Is there a way to directly solve this matrix equation: $XAX^{T} = B$

Question

$X^{T}$ is the transpose of $X$. $A$ is a $n$ x $n$ matrix and $B$ is a $m$ x $m$ matrix, $m$ > $n$, both of them are known, $A$ is positive definitive and $B$ is symmetric. I would like to find $X$.

Answer 1

EDIT. Note that if $A$ is real symmetric $\geq 0$, then necessarily $B$ is real symmetric $\geq 0$ with $rank(B)\leq rank(A)$.

Thus we assume that $A,B$ are symmetric $\geq 0$ matrices that satisfy $rank(B)=r\leq rank(A)=s\leq n<m$. There are invertible matrices $P\in M_m,Q\in M_n$ s.t. $A=Qdiag(I_s,0_{n-s})Q^T,B=Pdiag(I_r,0_{m-r})P^T$and therefore $P^{-1}XQdiag(I_s,0)(P^{-1}XQ)^T=diag(I_r,0)$. Since we know $P,Q$, It suffices to find the $Y=P^{-1}XQ\in M_{m,n}$ s.t. $Ydiag(I_r,I_{s-r},0_{n-s})Y^T=diag(I_r,0_{s-r},0_{n-s},0_{m-n})$. With respect to the previous block matrices, let $Y=[Y_{i,j}],1\leq i\leq 4,1\leq j\leq 3$. By identification of the diagonal elements of RHS and LHS , $Y_{2,1}Y_{2,1}^T+Y_{2,2}Y_{2,2}^T=0$; that implies $Y_{2,1}=0,Y_{2,2}=0$. In the same way, $Y_{3,1}=0,Y_{3,2}=0$ and $Y_{4,1}=0,Y_{4,2}=0$. Finally $Y=\begin{pmatrix}Y_{1,1}&Y_{1,2}&Y_{1,3}\\0&0&Y_{2,3}\\0&0&Y_{3,3}\\0&0&Y_{4,3}\end{pmatrix}$ where $Y_{1,1}Y_{1,1}^T+Y_{1,2}Y_{1,2}^T=I_r$.