On the subject of 2x2 matrices.

src/op/system/complete-upgrade.tex

@@ -73,7 +73,7 @@ The following is a slight generalisation of one form of the above example, but s
         \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]{lemma:schur-complement-positive}, 
+    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)^* \\

src/op/system/completely.tex

@@ -90,74 +90,6 @@
 
 % Omitted
 
-\begin{lemma}
-\label{lemma:norm-matrix-test}
-    Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if
-    \[
-    \begin{bmatrix}
-        1 &x \\
-        x^* &y
-    \end{bmatrix} \ge 0
-    \]
-
-    in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if
-    \[
-    \begin{bmatrix}
-        1 &x \\
-        x^* &1
-    \end{bmatrix} \ge 0
-    \]
-
-    in $M_2(A)$.
-\end{lemma}
-\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ]
-    Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, 
-    \[
-    \angles{ \begin{bmatrix}
-            1 &x \\
-            x^* &y
-        \end{bmatrix}\begin{bmatrix}
-            \xi \\
-            \eta
-        \end{bmatrix}, \begin{bmatrix}
-            \xi \\
-            \eta
-        \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H}
-    \]
-
-    ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
-    \begin{align*}
-    \angles{ \begin{bmatrix}
-            1 &x \\
-            x^* &y
-        \end{bmatrix}\begin{bmatrix}
-            \xi \\
-            \eta
-        \end{bmatrix}, \begin{bmatrix}
-            \xi \\
-            \eta
-        \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\
-        &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0
-    \end{align*}
-
-    ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, 
-    \begin{align*}
-        0 &\le \angles{ \begin{bmatrix}
-            1 &x \\
-            x^* &y
-        \end{bmatrix}\begin{bmatrix}
-            -x\eta \\
-            \eta
-        \end{bmatrix}, \begin{bmatrix}
-            -x\eta \\
-            \eta
-        \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\
-        &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H}
-    \end{align*}
-
-    so $y \ge x^*x$. 
-\end{proof}
-
 \begin{proposition}
 \label{proposition:unital-2-positive-contractive}
     Let $A, B$ be unital $C^*$-algebras, $S \subset A$ be an operator system, $\phi: S \to B$ be a unital $2$-positive map, then $\norm{\phi}_{L(S; B)} \le 1$. 
@@ -213,194 +145,6 @@
     By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \phi(x^*x)$. 
 \end{proof}
 
