Lemma 34.2.4.label Let $A$ be a unital $C^{*}$-algebra and $x, y \in A$ be unitarily equivalent, then:
- (1)
$\norm{x}_{A} = \norm{y}_{A}$.
- (2)
$\sigma_{A}(x) = \sigma_{A}(y)$.
Proof. (1): Let $u \in A$ be unitary such that $x = uyu^{*}$, then $\norm{x}_{A} \le \norm{u}_{A}\norm{x}_{A}\normn{u^*}_{A}$. By Lemma 34.2.2 and (1) of Proposition 34.1.3, $\norm{u}_{A} = \normn{u^*}_{A} = 1$, so $\norm{x}_{A} \le \norm{y}_{A}$. As the argument is symmetric, $\norm{x}_{A} = \norm{y}_{A}$.
(2): Let $\lambda \in \complex$, then
\[u(y - \lambda)u^{*} = uyu^{*} - \lambda uu^{*} = uyu^{*} - \lambda = x - \lambda\]
Since $u, u^{*} \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_{A}(x) = \sigma_{A}(y)$.$\square$
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