How to solve $A^{\frac 12} B A^{\frac 12} = C$ for $A$? Suppose that matrices $A,B,C$ are symmetric and positive definite. Then, $A$ has a unique, positive square root, which we call $A^{\frac 12}$. If $$A^{\frac 12} B A^{\frac 12} = C$$ then can we write an expression for $A$ in terms of $B, C$?
 A: The problem is: find the solutions of the equation
(*) $XBX=C$ where the unknown $X$ and the given matrices $B,C$ are $n\times n$ symmetric $>0$.
$\textbf{Proposition.}$ (*) has the unique solution
$X=B^{-1/2}S^{1/2}B^{-1/2}$, where $S=B^{1/2}CB^{1/2}$.
$\textbf{Proof}.$ (*) is equivalent to $(XB)^2=CB$, that is,
$(XB)^2=B^{-1/2}SB^{1/2}$, where $S$ is symmetric $>0$.
We put $XB=B^{-1/2}ZB^{1/2}$ where $Z^2=S$. 
Then $Z=B^{1/2}XB^{1/2}$ is symmetric $>0$ and $Z=S^{1/2}$.
Finally $X=B^{-1/2}S^{1/2}B^{-1/2}$ as required. $\square$
A: We have the following quadratic matrix equation in symmetric positive definite matrix $\mathrm X$
$$\mathrm X \mathrm B \mathrm X = \mathrm C$$
Since matrix $\rm B$ is symmetric and positive definite, it has a (symmetric and positive definite) square root. Left- and right-multiplying both sides of the matrix equation by $\mathrm B^{\frac 12}$, we get
$$\mathrm B^{\frac 12} \mathrm X \mathrm B^{\frac 12} \mathrm B^{\frac 12} \mathrm X \mathrm B^{\frac 12} = \mathrm B^{\frac 12} \mathrm C \mathrm B^{\frac 12}$$
Let $\mathrm Y := \mathrm B^{\frac 12} \mathrm X \mathrm B^{\frac 12}$. Hence, we have the following quadratic matrix equation in $\rm Y$
$$\mathrm Y^2 = \mathrm B^{\frac 12} \mathrm C \mathrm B^{\frac 12}$$
Since matrix $\rm C$ is symmetric and positive definite, the RHS of the matrix equation above is also symmetric and positive definite and, thus, it has a (symmetric and positive definite) square root. Hence, the solution is
$$\mathrm Y = \left( \mathrm B^{\frac 12} \mathrm C \mathrm B^{\frac 12} \right)^{\frac 12}$$
and, thus, the solution of the original quadratic matrix equation is
$$\mathrm X = \color{blue}{\mathrm B^{-\frac 12} \left( \mathrm B^{\frac 12} \mathrm C \mathrm B^{\frac 12} \right)^{\frac 12} \mathrm B^{-\frac 12}}$$ 
A: Since the matrices are symmetric and positive definite, you shall be able and express
$B$ and $C$ as
$$
B = U\;\Sigma \;\overline U \quad C = V\;\Lambda \;\overline V  = V\;\left( {\Lambda \Sigma ^{\, - 1} } \right)^{\,1/2} \;\Sigma \;\overline {\left( {\Lambda \Sigma ^{\, - 1} } \right)^{\,1/2} } \;\overline V 
$$
where the overbar indicates the transpose, and $\Sigma, \;\Lambda$ are diagonal.
Thereupon you get
$$
\eqalign{
  & A^{\,1/2} \,U\;\Sigma \;\overline U \,A^{\,1/2}  = A^{\,1/2} \,U\;\Sigma \;\overline U \,\overline {A^{\,1/2} }  =   \cr 
  &  = \left( {A^{\,1/2} \,U} \right)\;\Sigma \;\overline {\left( {A^{\,1/2} \,U} \right)}  = V\;\Lambda \;\overline V  =   \cr 
  &  = V\;\left( {\Lambda \Sigma ^{\, - 1} } \right)^{\,1/2} \;\Sigma \;\overline {\left( {\Lambda \Sigma ^{\, - 1} } \right)^{\,1/2} }  \cr} 
$$
i.e.:
$$
A^{\,1/2}  = V\;\left( {\Lambda \Sigma ^{\, - 1} } \right)^{\,1/2} U^{\, - 1} 
$$
But from here I don't see a way to express $A^{1/2}$ directly in function of $B$ and $C$.
