Let $P$ and $Q$ be positive definite matrices. Consider the following relation $$ XPX^\top = P-Q $$ where $X$ is any square real matrix.
My question: What is the most general form of $X$ which guarantees that the above relation is met?
Some observations. Since $P-Q$ is positive semi-definite, then I guess that the answer would be $X=(P-Q)^{1/2}TP^{-1/2}$, with $T$ being an arbitrary orthogonal matrix. Am I correct?
We consider the equation $XPX^T=R$ in $X\in M_n$ where $P\in M_n$ is symmetric $>0$ and $R\in M_n$ is symmetric. If $R$ is not $\geq 0$, then no solutions.
We assume that $R\geq 0$. Then $(XP^{1/2})(XP^{1/2})^T=R^{1/2}(R^{1/2})^T$. Then, by the SVD, $rank(XP^{1/2})=rank(R^{1/2})=r$ and there is $U\in O(n)$ s.t. $XP^{1/2}=R^{1/2}U$ and $X=R^{1/2}UP^{-1/2}$. Thus the set $Z$ of solutions $X$ is an algebraic set but, beware, its dimension is not, in general, equal to $dim(O(n))=n(n-1)/2$.
We may assume that $R=diag(\lambda_1,\cdots,\lambda_r,0,\cdots,0))=diag(\Lambda,0_{n-r})$ where $\lambda_i>0$. Thus $XP^{1/2}=diag(\Lambda^{1/2},0)\begin{pmatrix}V_r&W\\H&K_{n-r}\end{pmatrix}=\begin{pmatrix}\Lambda^{1/2}V&\Lambda^{1/2}W\\0&0\end{pmatrix}$, and the parameters are the $r^2+r(n-r)$ entries of $V,W$ linked by the $r(r+1)/2$ relations $VV^T+WW^T=I_r$ ($U$ is orthogonal).
We can prove that $dim(Z)=r^2+r(n-r)-r(r+1)/2=r(n-(r+1)/2)$.