34.2 Unitary Elements
Definition 34.2.1 (Unitary).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$, then $x$ is unitary if $x \in G(A)$ and $x^{*} = x^{-1}$.
Lemma 34.2.2.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be unitary, then $\norm{x}_{A} = 1$.
Proof. $\normn{x^2}_{A} = \norm{xx^*}_{A} = \norm{1}_{A} = 1$.$\square$
Definition 34.2.3 (Unitarily Equivalent).label Let $A$ be a unital $C^{*}$-algebra and $x, y \in A$, then $x$ and $y$ are unitarily equivalent if there exists a unitary element $u \in A$ such that $x = uyu^{*}$.
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
Since $u, u^{*} \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_{A}(x) = \sigma_{A}(y)$.$\square$
Proposition 34.2.5.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be unitary, then $\sigma_{A}(x) \subset \partial B_{\complex}(0, 1)$.
Proof, [Proposition II.8.2, Zhu93]. By Lemma 34.2.2, $\norm{x}_{A} = 1$, so $\sigma_{A}(x) \subset \ol{B_\complex(0, 1)}$. Thus
by (4) of Proposition 34.1.3. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_{A}(x) \subset \partial B_{\complex}(0, 1)$.$\square$
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