Updated draft.

src/fa/norm/hilbert.tex

@@ -369,4 +369,160 @@ A significant property of Hilbert spaces is that every closed subspace is comple
 \end{proof}
 
 
+\begin{definition}[Quadratic Function]
+\label{definition:hilbert-quadratic-function}
+    Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if:
+    \begin{enumerate}
+        \item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda x) = |\lambda|^2 q(x)$. 
+        \item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$, 
+        \[
+        q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta)
+        \]
+    \end{enumerate}
+
+    and \textbf{bounded} if
+    \begin{enumerate}[start=2]
+        \item There exists $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:quadratic-function-sesquilinear}
+    Let $H$ be a complex Hilbert space, $q: H \to \complex$ be a bounded \hyperref[quadratic function]{definition:hilbert-quadratic-function}, and
+    \[
+        H \times H \to \complex \quad (\xi, \eta) \mapsto \dpn{\xi, \eta}{q} := \frac{1}{4}\sum_{k = 0}^3i^kq(\xi + i^k\eta)
+    \]
+
+    then
+
+
+    \begin{enumerate}
+        \item $\dpn{\cdot, \cdot}{q}$ is the unique bounded sesquilinear form on $H$ such that $\dpn{\xi, \xi}{q} = q(\xi)$ for all $\xi \in H$. 
+        \item There exists a unique $T \in B(H)$ such that $q(\xi) = \dpn{T\xi, \xi}{H}$ for all $\xi \in H$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Uniqueness is given by \hyperref[polarisation]{proposition:polarisation-complex}. Existence requires a large number of checkpoints, which are detailed below. Let $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. 
+    \begin{enumerate}[label=(1.\roman*)]
+        \item For each $\xi, \eta, \theta \in H$, 
+        \[
+        \dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
+        \]
+
+        and
+        \[
+        \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2 \angles{\xi, \frac{\eta + \theta}{2}}_{q}
+        \]
+        \item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in [0, 1]$, 
+        \[
+        \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+        \]
+
+        \item For each $\xi, \eta, \theta \in H$, 
+        \[
+        \dpn{\xi + \theta, \eta}{H} = \dpn{\xi, \eta}{H} + \dpn{\theta, \eta}{H}
+        \]
+
+        and
+        \[
+        \dpn{\xi, \eta + \theta}{H} = \dpn{\xi, \eta}{H} + \dpn{\xi, \theta}{H}
+        \]
+
+        \item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in \mathbb{D}$, 
+        \[
+        \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+        \]
+
+        \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le M(\norm{\xi}_H^2 +\norm{\eta}_H^2)$.
+        \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le 3M \norm{\xi}_H\norm{\eta}_H$. 
+
+        \item $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form on $H$.
+        \item For all $\xi \in H$, $\dpn{\xi, \xi}{q} = q(\xi)$. 
+    \end{enumerate}
+    
+
+    (1.i): Let $\xi, \eta, \theta \in H$, then
+    \[
+        4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q}  = \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)}
+    \]
+
+    For each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $((\xi + \theta)/2 + i^k\eta, (\xi - \eta)/2)$, 
+    \[
+    q(\xi + i^k\eta) + q(\theta + i^k\eta) = 2\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}}
+    \]
+
+    so
+    \begin{align*}
+        4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q}  &= \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)} \\ 
+        &= 2\sum_{k = 0}^3 i^k\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}} \\
+        &= 2\sum_{k = 0}^3 i^k{q\paren{\frac{\xi + \theta}{2} + i^k\eta}} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
+    \end{align*}
+
