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/Part 6: Operator Algebras/Chapter 34: $C^{*}$-Algebras

34.4 Self-Adjoint Elements

Definition 34.4.1 (Self-Adjoint).label Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then $x$ is self-adjoint if $x = x^{*}$. The space $A_{sa}= \bracs{x \in A| x = x^*}$ is the self-adjoint part of $A$, and:

  1. (1)

    $A_{sa}$ is a $\real$ subspace of $A$.

  2. (2)

    $A = \complex(A_{sa})$, with equivalent norms.

  3. (3)

    For each $x \in A$, let

    \[\text{Re}(x) = \frac{x + x^{*}}{2}\quad \text{Im}(x) = \frac{x - x^{*}}{2i}\]

    then $\text{Re}(x), \text{Im}(x) \in A_{sa}^{2}$ and $x = \text{Re}(x) + i\text{Im}(x)$.

  4. (4)

    For each $x \in A$, $x^{*} = \text{Re}(x) - i\text{Im}(x)$.

Proof. By Proposition 11.2.3.$\square$

Definition 34.4.2 (Normal).label Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then the following are equivalent:

  1. (1)

    $\text{Re}(x)\text{Im}(x) = \text{Im}(x)\text{Re}(x)$.

  2. (2)

    $x^{*}x = xx^{*}$.

If the above holds, then $x$ is normal.

Theorem 34.4.3.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then $\norm{x}_{A} = [x]_{sp}$.

Proof, [Theorem 8.1, Zhu93]. First suppose that $x$ is self-adjoint. In this case,

\begin{align*}\normn{x^2}_{A}&= \normn{xx^*}_{A} = \norm{x}_{A}^{2} \\ \normn{x^{2^n}}_{A}&= \norm{x}_{A}^{2^n}\end{align*}

for all $n \in \natp$. Thus by the spectral radius formula,

\[[x]_{sp}= \limsup_{n \to \infty}\norm{x^{n}}_{A}^{1/n}\ge \limsup_{n \to \infty}\normn{x^{2^n}}_{A}^{1/2^n}= \norm{x}_{A}\]

Now suppose that $x$ is only normal. Since $x$ and $x^{*}$ commute, $[xx^{*}]_{sp}\le [x]_{sp}[x^{*}]_{sp}$ by Proposition 33.5.11. Thus

\[\norm{x}^{2}_{A} = \normn{xx^*}_{A} = [xx^{*}]_{sp}\le [x]_{sp}[x^{*}]_{sp}= [x]_{sp}^{2}\]

by (5) of Proposition 34.1.3.$\square$

Corollary 34.4.4.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then:

  1. (1)

    There exists $\lambda \in \sigma_{A}(x)$ such that $|\lambda| = \norm{x}_{A}$.

  2. (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 exists $\lambda \in \sigma_{A}(x)$ 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, $\norm{x}_{A} = [x]_{sp}= 0$.$\square$

Corollary 34.4.5.label Let $A$ be a unital $C^{*}$-algebra over $\complex$, then for each $x \in A$,

\[\norm{x}_{A}^{2} = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}\]

In particular, there exists at most one norm on $A$ making it a $C^{*}$-algebra.

Proof. Since $x^{*}x$ is self-adjoint, Theorem 34.4.3 implies that

\[\norm{x}_{A}^{2} = \norm{x^*x}_{A} = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}\]

which depends only on the algebraic structure of $A$.$\square$

Proposition 34.4.6.label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be self-adjoint, then $\sigma_{A}(x) \subset \real$.

Proof. Let

\[y = \exp(ix) = \sum_{n = 0}^{\infty} \frac{i^{n}x^{n}}{n!}\]

then

\[y^{*}= \sum_{n = 0}^{\infty} \frac{(-i)^{n} (x^{*})^{n}}{n!}= \exp(-ix^{*})\]

Since $x$ is normal, $y$ is also normal. By Proposition 33.6.3,,

\[y^{*}y = \exp(-ix^{*} + ix) = \exp(-ix + ix) = 1\]

so $y$ is unitary. By Proposition 34.3.5 and the Spectral Mapping Theorem, $\exp(i\sigma_{A}(x)) = \sigma_{A}(y) \subset \partial B_{\complex}(0, 1)$. Thus $i\sigma_{A}(x) \subset \bracs{\text{Re} = 0}$, and $\sigma_{A}(x) \subset \real$.$\square$

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