1
$\begingroup$

Let's say that in a unital C* algebra, we have $b \geq a \geq 0$ and $a$ is invertible. Then $b$ is also invertible. Can we conclude that $a^{-1} \geq b^{-1}$? If so, why? Can any related statement be made if we only assume $a \leq b$? (And maybe that $a$, and hence $b$ are self adjoint.)

  • 0
    This is an abstract version of this earlier question: http://math.stackexchange.com/questions/95862/positive-invertible-operators (Essentially the same answer applies. And it was applied by the same answerer.)2012-12-15

1 Answers 1

3

Yes. If $b \ge a \ge 0$ and $a$ is invertible, then $a^{-1/2} b a^{-1/2} \ge a^{-1/2} a a^{-1/2} = 1$ so $ a^{1/2} b^{-1} a^{1/2} = (a^{-1/2} b a^{-1/2})^{-1} \le 1$ and then $b^{-1} = a^{-1/2} a^{1/2} b^{-1} a^{1/2} a^{-1/2} \le a^{-1}$. Without the $\ge 0$ it's false, e.g. try $a = -1$ and $b = 1$.

  • 0
    @Jeff: You can also show that directly, just using the spectrum, if you defined $b ≥ a$ by $\mathrm{spec}(b-a) \subseteq [0, ∞)$, which I believe is how it is commonly done for C*-algebras.2015-07-03