Given a real matrix $A$ and a vector $v$, does anyone know the solution to the following optimization problem?
$$\begin{array}{ll} \text{supremize} & \mbox{Tr} (A^t X)\\ \text{subject to} & v^t X^{-1}v \leq 1\end{array}$$
where $X$ is a real symmetric positive definite matrix.
This is a convex problem, but the domain is not closed, which complicates a bit the situation.
On
Using the Schur complement test for positive semidefiniteness, the given problem can be rewritten as
$$\begin{array}{ll} \text{supremize} & \langle \mathrm A , \mathrm X \rangle\\ \text{subject to} & \begin{bmatrix} \mathrm X & \mathrm v\\ \mathrm v^\top & 1\end{bmatrix} \succeq \mathrm O_{n+1}\\ & \mathrm X \succ \mathrm O_n\end{array}$$
Using $\mathrm X \succeq \mathrm O_n$ instead of $\mathrm X \succ \mathrm O_n$, we obtain the following semidefinite program (SDP)
$$\begin{array}{ll} \text{maximize} & \langle \mathrm A , \mathrm X \rangle\\ \text{subject to} & \begin{bmatrix} \mathrm X & \mathrm v\\ \mathrm v^\top & 1\end{bmatrix} \succeq \mathrm O_{n+1}\\ & \mathrm X \succeq \mathrm O_n\end{array}$$
If the optimal $\rm X$ is positive definite, then we have solved the original problem.
If $A$ is positive definite, you can make $\operatorname{Tr}(A^tX)$ as big as you like by taking $X$ to be a huge multiple of the identity matrix.
In any case $A$ might as well be symmetric, and hence the difference of 2 p.s.d. matrices. Maybe you can then take $X$ to look like a huge multiple of the identity restricted to the subspace where $A$ looks p.s.d., to the same effect?