+    Similarly, for each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $(\xi + i^k(\eta + \theta)/2, i^k(\eta - \theta)/2)$,
+    \[
+    q(\xi + i^k\eta) + q(\xi + i^k\theta) = 2\braks{q\paren{\xi + i^k\frac{\eta + \theta}{2}} + q\paren{i^k\frac{\eta - \theta}{2}}}
+    \]
+
+    and
+    \[
+    \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2\angles{\xi, \frac{\eta + \theta}{2}}_q
+    \]
+    
+
+    so $\dpn{\cdot, \cdot}{q}: H \times H \to \complex$ is a $\real$-bilinear form. 
+
+    (1.ii): By definition of $\dpn{\cdot, \cdot}{q}$, the claim holds for $0$ and $1$. Let $n \in \natz$ and suppose inductively that the claim holds for all $\lambda \in \mathbb{D}_n \cap [0, 1]$. Take $\lambda \in \mathbb{D}_{n+1} \cap [0, 1]$, then there exists $\mu, \nu \in \mathbb{D}_n \cap [0, 1]$ such that $\lambda = (\mu + \nu)/2$. In which case, by (1.i),
+    \begin{align*}
+        \dpn{\lambda \xi, \eta}{q} &= \angles{\frac{\mu + \nu}{2}\xi, \eta}_{q} = \frac{1}{2}\dpn{\mu\xi, \eta}{q} + \frac{1}{2}\dpn{\nu\xi, \eta}{q} \\
+        &= \frac{\mu + \nu}{2}\dpn{\xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}
+    \end{align*}
+
+    By induction, $\dpn{\lambda \xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. 
+
+    Similarly, $\dpn{\xi, \lambda \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. 
+
+    (1.iii): Let $\xi, \eta, \theta \in H$, then by (1.i) and (1.ii), 
+    \[
+        \frac{1}{2}(\dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q}) = \angles{\frac{\xi + \theta}{2}, \eta}_q = \frac{1}{2}\dpn{\xi + \theta, \eta}{q}
+    \]
+    
+    (1.iv): By (1.ii) and repeated application of (1.iii), 
+    \[
+    \dpn{\lambda \xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+    \]
+
+    for all $\lambda \in \mathbb{D}$. 
+    
+    (1.v): Let $\xi, \eta \in H$, then
+    \begin{align*}
+        |\dpn{\xi, \eta}{q}| &\le \frac{1}{4}\sum_{k = 0}^3 |q(\xi + i^k\eta)| = \frac{M}{4}\sum_{k = 0}^3\normn{\xi + i^k\eta}_H \\
+        &\le \frac{M}{4}\sum_{k = 0}^3 [\norm{\xi}_H^2 + \norm{\eta}_H^2 + i^{-k}\dpn{\xi, \eta}{H}] = M(\norm{\xi}_H^2 + \norm{\eta}_H^2)
+    \end{align*}
+    
+    
+    
+    (1.vi): Let $\xi, \eta \in H$ and assume without loss of generality that $\xi, \eta \ne 0$. Since $\norm{\eta}_H/\norm{\xi}_H > 0$, there exists $n \in \integer$ such that $\norm{\eta}_H/\norm{\xi}_H \in [2^n, 2^{n+1}]$. In which case, by (1.iv), 
+    \begin{align*}
+    |\dpn{\xi, \eta}{q}| &= 2^{-n}|\dpn{2^n\xi, \eta}{q}| \le 2^{-n}M(4^n \norm{\xi}_H^2 + \norm{\eta}_H^2) = 2^nM\norm{\xi}_H^2 + 2^{-n}M\norm{\eta}_H^2 \\
+    &\le M\norm{\xi}_H\norm{\eta}_H + 2M\norm{\xi}_H\norm{\eta}_H = 3M\norm{\xi}_H\norm{\eta}_H
+    \end{align*}
+    
+    (1.vii): By (1.iv) and (1.vi), $\dpn{\cdot, \cdot}{q}$ is a $\real$-bilinear form on $H$. For any $\xi, \eta \in H$, by absolute homogeneity of $q$, 
+    \begin{align*}
+        \dpn{i\xi, \eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(i\xi + i^k\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k-1}\eta) \\
+        &= \frac{i}{4}\sum_{k = 0}^3 i^{k-1} q(\xi + i^{k-1}\eta) = \dpn{\xi, \eta}{q} \\
+        \dpn{\xi, i\eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k+1}\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(i^{-1}\xi + i^{k}\eta) \\
+        &= \dpn{-i\xi, \eta}{q} =  -i \dpn{\xi, \eta}{q}
+    \end{align*}
+
+    Therefore $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form. 
+    
+    (1.viii): Let $\xi \in H$, then by absolute homogeneity of $q$, 
+    \begin{align*}
+        \dpn{\xi, \xi}{q} &= \frac{1}{4}\braks{q(\xi + \xi) - q(\xi - \xi) + iq(\xi + i\xi) - iq(\xi - i\xi)} \\
+        &= \frac{1}{4}\braks{4q(\xi) + iq(\xi + i\xi) - iq(i\xi + \xi)} = q(\xi)
+    \end{align*}
+    
+    (2): By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}.
+\end{proof}
 

