From 5f19f6bea3f80c3f15f00bb4bee59dce83190199 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 19 Oct 2026 16:10:53 -0400 Subject: [PATCH] Typo fixes. --- src/op/system/dilation.tex | 42 +++++++++++++++++--------------------- 1 file changed, 19 insertions(+), 23 deletions(-) diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex index a00fa91..fe8766a 100644 --- a/src/op/system/dilation.tex +++ b/src/op/system/dilation.tex @@ -44,7 +44,7 @@ \label{lemma:bh-dilation-transitive} Let $H$ be a complex Hilbert space, $R \in B(H)$, $(K, S)$ be a dilation of $R$, and $(L, T)$ be a dilation of $S$, then $(L, T)$ is a dilation of $R$. - Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. + Similarly, if $(K, S)$ is a power dilation of $R$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. \end{lemma} @@ -128,7 +128,7 @@ \end{bmatrix} \] - Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. + Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, let $Q \in B(H^2)$ be the projection onto the first summand, then $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. \end{proof} \begin{theorem}[Von Neumann's Inequality] @@ -165,7 +165,7 @@ \begin{align*} \int_\gamma \frac{\ol{\lambda^n}}{\lambda - T} d\lambda &= \int_\gamma \frac{\lambda^{-n}}{\lambda - T} d\lambda = \int_\gamma \frac{\lambda^{-{(n+1)}}}{1 - T/\lambda} d\lambda \\ &= \sum_{k = 0}^\infty T^k\int_{\gamma} \lambda^{-(n+k+1)}d\lambda = \begin{cases} - 1 & n = 0 \\ + 2\pi i & n = 0 \\ 0 & n > 0 \end{cases} \end{align*} @@ -189,7 +189,7 @@ Therefore \begin{align*} - p(T) + q(T)^* + p(0) + \ol{q(0)} &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}](P(T; t) - 1)dt \\ + p(T) + q(T)^* + p(0) + \ol{q(0)} &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}](P(T; t) + 1)dt \\ p(T) + q(T)^* &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}]P(T; t)dt \end{align*} @@ -206,13 +206,13 @@ Let $M = \sup_{z \in \partial B_\complex(0, 1)}|p(z) + \ol{q(z)}|$. For each $\xi, \eta \in H$, let $\dpn{\xi, \eta}{t} = \dpn{P(T; t)\xi, \eta}{H}$, then $\dpn{\cdot, \cdot}{t}$ is a pseudo inner product on $H$. Hence by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, \begin{align*} - \dpn{[p(T) + q(T)^*]\xi, \eta}{H} &= \frac{1}{2\pi}\int_0^{2\pi} [p(T) + q(T)^*]\dpn{P(T; t)\xi, \eta}{H}dt \\ - &= \frac{1}{2\pi}\int_0^{2\pi} [p(T) + q(T)^*]\dpn{\xi, \eta}{t}dt \\ + \dpn{[p(T) + q(T)^*]\xi, \eta}{H} &= \frac{1}{2\pi}\int_0^{2\pi} [p(e^{it}) + \ol{q(e^{it})}]\dpn{P(T; t)\xi, \eta}{H}dt \\ + &= \frac{1}{2\pi}\int_0^{2\pi} [p(e^{it}) + \ol{q(e^{it})}]\dpn{\xi, \eta}{t}dt \\ |\dpn{[p(T) + q(T)^*]\xi, \eta}{H}| &\le \frac{1}{2\pi}\int_0^{2\pi} M\dpn{\xi, \xi}{t}^{1/2}\dpn{\eta, \eta}t^{1/2}dt\\ &\le M\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt}^{1/2}\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\eta, \eta}{t} dt}^{1/2} \end{align*} - Since $I_H(\xi) = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$, + Since $I_H = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$, \[ \frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt = \frac{1}{2\pi}\int_0^{2\pi} \dpn{P(T; t)\xi, \xi}{H}dt = \norm{\xi}_H^2 \] @@ -228,6 +228,7 @@ \begin{enumerate}[label=(\alph*)] \item $U, V$ are unitary. \item $S = \text{diag}(s_1, \cdots, s_n)$, where $\seqf{s_j} \subset [0, 1]$. + \item $T = USV$. \end{enumerate} Let $B = \ol{B_\complex(0, 1)}^n \subset \complex^n$ and @@ -242,8 +243,8 @@ f_{\xi, \eta}: B \to \complex \quad (z_1, \cdots, z_n) \mapsto \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n} \] - then $f_{\xi, \eta}$ achieves its maximum modulus on $\partial B$. - \item For any $(z_1, \cdots, z_n) \in B$, if + then $f_{\xi, \eta}$ achieves its maximum modulus on a point $(z_1, \cdots, z_n) \in \complex^n$ such that $|z_j| = 1$ for all $1 \le j \le n$. + \item For any $(z_1, \cdots, z_n) \in \complex$, if \[ \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)} \] @@ -252,10 +253,6 @@ \item $\norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)}$. \item $\norm{p(T)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}$. \end{enumerate} - - - For each $\seqf{z} \in \overline{B_{\complex^n}(0, 1)}$ - \end{theorem} \begin{proof}[Proof, {{\cite[Exercise 2.16]{Paulsen}}}, attributed to Wermer. ] (1): By the Maximum Modulus Principle. @@ -270,7 +267,7 @@ (3): Since $(s_1, \cdots, s_n) \in B$, (2) implies that \begin{align*} \norm{p(T)}_{M_n(\complex)} &\le \max_{z \in B}\norm{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} \\ - &\le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)} + &\le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)} \end{align*} (4): For any $W \in U(M_n(\complex))$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that @@ -280,7 +277,7 @@ As the above holds for all unitary matrices, \[ - \norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(U(W))}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)} + \norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)} \] by (3). @@ -300,7 +297,7 @@ such that: \begin{enumerate} \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. - \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. + \item For all $a \in A$, $\phi(a) = V^*\pi(a)V$. \end{enumerate} Moreover, if $\phi$ is unital, then: @@ -312,7 +309,7 @@ The triple $(\pi, V, K)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(\pi, V, K)$ is \textbf{minimal}. \end{theorem} \begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ] - (a): Let $A \otimes H$ and define the Hermitian form + (a): Let $A \otimes H$ be the algebraic tensor product and define the Hermitian form \[ \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H} \] @@ -331,10 +328,10 @@ Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, \[ - \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0 + |\dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0 \] - so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. + so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}$. Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. @@ -352,7 +349,7 @@ then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, \begin{align*} - \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, b_i \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\ + \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, (ab_i) \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\ &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2 \end{align*} @@ -411,7 +408,7 @@ \end{enumerate} \end{proposition} \begin{proof} - (1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = V^* \in B(K)$ be the orthogonal projection onto $H$. + (1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = VV^* \in B(K)$ be the orthogonal projection onto $H$. Let $a \in A$ and $\xi, \eta \in H$, then \begin{align*} @@ -419,7 +416,6 @@ \dpn{\phi(a^*a)\xi, \eta}{H} &= \dpn{\pi(a^*a)V\xi, V\eta}{K} = \dpn{\pi(a)^*\pi(a)P\xi, P\eta}{K} \end{align*} - By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*P\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$. - + By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$. \end{proof}