# 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.

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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.

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This does work for arbitrary C* algebras. –  Jonas Meyer Jul 20 '11 at 19:53
Thank you. You proof works perfectly well in the context of a general C*-algebra. –  André Henriques Jul 20 '11 at 19:55
@Jonas, @André: Thank you for the comment about general C*-algebra! –  Tsuyoshi Ito Jul 20 '11 at 19:56
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! –  Federico Poloni Jul 20 '11 at 20:04
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. –  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 $a$).

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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. –  Martin Brandenburg Jul 21 '11 at 11:58
@ 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! –  Mahmood Alaghmandan Jul 21 '11 at 23:25