Let $X_{n \times p}$ (rank $\leq p)$, $Y_{n \times q}$ is of rank $q$, $\Sigma_{q \times q}$ is of rank $q$ and $B_{q \times p}$ such that $Rank(J + B^TU^T) = q$ where $J = [0_{p\times q} | I_p]$ of rank $p$ and $U^T = [I_q | -C_{q \times p}]$ of rank $q$.
Can we solve the following equation for $B$ under two cases:
\begin{align} X B \Sigma^{-1} B^T = Y\Sigma^{-1}B^T \end{align}
where $\Sigma = A + CB + B^TC^T + B^TDB$, where $A$ is symmetric and $D$ is symmetric and positive definite.
I can see two solutions here:
$B$ is a null matrix.
$B$ is full column rank.
Are there any other possibilities?
$X B \Sigma^{-1} B^T- Y\Sigma^{-1}B^T $ is the gradient of some complicated objective function. Is that mean the objective function is non-convex?
Because it is hard to compute the second derivative of the matrix to see whether it is convex or not.
A mistake: $B\in M_{p,q}$.
Clearly, if the equation (1) $XB=Y$ has at least one solution, then the complete equation $(*)$ has at least a solution.
$(1)$ has a solution iff $XX^+Y=Y$ where $X^+$ is the Moore- Penrose inverse. cf.
https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_inverse#Obtaining_all_solutions_of_a_linear_system
Then the general solution of $(1)$ is $B=X^+Y+(I-X^+X)W$ where $W\in M_{p,q}$ is arbitrary.
Notice also that, if $rank(B)=q$, then $(*)$ is equivalent to $(1)$ but that not implies that $(1)$ has a solution.