Draft of today's classes.

src/op/system/complete-upgrade.tex

@@ -242,3 +242,25 @@ The following is a slight generalisation of one form of the above example, but s
     which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive. 
 \end{proof}
 
+\begin{proposition}[Choi-Kraus]
+\label{proposition:choi-kraus}
+    Let $\phi: M_n(\complex) \to M_m(\complex)$ be a completely positive map, then there exists $\bracsn{V_j}_1^{mn} \subset L(\complex^m; \complex^n)$ such that
+    \[
+    \phi(a) = \sum_{j = 1}^{nk}V_j^* a V_j
+    \]
+
+    for all $a \in M_n(\complex)$. 
+\end{proposition}
+\begin{proof}
+    Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\phi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \phi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$. 
+
+    As $M_n(\complex)$ is simple, the only (non-zero) irreducible representation of $M_n(\complex)$ is the identity. Since $(K, \pi)$ is a direct sum of irreducible representations, after appending additional factors, assume without loss of generality that $K = \bigoplus_{k = 1}^{nm}\complex^n$ and that $\pi(a) = I_{M_{mn}(\complex)} \otimes a$ for all $a \in M_n(\complex)$.
+
+    For each $1 \le k \le mn$, let $P_k \in B(K)$ be the orthogonal projection onto the subspace corresponding to the $k$-th summand, and $Q_k \in L(K; \complex^n)$ be the orthogonal projection onto the $k$-th summand itself, then
+    \begin{align*}
+    \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_j\pi(a)P_j)V = \sum_{k = 1}^{mn}V^*Q_j^*aQ_jV \\
+    &= \sum_{k = 1}^{mn} (Q_jV)^* a (Q_jV)
+    \end{align*}
+    
+\end{proof}
+

