Solving the equation $xax=b$ in a C*-algebra. 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.
 A: I do not know about general C*-algebras, but the statement is true for complex matrices.
Uniqueness: Assume b = xax.  Then a1/2ba1/2 = a1/2xaxa1/2 = (a1/2xa1/2)2, which implies that a1/2xa1/2 = (a1/2ba1/2)1/2.  Since a is invertible, x must be a−1/2(a1/2ba1/2)1/2a−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.
A: In the case of $\Bbb{M}_n(\Bbb{C})$, you should diagonalize $a$, say $d(\lambda_1,\cdots,\lambda_n)$. Then $a^{1/2}$ is $d(\sqrt{\lambda_1},\cdots,\sqrt{\lambda_n})$. Easily you can find $a^{-1/2}$ (since $a^{1/2}$ is of course invertible). The rest is just following the previous answer:
$x=a^{-1/2}(a^{1/2}ba^{1/2})^{1/2}a^{-1/2}.$
Then you can return the basis to the previous one (i.e. the basis before diagonalization of $a$).
