Is there any way to solve for $X$ the equation:
$$XAX^T=B$$
Where $A$ is an $n\times n$ matrix, $B$ is an $m\times m$ matrix, and $X$ is an $m\times n$ matrix, with $m < n$. The elements of $X$ can be real or complex, I'm looking for any solution at all.
So are there any methods, research or ideas for finding a solution or a set of solutions for this problem?
Necessarily $rank(A)\geq rank(B)$. Moreover $X(A+A^T)X^T=B+B^T$; then necessarily, $rank(A+A^T)\geq rank(B+B^T)$.
Assume that $A,B$ are real and $B+B^T$ is invertible; let $signature(B+B^T)=\{p\times +,(m-p)\times -\}$. Thus necessarily, $X^T$ is injective and $signature(A+A^T)$ is in the form $\{u\times +,v\times -,n-u-v\times 0\}$ with $u\geq p,v\geq m-p$. I think that the previous conditions suffice to the existence of $X$ s.t. $(*)$ $X(A+A^T)X^T=B+B^T$. Of course, a solution of $(*)$ is not necessarily solution of $XAX^T=B$.
EDIT . For the general solution of $(*)$ when $A+A^T\geq 0,B+B^T\geq 0$, see my answer in Is there a way to directly solve this matrix equation: $XAX^{T} = B$