I am interested in verifying that the feasibility of the following program:
For $A, B \in S_{n\times n}$, $i$.$e$. symmetric matrices.
$$ \begin{align} \underset{ Y \in S_{n \times n} }{ \text{Find} } \quad & Y \succeq 0 \\ s.t. \quad & \langle A,Y \rangle = a_{1} \\ & Y \succeq 0 \\ & \langle B,Y \rangle \leq b_{1} \\ &\operatorname{Rank} (Y) = s \end{align} $$ $s$ is just some number less than $n$, for simplicity, assume it is $n-1$ or $n - 2$.
For a weak alternative of the following program (at most one of primal or dual holds), is it possible to include the rank constraints somehow?
Ignoring the rank constraints. I think the dual is (might be wrong)
$$ \begin{align} \underset{v,\lambda \in \mathbb{R}, M \in S_{n\times n} }{\text{Find} } &(v,\lambda,M) \\ s.t. \quad & M \succeq 0 \\ & \lambda \geq 0 \\ & -va_{1}-\lambda b_{1} > 0 \\ & vA + \lambda B - M = 0 \end{align} $$ Where the $M$ and $\lambda$ can be combined into a large positive semidefinite matrix if needed.
Anyway, is there a way to encode the rank constraints into the dual program so that the weak alternative can be a little stronger?
My guess is that maybe we can factor $Y = UU^T$ and find the dual that way? But that only gets the inequality direction. As of bounding rank to be greater than something, maybe we can possibly add linear constraints or factor it differently? I don't mind a lot of extra linear constraints.