-\begin{lemma}[Schur Complement Condition for Positivity]
-\label{lemma:schur-complement-positive}
-    Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then
-    \begin{enumerate}[label=(P\arabic*)]
-        \item If $P$ is also positive, then
-        \[
-        \begin{bmatrix}
-            P &A \\
-            A^* &Q
-        \end{bmatrix} = \begin{bmatrix}
-            P^{1/2} &0 \\
-            A^*P^{-1/2} &I
-        \end{bmatrix}\begin{bmatrix}
-            I &0 \\
-            0 & Q - AP^{-1}A^*
-        \end{bmatrix}\begin{bmatrix}
-            P^{1/2} &P^{-1/2}A \\
-            0 &I
-        \end{bmatrix}
-        \]
-        
-        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$.
-    \end{enumerate}
-
-    The operator $Q - A^*P^{-1}A$ is the \textbf{Schur complement} of $P$. Similarly, if $Q$ is invertible, then
-    \begin{enumerate}[label=(Q\arabic*)]
-        \item If $Q$ is also positive, then
-        \[
-        \begin{bmatrix}
-            P &A \\
-            A^* &Q
-        \end{bmatrix} = \begin{bmatrix}
-            Q^{1/2} &Q^{-1/2}A \\
-            0 &I
-        \end{bmatrix}\begin{bmatrix}
-            I &0 \\
-            0 & P - A^*Q^{-1}A
-        \end{bmatrix}\begin{bmatrix}
-            Q^{1/2} &0 \\
-            A^*Q^{-1/2} &I
-        \end{bmatrix}
-        \]
-        
-        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - A^*Q^{-1}A \ge 0$.
-    \end{enumerate}
-    
-    Finally, let $P^{-1/2}: P^{1/2}(H) \to {P^{1/2}(H)}$ and $Q^{-1/2}: Q^{1/2}(H) \to {Q^{1/2}(H)}$ be inverses of $P^{1/2}$ and $Q^{1/2}$ on their images, respectively, $P^{-1} = (P^{-1/2})^2$, and $Q^{-1} = (Q^{-1/2})^2$, then the following are equivalent:
-    \begin{enumerate}
-        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
-        \item For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$.
-        \item $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - A^*P^{-1}A \ge 0$.
-        \item $Q \ge 0$, $A(H) \subset Q^{1/2}(H)$, and $P - AQ^{-1}A^* \ge 0$.
-    \end{enumerate}
-    
-\end{lemma}
-\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
-    (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus
-    \begin{align*}
-        \begin{bmatrix}
-            P &A \\
-            A^* &Q
-        \end{bmatrix} &= \begin{bmatrix}
-            P^{1/2}P^{1/2} &P^{1/2}P^{-1/2}A \\
-            A^*P^{-1/2}P^{1/2} &Q
-        \end{bmatrix} \\
-        &= \begin{bmatrix}
-            P^{1/2} &0 \\
-            A^*P^{-1/2} &I
-        \end{bmatrix}\begin{bmatrix}
-            P^{1/2} &P^{-1/2}A \\
-            0 &Q - A^*P^{-1}A
-        \end{bmatrix} \\
-        \begin{bmatrix}
-            P^{1/2} &P^{-1/2}A \\
-            0 &Q - A^*P^{-1}A
-        \end{bmatrix} &= \begin{bmatrix}
-            I &0 \\
-            0 &Q - A^*P^{-1}A
-        \end{bmatrix} \begin{bmatrix}
-            P^{1/2} &P^{-1/2}A \\
-            0 &I
-        \end{bmatrix} 
-    \end{align*}
-    
-    and
-    
-    \[
-    \begin{bmatrix}
-        P &A \\
-        A^* &Q
-    \end{bmatrix} = \begin{bmatrix}
-        P^{1/2} &0 \\
-        A^*P^{-1/2} &I
-    \end{bmatrix}\begin{bmatrix}
-        I &0 \\
-        0 & Q - A^*P^{-1}A
-    \end{bmatrix}\begin{bmatrix}
-        P^{1/2} &P^{-1/2}A \\
-        0 &I
-    \end{bmatrix}
-    \]
-
-    (P2): If $P$ is also positive and $Q - A^*P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right]$ is of the form $MDM^*$, where $D$ is positive. Thus the given matrix is also positive. 