src/op/system/complete-upgrade.tex

@@ -34,60 +34,6 @@ The following is a slight generalisation of one form of the above example, but s
     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. 
 \end{proposition}
 
-
-
-\begin{proposition}
-\label{proposition:completely-positive-norm}
-    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then
-    \[
-    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
-    \]
-\end{proposition}
-\begin{proof}
-    After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. 
-    
-    It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then
-    \[
-    \begin{bmatrix}
-            1_{M_n(A)} &T \\
-            T^* &1_{M_n(A)}
-    \end{bmatrix} \ge 0
-    \]
-
-    in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, 
-    \[
-    \phi_{2n}\begin{bmatrix}
-            1_{M_n(A)} &T \\
-            T^* &1_{M_n(A)}
-    \end{bmatrix} = \begin{bmatrix}
-            \phi_n(1_{M_n(A)}) &\phi_n(T) \\
-            \phi_n(T)^* & \phi_n(1_{M_n(A)})
-    \end{bmatrix} \ge 0
-    \]
-
-    First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then
-    
-    \begin{align*}
-        \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\
-        1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\
-        \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} 
-    \end{align*}
-    
-    By the \hyperref[Schur complement condition]{proposition:schur-complement-positive}, 
-    \begin{align*}
-    \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\
-    &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\
-    \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 
-    \end{align*}
-
-    and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$.
-    
-    Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, 
-    \[
-    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
-    \]
-\end{proof}
-
 \begin{proposition}
 \label{proposition:commutative-mn-facts}
     Let $X$ be a compact Hausdorff space, then:

src/op/system/completely.tex

@@ -180,7 +180,58 @@
 \end{proof}
 
 
-% Omitted
+\begin{proposition}
+\label{proposition:completely-positive-norm}
+    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then
+    \[
+    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
+    \]
+\end{proposition}
+\begin{proof}
+    After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. 
+    
+    It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then
+    \[
+    \begin{bmatrix}
+            1_{M_n(A)} &T \\
+            T^* &1_{M_n(A)}
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, 
+    \[
+    \phi_{2n}\begin{bmatrix}
+            1_{M_n(A)} &T \\
+            T^* &1_{M_n(A)}
+    \end{bmatrix} = \begin{bmatrix}
+            \phi_n(1_{M_n(A)}) &\phi_n(T) \\
+            \phi_n(T)^* & \phi_n(1_{M_n(A)})
+    \end{bmatrix} \ge 0
+    \]
+
+    First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then
+    
+    \begin{align*}
+        \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\
+        1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\
+        \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} 
+    \end{align*}
+    
+    By the \hyperref[Schur complement condition]{proposition:schur-complement-positive}, 
+    \begin{align*}
+    \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\
+    &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\
+    \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 
+    \end{align*}
+
+    and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$.
+    
+    Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, 
+    \[
+    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
+    \]
+\end{proof}
+
 
 \begin{proposition}
 \label{proposition:unital-2-positive-contractive}

src/op/system/numerical-radius.tex

@@ -51,7 +51,7 @@
 
     so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$. 
     
