diff --git a/src/op/system/complete-upgrade.tex b/src/op/system/complete-upgrade.tex index 13aa67d..48fb9a3 100644 --- a/src/op/system/complete-upgrade.tex +++ b/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} + diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex index fe8766a..80f3267 100644 --- a/src/op/system/dilation.tex +++ b/src/op/system/dilation.tex @@ -48,12 +48,175 @@ \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}}}. ] - Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements: +\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: \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 $V$. @@ -130,6 +293,25 @@ 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] \label{theorem:von-neumann-inequality-polynomial} @@ -284,138 +466,3 @@ \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 $a \in A$, $\phi(a) = V^*\pi(a)V$. - \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}. -\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}[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} - diff --git a/src/op/system/schur.tex b/src/op/system/schur.tex index a371b40..d9fc367 100644 --- a/src/op/system/schur.tex +++ b/src/op/system/schur.tex @@ -73,4 +73,51 @@ 0 &\le \braks{\sum_{j = 1}^n E_{j1} \otimes z^j}\braks{\sum_{j = 1}^n E_{j1} \otimes z^j}^* = \sum_{j, k = 1}^n E_{jk} \otimes z^{j-k} \\ 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} \ No newline at end of file +\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} +