-
-    On the other hand, if $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition, 
-    \[
-    \begin{bmatrix}
-            I &0 \\
-            0 & Q - A^*P^{-1}A
-        \end{bmatrix} = \begin{bmatrix}
-            P^{-1/2} &0 \\
-            -P^{-1}A^* &I
-        \end{bmatrix}\begin{bmatrix}
-        P &A \\
-        A^* &Q
-    \end{bmatrix} \begin{bmatrix}
-            P^{-1/2} &-P^{-1}A \\
-            0 &I
-        \end{bmatrix} 
-    \]
-
-    is also of the form $MDM^*$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^*P^{-1}A$ are also positive. 
-
-    (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2), $P + \eps \ge 0$ and $Q - A^*(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by (P2) and the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
-    \begin{align*}
-        |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\
-        &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\
-        &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\
-        &= \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2}
-    \end{align*}
-    
-    As the above holds for all $\eps > 0$, 
-    \[
-     |\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}
-    \]
-
-    (2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$, 
-    \[
-     |\dpn{A\xi, P^{-1/2}\eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{\eta, \eta}{H}^{1/2}
-    \]
-
-    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\theta \in \ol{P^{1/2}(H)}$ such that 
-    \[
-    \dpn{A\xi, P^{-1/2}\eta}{H} = \dpn{\theta, \eta}{H}
-    \]
-
-    for all $\eta \in P^{1/2}(H)$, so 
-    \[
-    \dpn{A\xi, P^{1/2}\eta}{H} = \dpn{\theta, P^{1/2}\eta}{H} = \dpn{P^{1/2}\theta, \eta}{H}
-    \]
-
-    for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$. In addition, for any $\xi \in H$, 
-    \begin{align*}
-        \dpn{(Q - A^*P^{-1}A)\xi, \xi}{H} &= \dpn{Q\xi, \xi}{H} - \dpn{P^{-1}A\xi, A\xi}{H} \\
-        &= \dpn{Q\xi, \xi}{H} - \dpn{A\xi, P^{-1}A\xi}{H} \\
-        &= \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H}^{1/2}\dpn{PP^{-1}A\xi, P^{-1}A\xi}{H}^{1/2} \\
-        &\ge \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H}^{1/2}\dpn{A^*P^{-1}A\xi, \xi}{H}^{1/2} \\
-        &\ge \dpn{Q\xi, \xi}{H} - \frac{\dpn{Q\xi, \xi}{H}}{2} - \frac{\dpn{A^*P^{-1}A\xi, \xi}{H}}{2} \\
-        &= \frac{1}{2}\dpn{(Q - A^*P^{-1}A)\xi, \xi}{H}
-    \end{align*}
-
-    by \hyperref[Young's Inequality]{lemma:young-inequality}. Therefore $Q - A^*P^{-1}A \ge 0$.
-
-    (3) $\Rightarrow$ (1): If $A(H) \subset P^{1/2}(H)$ and $Q - A^*P^{-1}A \ge 0$. For each $\xi \in H$, 
-    \[
-    \dpn{P^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{P^{-1}P\xi, \xi}{H} = \dpn{\xi, \xi}{H}
-    \]
-
-    and by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
-    \[
-    \dpn{(P+\eps)^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{(P+\eps)^{-1}P\xi, \xi}{H} \le \dpn{\xi, \xi}{H}
-    \]
-
-    Thus $\dpn{P^{-1}\xi, \xi}{H} \ge \dpn{(P+\eps)^{-1}\xi, \xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$, 
-    \begin{align*}
-    \dpn{P^{-1}A\xi, A\xi}{H} &\ge \dpn{(P+\eps)^{-1}A\xi, A\xi}{H} \\
-    \dpn{A^*P^{-1}A\xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H}
-    \end{align*}
-    
-    so
-
-    \[
-    0 \le Q - A^*P^{-1}A \le Q - A^*(P + \eps)^{-1}A 
-    \]
-
-    and $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
-    
-\end{proof}
 
