Corollary 34.4.4.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then:
- (1)
There exists $\lambda \in \sigma_{A}(x)$ such that $|\lambda| = \norm{x}_{A}$.
- (2)
If there exists $n \in \natp$ such that $x^{n} = 0$, then $x = 0$ as well.
Proof. (1): Since $\sigma_{A}(x)$ is compact, there exisst $\lambda \in \sigma_{A}(x)$ such that $|\lambda| = [x]_{sp}$. By Theorem 34.4.3, $|\lambda| = [x]_{sp}= \norm{x}_{A}$.
(2): By the Spectral Mapping Theorem,
\[\bracs{\lambda^n| \lambda \in \sigma_A(x)}= \bracs{0}\]
Thus $\sigma_{A}(x) = \bracs{0}$. By Theorem 34.4.3, $\norm{x}_{A} = [x]_{sp}= 0$.$\square$
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