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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.

edited Jul 20 2011 at 19:56

asked Jul 20 2011 at 19:10

André Henriques
14.9k●2●30●83

André Henriques · Accepted Answer · 2011-07-21 12:58:19Z UTC

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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 a^1/2ba^1/2 = a^1/2xaxa^1/2 = (a^1/2xa^1/2)², which implies that a^1/2xa^1/2 = (a^1/2ba^1/2)^1/2. Since a is invertible, x must be a^−1/2(a^1/2ba^1/2)^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.

edited Jul 21 2011 at 12:58

André Henriques

answered Jul 20 2011 at 19:46

Tsuyoshi Ito
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This does work for arbitrary C* algebras. – Jonas Meyer Jul 20 2011 at 19:53

Thank you. You proof works perfectly well in the context of a general C*-algebra. – André Henriques Jul 20 2011 at 19:55

@Jonas, @André: Thank you for the comment about general C*-algebra! – Tsuyoshi Ito Jul 20 2011 at 19:56

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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 2011 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 2011 at 20:14

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Mahmood Alaghmandan · Answer 2 · 2011-07-21 03:04:07Z UTC

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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 $a$).

edited Jul 21 2011 at 3:04

answered Jul 20 2011 at 23:03

Mahmood Alaghmandan
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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 2011 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 2011 at 23:25

Solving the equation $xax=b$ in a C*-algebra.

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