src/op/system/dilation.tex

@@ -44,16 +44,179 @@
 \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}
 
 
+\begin{theorem}[Stinespring's Dilation Theorem]
+\label{theorem:stinespring-dilation}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, where:
+    \begin{enumerate}[label=(\alph*)]
+        \item $K$ is a complex Hilbert space. 
+        \item $V: H \to K$ is a continuous linear map.
+        \item $\pi: A \to B(K)$ is a unital *-homomorphism.
+    \end{enumerate}
+
+    such that:
+    \begin{enumerate}
+        \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
+        \item For all $a \in A$, $\phi(a) = V^*\pi(a)V$. 
+    \end{enumerate}
+
+    Moreover, if $\phi$ is unital, then:
+    \begin{enumerate}[start=2]
+        \item $V$ is an isometry. 
+        \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
+    \end{enumerate}
+    
+    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}\footnote{$K$ can always be trimmed so that $(\pi, V, K)$ is minimal.}.
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
+    (a): Let $A \otimes H$ be the algebraic tensor product and define the Hermitian form
+    \[
+    \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
+    \]
+
+    For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and
+    \[
+    a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
+    \]
+
+    then since $\phi$ is completely positive, 
+    \[
+        \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
+    \]
+
+    and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. 
+    
+    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}, 
+    \[
+    |\dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
+    \]
+
+    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. 
+
+    (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
+    \]
+
+    so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. 
+
+    (c): For each $a \in A$, let
+    \[
+    \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
+    \]
+
+    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, (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*}
+
+    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. 
+
+    By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, 
+    \begin{align*}
+        \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
+        &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
+        &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
+    \end{align*}
+
+    By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. 
+
+    (1): For each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
+    \]
+
+    so
+    \[
+    \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
+    \]
+
+    by \autoref{proposition:bh-numerical-radius-gymnastics}. 
+
+    (2): For each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
+    \]
+
+    so $V^*\pi(a)V = \phi(a)$. 
+
+    (3): If $\phi$ is unital, then for each $\xi \in H$,
+    \[
+    \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
+    \]
+
+    so $V$ is an isometry. 
+
+    (4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
+    \]
+
+    Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$. 
+    
+\end{proof}
+
+\begin{proposition}[Uniqueness of Minimal Stinespring Representations]
+\label{proposition:stinespring-minimal-uniqueness}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a completely positive map, and $(\pi_1, V_1, K_1), (\pi_2, V_2, K_2)$ be minimal Stinespring representations for $\phi$, then there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$.
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2]{Paulsen}}}. ]
+    Suppose there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$, then for each $\seqf{a_j} \subset A$ and $\seqf{\xi_j} \subset H$, 
+    \[
+    U\braks{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j} = \sum_{j = 1}^n \pi_2(a_j)UV_1\xi_j = \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j
+    \]
+
+    By minimality, $\pi_1(A)V_1H$ is dense in $K_1$, and $\pi_2(A)V_2H$ is dense in $K_2$. As such, it is sufficient to show that the mapping $\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j \mapsto \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j$ is a well-defined isometry. To this end, note that
+    \begin{align*}
+        \norm{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j}_{K_1}^2 &= \sum_{i, j = 1}^n \dpn{\pi_1(a_j)V_1\xi_j, \pi_1(a_i)V_1\xi_j}{K_1} \\
+        &= \sum_{i, j = 1}^n \dpn{V_1^*\pi_1(a_i^*a_j)V_1\xi_j, \xi_j}{H} = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_j}{H} \\
+        &= \sum_{i, j = 1}^n \dpn{V_2^*\pi_2(a_i^*a_j)V_2\xi_j, \xi_j}{H} \\
+        &= \sum_{i, j = 1}^n \dpn{\pi_2(a_j)V_2\xi_j, \pi_2(a_i)V_2\xi_j}{K_2} = \norm{\sum_{j = 1}^n \pi_2(a_j)V_2\xi_j}_{K_2}^2
+    \end{align*}
+    
+    
+\end{proof}
+
+
+\begin{proposition}[Multiplicative Domain of Stinespring Representations]
+\label{proposition:stinespring-multiplicative-domain}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
+    \begin{enumerate}
+        \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$. 
+        \item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (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*} 
+    \dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
+    \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}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ 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}
+
+
+
+
 \begin{theorem}[Sz.-Nagy's Dilation Theorem]
 \label{theorem:sznagy-dilation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation $(K, U)$ of $T$ such that:
+    \begin{enumerate}
+        \item $K$ is the smallest closed reducing subspace for $U$ containing $H$ such that $T^n = P_HU^n|_H$\footnote{Where $P_H$ is the orthogonal projection onto $H$.} for all $n \in \natz$. 
+        \item[(U)] For any unitary power dilation $(K', U')$ satisfying (1), there exists a unitary $V: K \to K'$ such that $V|_{H} = I_H$ and $VUV^* = U'$. 
+    \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
+\begin{proof}[Constuctive proof without uniqueness, {{\cite[Theorem 1.1]{Paulsen}}}. ]
-    Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements:
+    Using \autoref{lemma:bh-dilation-transitive}, it is sufficient to show the following statements:
     \begin{enumerate}[label=(\roman*)]
         \item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$. 
         \item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$. 
@@ -128,7 +291,26 @@
         \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{proof}[Proof, using Stinespring's Dilation Theorem, {{\cite[Theorem 4.3]{Paulsen}}}. ]
+    (1): Let $\mathbb{T} = \partial B_\complex(0, 1)$. By \autoref{theorem:polynomial-map-positive}, the mapping
+    \[
+    \phi: \complex[z, \ol z] \to B(H) \quad \phi(p + \ol q) = p(T) + q(T)^*
+    \]
+
+    extends into a unital positive map $C(\mathbb{T}; \complex) \to B(H)$, which is completely positive by \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}. 
+
+    Let $(\pi, W, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$, 
+    \[
+    T^n = \phi(z^n) = P_H\pi(z^n)|_H = P_HU^n|_H
+    \]
+
+    so $(K, U)$ is a unitary power dilation of $T$. 
+    
+    (U): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $(\pi, V, K)$ is minimal if and only if $\bigcup_{n \in \integer}U^n H$ is dense in $K$, if and only if $U$ admits no non-trivial reducing subspace. 
+    
+    By \hyperref[uniqueness of minimal Stinespring representations]{proposition:stinespring-minimal-uniqueness}, such a dilation is unique. 
 \end{proof}
 
 \begin{theorem}[Von Neumann's Inequality]
@@ -165,7 +347,7 @@
     \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 +371,7 @@
 
     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 +388,13 @@
 
     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 +410,7 @@
     \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 +425,8 @@
         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 +435,6 @@
         \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 +449,7 @@
     (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
@@ -280,146 +459,10 @@
 
