Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 12 down vote accepted

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.

share|improve this answer
4  
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
4  
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 $a$).

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.