I have a question about a passage in the book C*-algebras and Operator Theory by Murphy, in the proof of Theorem 2.1.8.
Let $a$ be a hermitian element of a unital C*-algebra and let $\lambda\in\sigma(a)$, and let $b=\sum_{n=1}^\infty i^n(a-\lambda)^{n-1}/n!$, then $e^{ia}-e^{i\lambda}=(a-\lambda)be^{i\lambda}$. Then he says: Since $b$ commutes with $a$, and since $a-\lambda$ is non-invertible, $e^{ia}-e^{i\lambda}$ is non-invertible. I don't see how the last statement follows.
I tried to proof the invertibility of $a-\lambda$ assuming $(a-\lambda)be^{ia}$ is invertible, but I couldn't. Can you help me? I don't see how the fact that $a$ and $b$ commute are of any help, since that still doesn't imply $a$ commutes with the inverse of $(a-\lambda)be^{i\lambda}$.
Note that $a-\lambda$ commutes with $be^{i\lambda}$. In any ring, if $x$ and $y$ commute and the product is invertible, then each is invertible.
Proof: suppose that $zxy=xyz=I$. Then $zyx=zxy-I$, so $x$ has a left inverse. From $xyz=I$ we know that $x$ has a right inverse. Then $x$ is invertible.