Typo fixes.

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 \\
+        \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(T) + q(T)^*]\dpn{\xi, \eta}{t}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 $\partial B$. 
+        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 $(z_1, \cdots, z_n) \in B$, if
+        \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}