     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).
 \end{proof}
 
 
-
-\begin{theorem}[Stinespring's Dilation Theorem]
-\label{theorem:stinespring-dilation}
-    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, where:
-    \begin{enumerate}[label=(\alph*)]
-        \item $K$ is a complex Hilbert space. 
-        \item $V: H \to K$ is a continuous linear map.
-        \item $\pi: A \to B(K)$ is a unital *-homomorphism.
-    \end{enumerate}
-
-    such that:
-    \begin{enumerate}
-        \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
-        \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. 
-    \end{enumerate}
-
-    Moreover, if $\phi$ is unital, then:
-    \begin{enumerate}[start=2]
-        \item $V$ is an isometry. 
-        \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
-    \end{enumerate}
-    
-    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
-    \[
-    \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
-    \]
-
-    For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and
-    \[
-    a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
-    \]
-
-    then since $\phi$ is completely positive, 
-    \[
-        \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
-    \]
-
-    and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. 
-    
-    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}, 
-    \[
-    \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
-    \]
-
-    so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. 
-
-    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. 
-
-    (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, 
-    \[
-    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
-    \]
-
-    so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. 
-
-    (c): For each $a \in A$, let
-    \[
-    \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
-    \]
-
-    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} \\
-        &\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*}
-
-    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. 
-
-    By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, 
-    \begin{align*}
-        \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
-        &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
-        &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
-    \end{align*}
-
-    By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. 
-
-    (1): For each $\xi \in H$, 
-    \[
-    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
-    \]
-
-    so
-    \[
-    \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
-    \]
-
-    by \autoref{proposition:bh-numerical-radius-gymnastics}. 
-
-    (2): For each $a \in A$ and $\xi, \eta \in H$, 
-    \[
-        \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
-    \]
-
-    so $V^*\pi(a)V = \phi(a)$. 
-
-    (3): If $\phi$ is unital, then for each $\xi \in H$,
-    \[
-    \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
-    \]
-
-    so $V$ is an isometry. 
-
-    (4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$, 
-    \[
-        \dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
-    \]
-
-    Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$. 
-    
-\end{proof}
-
-\begin{proposition}[Multiplicative Domain of Stinespring Representations]
-\label{proposition:stinespring-multiplicative-domain}
-    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
-    \begin{enumerate}
-        \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$. 
-        \item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$. 
-    \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 $H$. 
-    
-    Let $a \in A$ and $\xi, \eta \in H$, then
-    \begin{align*} 
-    \dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
-    \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 $\pi(a)$. 
-    
-\end{proof}
-

src/op/system/schur.tex

@@ -74,3 +74,50 @@
     0 &\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}} = \sum_{j, k = 1}^n E_{jk} \otimes A_{j - k}
     \end{align*}
 \end{proof}
+
+\begin{theorem}
+\label{theorem:polynomial-map-positive}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, and $\mathbb{T} = \partial B_\complex(0, 1)$, then the mapping
+    \[
+    \phi: \complex[z, \ol z] \to B(H) \quad p \mapsto p(T)
+    \]
+
+    extends into a completely positive map on $C(\mathbb{T}; \complex)$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 2.6]{Paulsen}}}. ]
+    Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Using \autoref{proposition:trig-moment-sequence-operator}, it is sufficient to show that the Toeplitz matrix
+    \[
+    A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix}
+        I & T^* &\cdots & (T^*)^n \\
+        T &\ddots &\ddots &\vdots \\
+        \vdots &\ddots &\ddots & T^* \\
+        T^n &\cdots &T &I
+    \end{bmatrix}
+    \]
+
+    is positive for all $n \in \natp$. To this end, let
+    \[
+    R = \begin{bmatrix}
+        0 &\cdots &\cdots &\cdots &0 \\
+        T & \ddots & & &\vdots \\
+        0 & \ddots &\ddots & &\vdots \\
+        \vdots &\ddots &\ddots &\ddots &\vdots \\
+        0 &\cdots &0 & T & 0
+    \end{bmatrix} = \sum_{j = 1}^{n - 1}E_{(j+1)j} \otimes T
+    \]
+
+    then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^n$. In particular, $R^{n+1} = 0$ and
+    \[
+    A = I + \sum_{k = 1}^n R^k + \sum_{k = 1}^n (R^*)^k = \sum_{k = 0}^\infty R^k + \sum_{k = 0}^\infty (R^*)^k - I
+    \]
+    
+    
+    As $\norm{R}_{M_n(B(H))} = \norm{T}_{B(H)} < 1$, by recognising the \hyperref[Neumann series]{lemma:neumann-series} and using the resolvent identity, 
+    \begin{align*}
+        A &= (I - R)^{-1} + (I - R^*)^{-1} - I \\
+        &= (I - R)^{-1}(I - R^* + I - R + (I - R)(I - R)^*)(1 - R^*)^{-1} \\
+        &= (I - R)^{-1}(I - RR^*)(1 - R^*)^{-1} \ge 0
+    \end{align*}
+    
+\end{proof}
+