 
 \begin{example}

src/op/system/index.tex

@@ -3,6 +3,7 @@
 
 \input{./dilation.tex}
 \input{./os.tex}
+\input{./two.tex}
 \input{./completely.tex}
 \input{./complete-upgrade.tex}
 \input{./module.tex}

src/op/system/module.tex

@@ -104,3 +104,33 @@
 
     By \autoref{lemma:matrix-positive-topleft-corner-zero}, $\phi(b)\phi(a) - \phi(ba) = 0$. 
 \end{proof}
+
+\begin{definition}[Conditional Expectation]
+\label{definition:cstar-conditional-expectation}
+    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{conditional expectation} if:
+    \begin{enumerate}
+        \item $\phi$ is unital and completely positive. 
+        \item $\phi|_B$ is the identity. 
+    \end{enumerate}
+\end{definition}
+
+
+\begin{example}[Conditional Expectation]
+    \label{example:matrix-conditional-expectation}
+    Let $\phi: M_n(\complex) \to \complex^n$ be defined by $\phi(A)_k = a_{kk}$, then:
+    \begin{enumerate}
+        \item $\phi$ is unital and completely positive. 
+        \item The multiplicative domain of $\phi$ is the diagonal matrices on $M_n(\complex)$. 
+    \end{enumerate}
+\end{example}
+\begin{proof}
+    (1): Identify $\complex^n$ as the diagonal matrices in $M_n(\complex)$, then for each $A \in M_n(\complex)$, 
+    \[
+    \phi(A) = \sum_{j = 1}^n E_{jj}AE_{jj}
+    \]
+
+    By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive. 
+
+    (2): ???
+\end{proof}
+

