Self-adjoint element of $C^*$-algebra has real spectrum I'm trying to understand the proof in $\S$3.9 on p.23 of these notes: http://strung.me/karen/CStarIntroDraft.pdf. The argument is as follows:
Let $A$ be a unital $C^*$-algebra, let $a \in A$ be self-adjoint ($a = a^*$), and let $\lambda \in \sigma(a)$. We want to show that $\lambda \in \mathbb{R}$, by showing that $e^{i\lambda} \in \sigma(e^{ia})$, where $e^{ia}$ has the usual power series definition; since $e^{ia}$ is unitary, any $\mu \in \sigma(e^{ia})$ has $|\mu| = 1$. I understand things so far.
We now write
$$ e^{ia} - e^{i\lambda} = e^{i\lambda}( e^{i(a - \lambda)} - 1) = e^{i\lambda}(a-\lambda)b  $$
where $b = \sum_{n=1}^{\infty} i^n(a-\lambda)^{n-1}/n!$ (in the notes, the sum starts from $n = 2$, but I think this is a typo). Then $b$ commutes with $a-\lambda$, so $e^{ia} - e^{i\lambda}$ cannot be invertible, since $a-\lambda$ is not.
My understanding is that the implication $(xy = yx \text{ and } (xy)^{-1} \text{ exists}) \implies (x^{-1} \text{ and } y^{-1} \text{ exist})$ is valid in a ring without zero divisors, but I don't see how it's valid here.
 A: The implication $(xy = yx \text{ and } (xy)^{-1} \text{ exists}) \implies (x^{-1} \text{ and } y^{-1} \text{ exist})$ is true in any ring (with unity).
Indeed, you have 
$$x  (y  (xy)^{-1}) =1$$
It is a bit trickier, but we can also show that
$$(y  (xy)^{-1})x =1$$ 
Indeed, let
$$t:= (y  (xy)^{-1})x$$
Then 
$$ty=(y  (xy)^{-1})xy=y \\
(t-1)y=0\\
(t-1)yx=0 \\
(t-1)xy=0\\
(t-1)xy(xy)^{-1}=0\\
t-1=0
$$ 
Note I am not sure but I think that the claim is not true if you don't know that $x,y$ commute, there is probably where you need the no zero divisors.
A: There is a slightly simpler argument: if $z:=xy$ is invertible, then $x$ has a right inverse, and $y$ has a left inverse, namely $z^{-1}$. And similarly, if $w:=yx$ is invertible, then $y$ has a right inverse and $x$ has a left inverse, namely $w^{-1}$. Now use that an element is invertible iff it is left and right invertible. This argument shows that it is enough to assume $xy$ and $yx$ to be invertible to ensure that $x$ and $y$ are invertible, they need not commute.
I can happen, e.g., that $xy=1$ and $yx=p$ is a not trivial projection, so not invertible. For instance, take $x$ to be the unilateral shift operator acting on $\ell^2(\mathbb{N})$ and $y$ is adjoint operator.