-    (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$ with $\dpn{T\xi, \eta}{H} \in \real$, 
+    (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_{B(H)}$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$ with $\dpn{T\xi, \eta}{H} \in \real$, 
     \begin{align*}
         \dpn{T\xi, \eta}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k\dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} \\
         &= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\
@@ -65,7 +65,7 @@
     \norm{T\xi}_H^2 &\le \norm{\xi}_H^2
     \end{align*}
 
-    (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. 
+    (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_{B(H)}$. The sharp case is given by $I_H$. 
 
     On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. 
 
@@ -97,7 +97,7 @@
 
 \begin{proposition}
 \label{proposition:cyclic-forward-shift-radius}
-    Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 2}^nE_{(j+1)j} \otimes T \in M_n(B(H))$, $\bracsn{e_j}_0^{n-1} \subset \complex^n$ be the canonical basis for $\complex^n$, and
+    Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 1}^{n-1}E_{(j+1)j} \otimes T \in M_n(B(H))$, $\bracsn{e_j}_0^{n-1} \subset \complex^n$ be the canonical basis for $\complex^n$, and
     \[
     S_n: \complex^n \to \complex^n \quad S_ne_j = e_{j+1 \mod n}
     \]
@@ -127,7 +127,7 @@
     (2): Let $\bracsn{f_j}_{0}^{n-1} \subset \complex^n$ be an orthonormal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$. For any $\bracsn{\xi_j}_0^{n-1} \subset H$, let $\xi = \sum_{j = 0}^{n-1}\xi_j \otimes f_j$, then 
     \begin{align*}
         \dpn{(T \otimes S_n)\xi, \xi}{H^n} &= \sum_{j, k = 0}^{n-1}\dpn{(T \otimes S_n)(\xi_j \otimes f_j), \xi_k \otimes f_k}{H^n} \\
-        &= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H} \dpn{S_nf_j, f_k}{H} = \sum_{j = 0}^{n-1}\omega^j\dpn{T\xi_j, \xi_j}{H} \\
+        &= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H} \dpn{S_nf_j, f_k}{\complex^n} = \sum_{j = 0}^{n-1}\omega^j\dpn{T\xi_j, \xi_j}{H} \\
         |\dpn{(T \otimes S_n)\xi, \xi}{H^n}| &\le \sum_{j = 0}^{n-1}|\omega^j| \cdot w(T) \cdot \norm{\xi_j}_{H}^2 = w(T)\norm{\xi}_{H^n}^2
     \end{align*}
 
@@ -140,7 +140,7 @@
 
     (3): Let $\xi = (\xi_1, \cdots, \xi_{n}) \in H^n$, then 
     \begin{align*}
-        \dpn{R\xi, \xi}{H^n} &= \sum_{j = 2}^{n}\dpn{(E_{(j+1)j}\otimes T)\xi, \xi}{H^n} = \sum_{j = 1}^{n-1}  \dpn{T\xi_j, \xi_{j+1}}{H} \\
+        \dpn{R\xi, \xi}{H^n} &= \sum_{j = 1}^{n-1}\dpn{(E_{(j+1)j}\otimes T)\xi, \xi}{H^n} = \sum_{j = 1}^{n-1}  \dpn{T\xi_j, \xi_{j+1}}{H} \\
         &= \dpn{T0, \xi_1}{H} + \dpn{T\xi_n, 0}{H} + \sum_{j = 1}^{n-1}  \dpn{T\xi_j, \xi_{j+1}}{H} \\
         &= \dpn{(T \otimes S_{n+1})(\xi_1, \cdots, \xi_n, 0), (\xi_1, \cdots, \xi_n, 0)}{H^{n+1}}
     \end{align*}

