle\\ \text{subject to} & \| \mathrm Y \|_{\text{F}}^2 = 1\end{array}$$

Note that

$$\langle \mathrm Q^{-\top}\mathrm C , \mathrm Y \rangle \leq \| \mathrm Q^{-\top} \mathrm C \|_{\text{F}} \underbrace{\| \mathrm Y \|_{\text{F}}}_{=1} = \| \mathrm Q^{-\top} \mathrm C \|_{\text{F}} = \color{blue}{\sqrt{\mbox{tr} (\mathrm C^{\top} \mathrm A^{-1} \mathrm C)}}$$

where the maximum is attained at $\mathrm Y_{\max} := \mathrm Q^{-\top} \mathrm C$, or, $\color{blue}{\mathrm X_{\max} := \mathrm A^{-1} \mathrm C}$. We can try to use a similar approach to handle the case where $\mathrm A$ is symmetric and but only positive semidefinite.