122 lines
4.6 KiB
TeX
122 lines
4.6 KiB
TeX
\section{Self-Adjoint Elements}
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\label{section:c-star-self-adjoint}
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\begin{definition}[Self-Adjoint]
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\label{definition:self-adjoint}
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Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then $x$ is \textbf{self-adjoint} if $x = x^*$. The space $A_{sa} = \bracs{x \in A| x = x^*}$ is the \textbf{self-adjoint part} of $A$, and:
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\begin{enumerate}
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\item $A_{sa}$ is a $\real$ subspace of $A$.
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\item $A = \complex(A_{sa})$ as a vector space.
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\item For each $x \in A$, let
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\[
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\text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
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\]
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then $\text{Re}(x), \text{Im}(x) \in A_{sa}^2$ and $x = \text{Re}(x) + i\text{Im}(x)$.
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\item For each $x \in A$, $x^* = \text{Re}(x) - i\text{Im}(x)$.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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By \autoref{proposition:complex-conjugation-properties}.
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\end{proof}
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\begin{definition}[Normal]
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\label{definition:c-star-normal}
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Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then the following are equivalent:
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\begin{enumerate}
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\item $\text{Re}(x)\text{Im}(x) = \text{Im}(x)\text{Re}(x)$.
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\item $x^*x = xx^*$.
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\end{enumerate}
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If the above holds, then $x$ is \textbf{normal}.
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\end{definition}
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\begin{theorem}
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\label{theorem:c-star-normal-spectral-radius}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $\norm{x}_A = [x]_{sp}$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem II.8.1]{Zhu}}}. ]
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First suppose that $x$ is self-adjoint. In this case,
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\begin{align*}
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\normn{x^2}_A &= \normn{xx^*}_A = \norm{x}_A^2 \\
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\normn{x^{2^n}}_A &= \norm{x}_A^{2^n}
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\end{align*}
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for all $n \in \natp$. Thus by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard},
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\[
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[x]_{sp} = \limsup_{n \to \infty}\norm{x^{n}}_A^{1/n} \ge \limsup_{n \to \infty}\normn{x^{2^n}}_A^{1/2^n} = \norm{x}_A
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\]
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Now suppose that $x$ is only normal. Since $x$ and $x^*$ commute, $[xx^*]_{sp} \le [x]_{sp}[x^*]_{sp}$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus
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\[
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\norm{x}^2_A = \normn{xx^*}_A = [xx^*]_{sp} \le [x]_{sp}[x^*]_{sp} = [x]_{sp}^2
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\]
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by (5) of \autoref{proposition:c-star-algebra-gymnastics}.
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\end{proof}
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\begin{corollary}
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\label{corollary:c-star-normal-spectral-radius-corollary}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
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\begin{enumerate}
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\item There exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = \norm{x}_A$.
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\item If there exists $n \in \natp$ such that $x^n = 0$, then $x = 0$ as well.
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\end{enumerate}
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\end{corollary}
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\begin{proof}
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(1): Since $\sigma_A(x)$ is compact, there exisst $\lambda \in \sigma_A(x)$ such that $|\lambda| = [x]_{sp}$. By \autoref{theorem:c-star-normal-spectral-radius}, $|\lambda| = [x]_{sp} = \norm{x}_A$.
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(2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic},
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\[
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\bracs{\lambda^n| \lambda \in \sigma_A(x)} = \bracs{0}
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\]
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Thus $\sigma_A(x) = \bracs{0}$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{x}_A = [x]_{sp} = 0$.
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\end{proof}
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\begin{corollary}
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\label{corollary:c-star-unique-norm}
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Let $A$ be a unital $C^*$-algebra over $\complex$, then for each $x \in A$,
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\[
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\norm{x}_A^2 = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
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\]
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In particular, there exists at most one norm on $A$ making it a $C^*$-algebra.
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\end{corollary}
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\begin{proof}
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Since $x^*x$ is self-adjoint, \autoref{theorem:c-star-normal-spectral-radius} implies that
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\[
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\norm{x}_A^2 = \norm{x^*x}_A = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
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\]
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which depends only on the algebraic structure of $A$.
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\end{proof}
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\begin{proposition}
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\label{proposition:self-adjoint-spectrum}
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Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then $\sigma_A(x) \subset \real$.
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\end{proposition}
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\begin{proof}
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Let
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\[
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y = \exp(ix) = \sum_{n = 0}^\infty \frac{i^nx^n}{n!}
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\]
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then
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\[
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y^*= \sum_{n = 0}^\infty \frac{(-i)^n (x^*)^n}{n!} = \exp(-ix^*)
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\]
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Since $x$ is normal, $y$ is also normal. By \autoref{proposition:functional-calculus-exp},,
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\[
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y^*y = \exp(-ix^* + ix) = \exp(-ix + ix) = 1
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\]
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so $y$ is unitary. By \autoref{proposition:unitary-spectrum} and the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, $\exp(i\sigma_A(x)) = \sigma_A(y) \subset \partial B_\complex(0, 1)$. Thus $i\sigma_A(x) \subset \bracs{\text{Re} = 0}$, and $\sigma_A(x) \subset \real$.
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\end{proof}
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