Geometric mean of matrix

Question

Edited:

I will give some definitions.

Definition 1 A matrix $C$ is called the square root of matrix $A$ if it satisfies $C^2=A$. It can be written as $C=A^{\frac{1}{2}}$.

Definition 2 Given two positive definite matrices $A,B$. The geometric mean of $A$ and $B$ is defined as $$A\#B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}A^{\frac{1}{2}}.$$

I want to proof that $A\#B$ is the only positive definite solution of the equation $$XA^{-1}X=B.$$ First, I must proof that $A\#B$ is the solution for the equation above. The proof is already given by Mr.@Tamshin Dion in the answer below. Next, I must proof that $A\#B$ is the only positive definite solution for the equation.

Let $X=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}A^{\frac{1}{2}}$ satisfying $$XA^{-1}X=B.$$ Let $Y$ is another positive definite matrix satisfying $$YA^{-1}Y=B.$$ Since $Y$ is positive definite, then it is also hermitian. By Spectral Theorem, there exist a unitary matrix $U$ such that $$Y=UDU^*$$ where $D$ is a diagonal matrix. Then, we have \begin{align*} YA^{-1}Y=B &\iff (UDU^*)A^{-1}(UDU^*)=B\\ &\iff DU^*A^{-1}UD=U^*BU\\ &\iff (UD)^*A^{-1}UD=U^*BU\\ &\iff (UD)^*A^{-1}UD=U^*XA^{-1}XU\\ &\iff (UD)^*A^{-1}UD=(X^*U)^*A^{-1}(XU)\\ &\iff (UD)^*A^{-1}UD=(XU)^*A^{-1}(XU) \end{align*} Now, Iam stuck in here. If I can show that $UD=XU$, then $UDU^*=X$ so that $Y=X$. But, how to proof it?

Notes :

1. For the definition of the geometric mean, I use Rajendra Bhatia's book "Positive Definite Matrices".

2. $U^*$ denotes conjugate transpose.

Answer 1

Since $XA^{-1}X=B$, we have $$\boxed{(A^{-\frac{1}{2}}XA^{-\frac{1}{2}})(A^{-\frac{1}{2}}XA^{-\frac{1}{2}})=A^{-\frac{1}{2}}BA^{-\frac{1}{2}}}$$ Then $X=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}A^{\frac{1}{2}}.$