From 60544ea6a0974901b6f97cad0c05d58097e421c0 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 30 Jun 2026 16:12:33 -0400 Subject: [PATCH] Added some spectrum gruntwork. --- src/op/banach/spectrum.tex | 4 +- src/op/c-star/index.tex | 1 + src/op/c-star/sa.tex | 85 ++++++++++++++++++++++++++++++++++++++ src/op/c-star/unitary.tex | 55 ++++++++++++++++++++++++ 4 files changed, 143 insertions(+), 2 deletions(-) create mode 100644 src/op/c-star/unitary.tex diff --git a/src/op/banach/spectrum.tex b/src/op/banach/spectrum.tex index f5a9230..f47fcc8 100644 --- a/src/op/banach/spectrum.tex +++ b/src/op/banach/spectrum.tex @@ -77,7 +77,7 @@ Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. \end{proof} -\begin{proposition}[Spectral Radius Formula] +\begin{proposition}[Beurling's Spectral Radius Formula] \label{proposition:spectral-radius-hadamard} Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. \end{proposition} @@ -128,7 +128,7 @@ \begin{proposition} \label{proposition:commutative-spectrum-gymnastics} - Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then + Let $A$ be a commutative unital Banach algebra and $x, y \in A$ with $x = y$, then \begin{enumerate} \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$. \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$. diff --git a/src/op/c-star/index.tex b/src/op/c-star/index.tex index bd63399..9586e09 100644 --- a/src/op/c-star/index.tex +++ b/src/op/c-star/index.tex @@ -2,5 +2,6 @@ \label{chap:c-star-algebras} \input{./involution.tex} +\input{./unitary.tex} \input{./sa.tex} \input{./order.tex} diff --git a/src/op/c-star/sa.tex b/src/op/c-star/sa.tex index 7bb1478..f1e8084 100644 --- a/src/op/c-star/sa.tex +++ b/src/op/c-star/sa.tex @@ -31,6 +31,91 @@ If the above holds, then $x$ is \textbf{normal}. \end{definition} +\begin{theorem} +\label{theorem:c-star-normal-spectral-radius} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $\norm{x}_A = [x]_{sp}$. +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem II.8.1]{Zhu}}}. ] + 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 \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard}, + \[ + [x]_{sp} = \limsup_{n \to \infty}\norm{x^{n}}_A^{1/n} \ge \limsup_{n \to \infty}\normn{x^{2^n}}_A^{1/2^n} = \norm{x}_A + \] + + Now suppose that $x$ is only normal. Since $x$ and $x^*$ commute, $[xx^*]_{sp} \le [x]_{sp}[x^*]_{sp}$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus + \[ + \norm{x}^2_A = \normn{xx^*}_A = [xx^*]_{sp} \le [x]_{sp}[x^*]_{sp} = [x]_{sp}^2 + \] + + by (5) of \autoref{proposition:c-star-algebra-gymnastics}. +\end{proof} + +\begin{corollary} +\label{corollary:c-star-normal-spectral-radius-corollary} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then: + \begin{enumerate} + \item There exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = \norm{x}_A$. + \item If there exists $n \in \natp$ such that $x^n = 0$, then $x = 0$ as well. + \end{enumerate} +\end{corollary} +\begin{proof} + (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}, $|\lambda| = [x]_{sp} = \norm{x}_A$. + + (2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, + \[ + \bracs{\lambda^n| \lambda \in \sigma_A(x)} = \bracs{0} + \] + + Thus $\sigma_A(x) = \bracs{0}$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{x}_A = [x]_{sp} = 0$. +\end{proof} + + +\begin{corollary} +\label{corollary:c-star-unique-norm} + 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. +\end{corollary} +\begin{proof} + Since $x^*x$ is self-adjoint, \autoref{theorem:c-star-normal-spectral-radius} 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 $A$. +\end{proof} + +\begin{proposition} +\label{proposition:self-adjoint-spectrum} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then $\sigma_A(x) \subset \real$. +\end{proposition} +\begin{proof} + Let + \[ + y = \exp(ix) = \sum_{n = 0}^\infty \frac{i^nx^n}{n!} + \] + + then + \[ + y^*= \sum_{n = 0}^\infty \frac{(-i)^n (x^*)^n}{n!} = \exp(-ix^*) + \] + + Since $x$ is normal, $y$ is also normal. By \autoref{proposition:functional-calculus-exp},, + \[ + y^*y = \exp(-ix^* + ix) = \exp(-ix + ix) = 1 + \] + + 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 $\sigma_A(x) \subset \real$. +\end{proof} + diff --git a/src/op/c-star/unitary.tex b/src/op/c-star/unitary.tex new file mode 100644 index 0000000..ab65824 --- /dev/null +++ b/src/op/c-star/unitary.tex @@ -0,0 +1,55 @@ +\section{Unitary Elements} +\label{section:unitary-c-star} + +\begin{definition}[Unitary] +\label{definition:unitary-element} + Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$. +\end{definition} + +\begin{lemma} +\label{lemma:unitary-unit} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$. +\end{lemma} +\begin{proof} + $\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$. +\end{proof} + +\begin{definition}[Unitarily Equivalent] +\label{definition:unitary-equivalent} + Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$. +\end{definition} + +\begin{lemma} +\label{lemma:unitary-equivalent-same-stuff} + Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then: + \begin{enumerate} + \item $\norm{x}_A = \norm{y}_A$. + \item $\sigma_A(x) = \sigma_A(y)$. + \end{enumerate} +\end{lemma} +\begin{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 \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, 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 $\sigma_A(x) = \sigma_A(y)$. +\end{proof} + +\begin{proposition} +\label{proposition:unitary-spectrum} + Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$. +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition II.8.2]{Zhu}}}. ] + By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus + \[ + \bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)} + \] + + by (4) of \autoref{proposition:c-star-algebra-gymnastics}. 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 $\sigma_A(x) \subset \partial B_\complex(0, 1)$. +\end{proof} + + +