diff --git a/src/op/system/complete-upgrade.tex b/src/op/system/complete-upgrade.tex index 48fb9a3..8e372cd 100644 --- a/src/op/system/complete-upgrade.tex +++ b/src/op/system/complete-upgrade.tex @@ -31,7 +31,7 @@ The following is a slight generalisation of one form of the above example, but s \begin{proposition} \label{proposition:multiply-completely-bounded} - Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = x\phi(s)x^*$ for all $s \in S$, then $\phi$ is completely positive. + Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = xsx^*$ for all $s \in S$, then $\phi$ is completely positive. \end{proposition} \begin{proposition} @@ -144,7 +144,7 @@ The following is a slight generalisation of one form of the above example, but s &= \sum_{i, j, k, l = 1}^n (E_{ij} \otimes a_{i, j})^*(E_{kl} \otimes a_{k, l}) = \sum_{i, j, k, l = 1}^n (E_{ji}E_{kl}) \otimes (a_{i, j}^* a_{k, l}) \end{align*} - For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{il}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, + For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{jl}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, \[ P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) \] @@ -193,7 +193,7 @@ The following is a slight generalisation of one form of the above example, but s \sum_{k = 1}^n E_{kk} \otimes \phi(T) = \sum_{k = 1}^n S_k(T \otimes 1_B)S_k^* \] - What the above shows, is that, under the identification that + What the above shows is that, under the identification that \[ M_n(\complex) \iso M_n(\complex) \otimes 1_B \subset M_n(B) \] @@ -233,7 +233,7 @@ The following is a slight generalisation of one form of the above example, but s (Not Completely Positive): For each $1 \le i, j \le n$, $\text{Tr}(E_{ij}) = 1$ if and only if $i = j$. So the Choi matrix is given by \begin{align*} - C_{ij} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\ + C_{\phi} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\ &= \sum_{i, j = 1}^n E_{ij} \otimes (\one_{\bracsn{i = j}} \otimes 1_{M_n(\complex)} - E_{ij}) \\ &= \sum_{i = 1}^n E_{ii} \otimes 1_{M_n(\complex)} - \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} \\ &= 1_{M_n(\complex)} \otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} @@ -246,20 +246,20 @@ The following is a slight generalisation of one form of the above example, but s \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 \[ - \phi(a) = \sum_{j = 1}^{nk}V_j^* a V_j + \phi(a) = \sum_{j = 1}^{mn}V_j^* a V_j \] 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 $\dim(K) \le n^2m$. + Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\pi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \pi(M_n(\complex))V\complex^m$, 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) + \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_k\pi(a)P_k)V = \sum_{k = 1}^{mn}V^*Q_k^*aQ_kV \\ + &= \sum_{k = 1}^{mn} (Q_kV)^* a (Q_kV) \end{align*} \end{proof} diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index 52e4800..41b9823 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -84,7 +84,15 @@ \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)} \] - The sharp case is given by operators of the form $\sum_{i, j = 1}^n E_{ij} \otimes b$ for $b \in A$. + The sharp case is given by operators of the form $E = \sum_{i, j = 1}^n E_{ij} \otimes E_{ji}$\footnote{A minimum "complexity" of $M_n$ is required on $A$ for the sharp case to hold.}. As + \begin{align*} + \braks{\sum_{i, j = 1}^n E_{ij} \otimes E_{ji}}^*\braks{\sum_{i, j = 1}^n E_{ij} \otimes E_{ji}} &= \sum_{i, j, k, l = 1}^n E_{ij}E_{kl} \otimes E_{ji}E_{lk} \\ + &= \sum_{i, j, l = 1}^n E_{il} \otimes E_{ji}E_{lj} = \sum_{i, j = 1}^n E_{ii} \otimes E_{jj} \\ + &= I_{M_n(\complex)} \otimes I_{M_n(\complex)} = I_{M_{n \times n}(\complex)} + \end{align*} + + the operator $E$ is of norm $1$, but $\sum_{i, j = 1}^n \norm{E_{ji}}_{M_n(\complex)}^2 = n^2$. + (2): For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(S)$, \begin{align*} diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex index 80f3267..825ed61 100644 --- a/src/op/system/dilation.tex +++ b/src/op/system/dilation.tex @@ -116,7 +116,7 @@ &\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. + 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 $B(K)$, and $\pi: A \to B(K)$ 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*} @@ -148,7 +148,7 @@ (3): If $\phi$ is unital, then for each $\xi \in H$, \[ - \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} + \dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} \] so $V$ is an isometry. @@ -174,10 +174,10 @@ 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 + \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_i}{K_1} \\ + &= \sum_{i, j = 1}^n \dpn{V_1^*\pi_1(a_i^*a_j)V_1\xi_j, \xi_i}{H} = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} \\ + &= \sum_{i, j = 1}^n \dpn{V_2^*\pi_2(a_i^*a_j)V_2\xi_j, \xi_i}{H} \\ + &= \sum_{i, j = 1}^n \dpn{\pi_2(a_j)V_2\xi_j, \pi_2(a_i)V_2\xi_i}{K_2} = \norm{\sum_{j = 1}^n \pi_2(a_j)V_2\xi_j}_{K_2}^2 \end{align*} @@ -215,7 +215,7 @@ \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}[Constuctive proof without uniqueness, {{\cite[Theorem 1.1]{Paulsen}}}. ] +\begin{proof}[Constructive 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 $T$. @@ -299,9 +299,9 @@ \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}. + extends to 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 $n \in \natz$, + Let $(\pi, V, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$, \[ T^n = \phi(z^n) = P_H\pi(z^n)|_H = P_HU^n|_H \] @@ -390,7 +390,7 @@ \begin{align*} \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\\ + |\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*} @@ -425,8 +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 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 + then $f_{\xi, \eta}$ achieves its maximum modulus at 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 $(z_1, \cdots, z_n) \in \complex^n$, if \[ \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)} \] diff --git a/src/op/system/schur.tex b/src/op/system/schur.tex index 390eaa8..f5f30c7 100644 --- a/src/op/system/schur.tex +++ b/src/op/system/schur.tex @@ -88,10 +88,10 @@ 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 \[ A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix} - I & T^* &\cdots & (T^*)^n \\ + I & T^* &\cdots & (T^*)^{n-1} \\ T &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots & T^* \\ - T^n &\cdots &T &I + T^{n-1} &\cdots &T &I \end{bmatrix} \] @@ -106,7 +106,7 @@ \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 + then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^k$. In particular, $R^{n+1} = 0$ and \[ A = I + \sum_{k = 1}^n R^k + \sum_{k = 1}^n (R^*)^k = \sum_{k = 0}^\infty R^k + \sum_{k = 0}^\infty (R^*)^k - I \] @@ -115,7 +115,7 @@ 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 - R^* + I - R - (I - R)(I - R)^*)(I - R^*)^{-1} \\ &= (I - R)^{-1}(I - RR^*)(1 - R^*)^{-1} \ge 0 \end{align*}