src/op/system/schur.tex

@@ -7,15 +7,15 @@
     Let $H$ be a complex Hilbert space, $A = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(B(H))$ be positive, and $\Lambda = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \in M_n(\complex)$ be positive, then $S_A(\Lambda) = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \otimes a_{ij}$ is also positive. 
 \end{lemma}
 \begin{proof}
-    For any $\xi \in \complex^n$, the projection map $P_\xi$ onto its span is given by $\sum_{i, j = 1}^n \xi_i E_{ij} \ol{\xi_j}$. In which case, 
+    For any unit vector $\xi \in \complex^n$, the projection map $P_\xi$ onto its span is given by $\sum_{i, j = 1}^n \xi_i E_{ij} \ol{\xi_j}$. In which case, 
     \begin{align*}
         S_A(P_\xi) &= \sum_{i, j = 1}^n \xi_i \ol{\xi_j} E_{ij} \otimes a_{ij} \\
-        &= \braks{\sum_{i = 1}^n \xi_{i}E_{ii}}\braks{\sum_{i, j = 1}^n \xi_i \ol{\xi_j} E_{ij} \otimes a_{ij}}\braks{\sum_{j = 1}^n {\xi_j}E_{jj}}^* \ge 0
+        &= \braks{\sum_{i = 1}^n \xi_{i}E_{ii}}\braks{\sum_{i, j = 1}^n E_{ij} \otimes a_{ij}}\braks{\sum_{j = 1}^n {\xi_j}E_{jj}}^* \ge 0
     \end{align*}
     
 
 
-    Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, let $\bracsn{\xi^{(j)}}_1^n \subset \complex^n$ be an orthonormal eigenbasis for $\Lambda$ with eigenvectors $\seqf{\lambda_j} \subset \complex$. For each $1 \le j \le n$, let $P_j \in M_n(\complex)$ be the rank one projection onto the span of $\xi^{(j)}$, then 
+    Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, let $\bracsn{\xi^{(j)}}_1^n \subset \complex^n$ be an orthonormal eigenbasis for $\Lambda$ with eigenvalues $\seqf{\lambda_j} \subset \complex$. For each $1 \le j \le n$, let $P_j \in M_n(\complex)$ be the rank one projection onto the span of $\xi^{(j)}$, then 
     \[
         S_A(\Lambda) = \sum_{j = 1}^n \lambda_j S_A(P_{j}) \ge 0
     \]
@@ -45,7 +45,7 @@
 
     (2) $\Leftrightarrow$ (3): Let $q = \sum_{-N \le j \le N}\mu_j z^j$ and $\xi = (\mu_{-N}, \cdots, \mu_{N}) \in \complex^{2N+1}$, then
     \begin{align*}
-        \dpn{Pq, q}{L^2(\mathbb{T}; \complex)} &= \sum_{-N \le j, k, l \le N}\dpn{\lambda_j z^j \cdot \mu_k z^k, \mu_l z^l}{L^2(\mathbb{T}; \complex)} \\
+        \dpn{Pq, q}{L^2(\mathbb{T}; \complex)} &= \sum_{-n \le j \le n}\sum_{-N \le k, l \le N}\dpn{\lambda_j z^j \cdot \mu_k z^k, \mu_l z^l}{L^2(\mathbb{T}; \complex)} \\
         &= \sum_{-N \le k, l \le N}\dpn{\lambda_{l-k}\mu_k z^l, \mu_l z^l}{L^2(\mathbb{T}; \complex)} \\
         &= \sum_{-N \le k, l \le N} \lambda_{l - k}\mu_k\ol{\mu_l}= \sum_{k, l = 1}^{2N+1} \lambda_{l - k}\xi_{k}\ol{\xi_{l}} \\
         &= \angles{\paren{\sum_{i, j = 1}^{2N+1} E_{ij}\lambda_{i - j}}\xi, \xi} \\