src/op/system/os.tex

@@ -147,6 +147,7 @@
     \begin{enumerate}
         \item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$. 
         \item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. 
+        \item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$.
         \item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general. 
     \end{enumerate}
 \end{proposition}
@@ -179,7 +180,21 @@
 
     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 $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. 
+    (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 $\xi, \eta \in H$, 
+    \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}) \\
+        &\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\
+        &= \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2)
+    \end{align*}
+
+    by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular, 
+    \begin{align*}
+    \norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\
+    \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 $I_H$. 
 
     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 $\norm{T}_{B(H)} \le 2w(T)$. 
 

src/op/system/two.tex

@@ -0,0 +1,349 @@
+\section{On the Subject of 2×2 Matrices}
+\label{section:op-sys-2by2}
+
+\begin{lemma}
+\label{lemma:norm-matrix-test}
+    Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if
+    \[
+    \begin{bmatrix}
+        1 &x \\
+        x^* &y
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if
+    \[
+    \begin{bmatrix}
+        1 &x \\
+        x^* &1
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_2(A)$.
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ]
+    Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, 
+    \[
+    \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H}
+    \]
+
+    ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+    \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\
+        &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0
+    \end{align*}
+
+    ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, 
+    \begin{align*}
+        0 &\le \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            -x\eta \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            -x\eta \\
+            \eta
+        \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\
+        &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H}
+    \end{align*}
+
+    so $y \ge x^*x$. 
+\end{proof}
+
+
+
+\begin{proposition}[Schur Complement Condition for Positivity, Invertible Case]
+\label{proposition:schur-complement-positive}
+    Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then
+    \begin{enumerate}[label=(P\arabic*)]
+        \item If $P$ is also positive, then
+        \[
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} = \begin{bmatrix}
+            P^{1/2} &0 \\
+            A^*P^{-1/2} &I
+        \end{bmatrix}\begin{bmatrix}
+            I &0 \\
+            0 & Q - A^*P^{-1}A
+        \end{bmatrix}\begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
+        \end{bmatrix}
+        \]
+        
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$.
+    \end{enumerate}
+
+    The operator $Q - A^*P^{-1}A$ is the \textbf{Schur complement} of $P$. Similarly, if $Q$ is invertible, then
+    \begin{enumerate}[label=(Q\arabic*)]
+        \item If $Q$ is also positive, then
+        \[
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} = \begin{bmatrix}
+            I &AQ^{-1/2} \\
+            0 &Q^{1/2}
+        \end{bmatrix}\begin{bmatrix}
+            P - AQ^{-1}A^* &0 \\
+            0 & I
+        \end{bmatrix}\begin{bmatrix}
+            I &0 \\
+            Q^{-1/2}A^* &Q^{1/2}
+        \end{bmatrix}
+        \]
+        
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - AQ^{-1}A^* \ge 0$.
+    \end{enumerate}    
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus
+    \begin{align*}
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} &= \begin{bmatrix}
+            P^{1/2}P^{1/2} &P^{1/2}P^{-1/2}A \\
+            A^*P^{-1/2}P^{1/2} &Q
+        \end{bmatrix} \\
+        &= \begin{bmatrix}
+            P^{1/2} &0 \\
+            A^*P^{-1/2} &I
+        \end{bmatrix}\begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} \\
+        \begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} &= \begin{bmatrix}
+            I &0 \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} \begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
+        \end{bmatrix} 
+    \end{align*}
+    
+    and
+    
+    \[
+    \begin{bmatrix}
+        P &A \\
+        A^* &Q
+    \end{bmatrix} = \begin{bmatrix}
+        P^{1/2} &0 \\
+        A^*P^{-1/2} &I
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        0 & Q - A^*P^{-1}A
+    \end{bmatrix}\begin{bmatrix}
+        P^{1/2} &P^{-1/2}A \\
+        0 &I
+    \end{bmatrix}
+    \]
+
+    (P2): If $P$ is also positive and $Q - A^*P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right]$ is of the form $MDM^*$, where $D$ is positive. Thus the given matrix is also positive. 
+
+    On the other hand, if $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition, 
+    \[
+    \begin{bmatrix}
+            I &0 \\
+            0 & Q - A^*P^{-1}A
+        \end{bmatrix} = \begin{bmatrix}
+            P^{-1/2} &0 \\
+            -A^*P^{-1} &I
+        \end{bmatrix}\begin{bmatrix}
+        P &A \\
+        A^* &Q
+    \end{bmatrix} \begin{bmatrix}
+            P^{-1/2} &-P^{-1}A \\
+            0 &I
+        \end{bmatrix} 
+    \]
+
+    is also of the form $MDM^*$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^*P^{-1}A$ are also positive. 
+\end{proof}
+
+
+\begin{proposition}[Schur Complement Condition for Positivity]
+\label{proposition:schur-complement-positive-general}
+    Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$, $P^{-1/2}: P^{1/2}(H) \to \ker(P)^\perp$ and $Q^{-1/2}: Q^{1/2}(H) \to \ker(Q)^\perp$ be partial inverses of $P^{1/2}$ and $Q^{1/2}$, respectively, then the following are equivalent:
