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.
I do not know about general C^{*}algebras, but the statement is true for complex matrices.
Uniqueness: Assume b = xax. Then a^{1/2}ba^{1/2} = a^{1/2}xaxa^{1/2} = (a^{1/2}xa^{1/2})^{2}, which implies that a^{1/2}xa^{1/2} = (a^{1/2}ba^{1/2})^{1/2}. Since a is invertible, x must be a^{−1/2}(a^{1/2}ba^{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.

4$\begingroup$ This does work for arbitrary C* algebras. $\endgroup$ – Jonas Meyer Jul 20 '11 at 19:53

$\begingroup$ Thank you. You proof works perfectly well in the context of a general C*algebra. $\endgroup$ – André Henriques Jul 20 '11 at 19:55

$\begingroup$ @Jonas, @André: Thank you for the comment about general C*algebra! $\endgroup$ – Tsuyoshi Ito Jul 20 '11 at 19:56

4$\begingroup$ Quick comment: in the matrix case, this is known as the matrix geometric mean $x=GM(A^{1},B)$ of $A^{1}$ and $B$; it has an interesting interpretation in terms of Riemannian geometry, and it is a nontrivial problem to generalize it to a "mean" of more than two matrices satisfying some basic properties. References: Bhatia, Holbrook Riemannian Geometry and Geometric Means, LAA '06, and Noncommutative geometric means, Math. Intelligencer, for a less technical exposition. Where does this problem arise for $\mathbb{C}^*$ algebras? I'm interested! $\endgroup$ – Federico Poloni Jul 20 '11 at 20:04

$\begingroup$ The proof uses that $\sqrt{a} b \sqrt{a} = (\sqrt{b} \sqrt{b})^{*} (\sqrt{b} \sqrt{a})$, which is therefore positive and has a root. $\endgroup$ – Martin Brandenburg Jul 20 '11 at 20:14
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 $a$).

1$\begingroup$ Well this just provides a computation of $\sqrt{a}$ in the case of matrix algebras. Else this is just a copy of the accepted answer. $\endgroup$ – Martin Brandenburg Jul 21 '11 at 11:58

$\begingroup$ @ Martin: In the last part of the question Andre says:"Already in the case $A=M 2 (\Bbb{C})$ , I don't know how to solve this." This is the answer: " using unitary matrices switch the basis to one appropriate one and the follow the previous answer, then using the inverse of that unitary matrix return the base: The $x$." So I used the accepted answer in Matrix Analysis to find $x$! Nothing more! $\endgroup$ – Mahmood Alaghmandan Jul 21 '11 at 23:25