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Let $A$ and $B$ be symmetric, positive definite matrices. Consider the following equations: $$ B = XAX, \qquad A = YBY,$$ where $X$ and $Y$ are also symmetric and positive definite matrices. These equations admit the solutions $$ \begin{aligned} X &= A^{-1/2}(A^{1/2} B A^{1/2})^{1/2} A^{-1/2}, \\ Y &= B^{-1/2}(B^{1/2} A B^{1/2})^{1/2} B^{-1/2}, \end{aligned} $$ see for example here. Given the reasoning in the linked post and the uniqueness of the matrix square root, it should hold that $Y$ is equal to $X^{-1}$. Is there a more direct way to see this, from the expressions of $X$ and $Y$?

Roberto Rastapopoulos
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    As evidence of this claim, I generated a bunch of PD matrices $A,B$ in Mathematica and confirmed that all such examples satisfied $XY=I$. – Semiclassical Aug 21 '19 at 20:18

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Denote $C = B^{1/2} A^{1/2}$ and get its SVD: $$ C = U S V^T $$ Then we have $$ A^{1/2} B A^{1/2} = C^T C = V S^2 V^T = ( V S V^T )^{2} $$ and $$ B^{1/2} A B^{1/2} = C C^T = U S^2 U^T = ( U S U^T )^{2} $$ Next we write $$ XY = A^{-1/2} (C^T C)^{1/2} C^{-1} (C C^T)^{1/2} B^{-1/2} $$ or $$ XY = A^{-1/2} ( V S V^T ) V S^{-1} U^T ( U S U^T ) B^{-1/2} = A^{-1/2} V S U^T B^{-1/2} = A^{-1/2} C^T B^{-1/2} = I $$

user7440
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