$\operatorname{Sp}(a^{*}) = \operatorname{Sp}(a)^{*}$ and the spectrum of a positive element Suppose $\mathcal{A}$ is a $C^{*}$-algebra. Let $a \in \mathcal{A}$ be fixed and define the spectrum:
$$\operatorname{Sp}(a) := \{z \in \mathbb{C}: \mbox{$a - z$ is not invertible}\}$$
and:
$$\operatorname{Sp}(a)^{*} := \{z^{*} \in \mathbb{C}: z \in \operatorname{Sp}(a)\}$$
I have to prove the following properties:

*

*$\operatorname{Sp}(a^{*}) = \operatorname{Sp}(a)^{*}$


*If $a$ is positive, that is, if $a = bb^{*}$ for some $b \in \mathcal{A}$, then $\operatorname{Sp}(a)$ is the positive real axis.
For property 1. this is my reasoning. First, if $a$ is invertible then $a^{*}$ is invertible and $(a^{*})^{-1} = (a^{-1})^{*}$. Changing the roles of $a$ and $a^{*}$, we get that if $a^{*}$ is invertible, so is $a$. Hence, $a$ is invertible iff $a^{*}$ is invertible. Thus, we get:
$$z \in \operatorname{Sp}(a^{*}) \iff \quad \mbox{$a^{*}-z$ is not invertible} \quad \iff \mbox{$a-z^{*}$ is not invertible} \quad \iff z^{*} \in \operatorname{Sp}(a)$$
I have no idea how to prove property 2. Is my proof of property 1. correct? Any help on property 2.?
 A: The first part of the argument is fine.
For the second, I'm not sure if there is an entirely elementary approach (I'll be really happy if I'm shown one!). Usually one defines things the other way around, defining positive as "selfadjoint with nonnegative spectrum" and then proving that any positive is of the form $x^*x$.
As mentioned by Matsmir in the comments, Murphy offers a decently simple argument. There is theory hidden in it, though, when he uses that a selfadjoint is a difference of positive elements; this requires functional calculus.
And if one is going to use functional calculus, here is a fairly conceptual argument. If $a=bb^*$, then $a^*=(bb^*)^*=b^{**}b^*=bb^*=a$. So $C^*(a)$ is an abelian C$^*$-algebra, isomorphic to $C(\operatorname{Sp}(a))$ via a $*$-isomorphism that maps $a$ to the identity function; because of this and the fact that $a=a^*$, we get $\operatorname{Sp}(a)\subset\mathbb R$.
Looking at the point-evaluations in $C(\operatorname{Sp}(a))$, we find for each $\lambda\in\operatorname{Sp}(a)$ a state $\varphi_\lambda$ such that $\varphi_\lambda^\vphantom\lambda(a)=\lambda$ (recall that $a$ corresponds to the identity function on  $C(\operatorname{Sp}(a))$, and $\varphi_\lambda^\vphantom\lambda$ is the map $\varphi_\lambda^\vphantom\lambda (f)=f(\lambda)$). By abuse of notation, I'm omitting to write the isomorphism $C^*(a)\simeq C(\operatorname{Sp}(a))$, and use the same notation for corresponding states on both sides.
Using the characterization of states as those linear functionals $\varphi$ such that $\|\varphi\|=\varphi(1)$, we can extend the states $\varphi_\lambda^\vphantom\lambda$ to all of $\mathcal A$. Then, using Cauchy-Schwarz, for any $\lambda\in\operatorname{Sp}(a)$ we have
$$
\lambda=\varphi_\lambda^\vphantom\lambda(a)=\varphi_\lambda^\vphantom\lambda(bb^*)\geq|\varphi_\lambda^\vphantom\lambda(b)|^2\geq0.
$$
Finally, the spectrum can never be $[0,\infty)$. The spectrum of any element in a Banach algebra is compact. For different examples of C$^*$-algebras $A$, any compact subset of $[0,\infty)$ can appear as the spectrum of a positive element.
