Let $a, b\in A\_+$ be positive elements of some *C*\*-algebra $A$.
Assume furthermore that $a$ is invertible.  

Is it true that
$$
\exists! x\in A\_+\quad:\quad xax=b\quad ?
$$
  
  

Already in the case $A=M\_2(\mathbb C)$, I don't know how to solve this.



Solving the equation $xax=b$ in a C*-algebra.
I do not know about general C\*-algebras, but the statement is true for complex matrices.


**Uniqueness:** Assume *b* = *x**a**x*. Then *a*1/2*b**a*1/2 = *a*1/2*x**a**x**a*1/2 = (*a*1/2*x**a*1/2)2, which implies that *a*1/2*x**a*1/2 = (*a*1/2*b**a*1/2)1/2. Since *a* is invertible, *x* must be *a*−1/2(*a*1/2*b**a*1/2)1/2*a*−1/2.


**Existence:** It is easy to check that 
$x=\sqrt{a}^{-1}\sqrt{\sqrt{a}\hspace{.15cm}b\sqrt{a}\quad}\sqrt{a}^{-1}$
 satisfies the condition.


I do not think that anything in this argument depends on the fact that we are considering matrices, but let me avoid claiming things about the subject which I do not know well.


