Solve $PV = PVP^TS^T$ when $PS = I$ Suppose $PV = PVP^TS^T$ such that $PS = I$. How can I solve for $P$. 
Note: $V$ is positive definite. $S$ is full column rank.
I can see that the solution might looks like: $P = (S^TV^{-1}S)^{-1}S^TV^{-1}$. But I can't reach this solution!
Any tips are greatly appreciated.
 A: Yes, that is the solution.
Say $P$ is $m \times n$, so $S$ is $n \times m$, and $m \le n$.  Of course if $m=n$, you simply have $P = S^{-1}$, so let's suppose $m < n$.
Let $U = \text{Ran}(S)$, which is an $m$-dimensional subspace of $\mathbb R^n$.  If $u \in U$, $u = S v$ for some $v \in \mathbb R^m$, and $P u = PS v = v$.  $P$ must have a kernel $W$ of dimension $n-m$, with $W \cap U = \{0\}$.  Thus $\mathbb R^n = U \oplus W$.  $SP$ is the projection of $\mathbb R^n$ on $U$, and $I-SP$ the projection on $W$, corresponding to this.  
Now you want $(SP-I) V P^T = 0$.  This says  $\text{Ran}(V P^T) \subseteq \text{Ker}(I-SP) = U$, i.e. $\text{Ran}(P^T) \subseteq V^{-1} U$, i.e.
$(V^{-1} U)^\perp \subseteq \text{Ker}(P) = W$.  But these have the same dimension, so they are equal.  So now we know $W = (V^{-1} U)^\perp$.
Note that this is possible, since if $0 \ne u \in U$, $u^T V^{-1} u > 0$ so
$u \notin (V^{-1} U)^\perp$.  Having determined the action of $P$ on $U$ and 
on $(V^{-1} U)^\perp$, we have determined $P$.
Now let's examine your candidate $C = (S^T V^{-1} S)^{-1} S^T V^{-1}$.
If $u \in U$, so $u = S v$ for some $v$, $C u = (S^T V^{-1} S)^{-1} (S^T V^{-1} S) v = v$, which is what it is supposed to be.
If $w \in W = (V^{-1} U)^\perp$, $V^{-1} w \in U^\perp = \text{Ker}(S^T)$, so
$S^T V^{-1} w = 0$ and $C w = 0$.  Thus it has the correct action on both $U$ and $(V^{-1} U)^\perp$, so it must be $P$.
A: Let $Y=PV^{1/2}$ and $X=V^{-1/2}S$. Then the equations $PS=I$ and $PV=PVP^TS$ can be rewritten as
\begin{align}
&YX=I,\tag{1}\\
&Y=YY^TX^T.\tag{2}
\end{align}
From $(1)$, we see that (a) $XYX=X$, (b) $YXY=Y$ and (c) $YX$ is symmetric. Also, from $(2)$ we get $XY=(XY)(XY)^T$. Since the RHS is symmetric, we see that (d) $XY$ is symmetric. Therefore, $Y$ satisfies the four defining properties of Moore-Penrose pseudoinverse of $X$. As Moore-Penrose pseudoinverse is unique, we conclude that $Y=X^+$. Finally, in general, it is known that $X^+=(X^TX)^{-1}X^T$ when $X$ has full column rank. Since this is the case here (because $X=V^{-1/2}S$), we conclude that
$$
Y=X^+=(X^TX)^{-1}X^T=(SV^{-1}S)^{-1}S^TV^{-1/2}
$$
and $P=YV^{-1/2}=(SV^{-1}S)^{-1}S^TV^{-1}$.
