$a^*a$ has a non-negative spectrum

Question

I am learning $C^*$-algebra, especially I work on the proof of the Gelfand-Naimark theorem. In many books such as the one of Arveson, it looks that the following lemma is the key stone of the proof:

Lemma : if $a\in \mathcal{A}$, then $a^*a$ has a non-negative spectrum.

And the proof use strongly the commutative version of the Gelfand-Naimark theorem.

But, let $\mathcal{B}$ be the closed sub-algebra generated by $a^*a$ and $e$. Since $\mathcal{B}$ is commutative $\sigma_{\mathcal{B}}(a^*a)=\{\phi(a^*a)\, \vert\, \phi\in \mathcal{M}_\mathcal{B}\}$ where $\mathcal{M}_\mathcal{B}$ is the space of characters of $\mathcal{B}$. But since $\phi(a^*)=\overline{\phi(a)}$ the spectrum is clearly include in $\mathbb{R}^+$. Finally we conclude using the spectral invariance.

Is this proof is correct, or I miss something important?

Answer 1

The proof is not correct because $a\notin B$ and you cannot write $\phi(a^*a)=|\phi(a)|^2.$

Unfortunately I don't have the book by Arveson. A proof that $a^*a$ is positive is not trivial and can be found e.g. in the book by Davidson, Theorem I.4.5.

Answer 2

If by "spectral invariance" you mean $\sigma_A(a^{\ast}a) \subset \sigma_B(a^{\ast}a)$, then yes, this proof is correct