+    \begin{enumerate}
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
+        \item For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$.
+        \item $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^*(P^{-1/2}A) \ge 0$.
+        \item $Q \ge 0$, $A^*(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A)^*(Q^{-1/2}A) \ge 0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2) of \autoref{proposition:schur-complement-positive}, $P + \eps \ge 0$ and $Q - A^*(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+        |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\
+        &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\
+        &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\
+        &\le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2}
+    \end{align*}
+    
+    As the above holds for all $\eps > 0$, 
+    \[
+     |\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}
+    \]
+
+    (2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$, 
+    \[
+     |\dpn{A\xi, P^{-1/2}\eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{\eta, \eta}{H}^{1/2}
+    \]
+
+    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\theta \in \ol{P^{1/2}(H)}$ such that 
+    \[
+    \dpn{A\xi, P^{-1/2}\eta}{H} = \dpn{\theta, \eta}{H}
+    \]
+
+    for all $\eta \in P^{1/2}(H)$, so 
+    \[
+    \dpn{A\xi, \eta}{H} = \dpn{\theta, P^{1/2}\eta}{H} = \dpn{P^{1/2}\theta, \eta}{H}
+    \]
+
+    for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$. 
+    
+    Now, for each Cauchy sequence $\seq{\xi_n} \subset H$ such that $\bracsn{P^{-1/2}A\xi_n}_1^\infty \subset \ker(P)^\perp$ is Cauchy, then there exists $\eta \in \ker(P)^\perp$ such that $P^{-1/2}A\xi_n \to \eta$ as $n \to \infty$. By continuity of $P^{1/2}$, $P^{1/2}P^{-1/2}A\xi_n \to P^{1/2}\eta$ as $n \to \infty$. Since $\eta$ is the unique element of $\ker(P)^\perp$ with $P^{1/2}\eta = A\xi = A\limv{n}\xi_n$, $\eta = P^{-1/2}A\limv{n}\xi_n$. By the \hyperref[Closed Graph Theorem]{theorem:closed-graph}, $P^{-1/2}A \in B(H)$, and the second part of the condition is well-defined.
+    
+    
+    Now, for any $\xi \in H$, 
+    \[
+        \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} = \dpn{Q\xi, \xi}{H} - \normn{P^{-1/2}A\xi}_H^2
+    \]
+
+    As $P^{-1/2}A\xi \in \ker(P)^\perp = \ol{P(H)}$, 
+    \begin{align*} 
+    \normn{P^{-1/2}A\xi}_H &= \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{P^{-1/2}A\xi, P^{1/2}\eta}{H}| = \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{A\xi, \eta}{H}| \\
+    &\le \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}\dpn{Q\xi, \xi}{H}^{1/2}\dpn{P^{1/2}\eta, P^{1/2}\eta}{H}^{1/2} = \dpn{Q\xi, \xi}{H}^{1/2}
+    \end{align*}
+
+    so
+    \[
+    \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} \ge \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H} = 0
+    \]
+    
+
+    (3) $\Rightarrow$ (1): For each $\xi \in H$, 
+    \[
+    \dpn{P^{-1/2}P^{1/2}\xi, P^{-1/2}P^{1/2}\xi}{H} = \dpn{\xi, \xi}{H}
+    \]
+
+    and by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    \dpn{(P+\eps)^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{(P+\eps)^{-1}P\xi, \xi}{H} \le \dpn{\xi, \xi}{H}
+    \]
+
+    Thus $\dpn{P^{-1/2}\xi, P^{-1/2}\xi}{H} \ge \dpn{(P+\eps)^{-1/2}\xi, (P+\eps)^{-1/2}\xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$, 
+    \begin{align*}
+    \dpn{P^{-1/2}A\xi, P^{-1/2}A\xi}{H} &\ge \dpn{(P+\eps)^{-1/2}A\xi, (P+\eps)^{-1/2}A\xi}{H} \\
+    \dpn{(P^{-1/2}A)^*(P^{1/2}A) \xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H}
+    \end{align*}
+    
+    so
+
+    \[
+    0 \le Q - (P^{-1/2}A)^*(P^{1/2}A) \le Q - A^*(P+\eps)^{-1}A
+    \]
+
+    and $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ by (P2) of \autoref{proposition:schur-complement-positive}. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:schur-complement-positive-estimate}
+    Let $H$ be a complex Hilbert space and $P, Q, A \in B(H)$ such that
+    \[
+    \begin{bmatrix}
+        P & A \\
+        A^* & Q
+    \end{bmatrix} \ge 0
+    \]
+
+    then
+    \begin{enumerate}
+        \item For each $\xi \in H$, 
+        \[
+        0 \le \dpn{(P + A + A^* + Q)\xi, \xi}{H} \le (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2
+        \]
+
+        \item $\normn{P + A + A^* + Q}_{B(H)} \le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2$. 
+        \item If $P = Q$, then $A^*A \le \norm{P}_{B(H)}P$ and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (1): For each $\xi \in H$, 
+    \[
+    0 \le \angles{\begin{bmatrix}
+        P & A \\
+        A^* & Q
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\ \xi
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \xi
+    \end{bmatrix}}_{H^2} = \dpn{(P + A + A^* + Q)\xi, \xi}{H}
+    \]
+
+    and
+    \begin{align*}
+        \dpn{A + A^*\xi, \xi}{H} &= 2\text{Re}\dpn{A + A^*\xi, \xi}{H} \\
+        |\dpn{A + A^*\xi, \xi}{H}| &\le 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} \\
+        \dpn{(P + A + A^* + Q)\xi, \xi}{H} &\le \dpn{P\xi, \xi}{H} + 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H} \\
+        &= (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\
+    \end{align*}
+
+    (2): By (3) of \autoref{proposition:bh-numerical-radius-gymnastics} and (1),
+    \begin{align*}
+    \norm{P + A + A^* + Q}_{B(H)} &= \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{(P + A + A^* + Q)\xi, \xi}{H} \\
+    &\le \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}(\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\
+    &\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2
+    \end{align*}
+    
+    (3): Using (2) of \autoref{proposition:schur-complement-positive}, assume without loss of generality that $P \ne 0$. For each $\xi, \eta \in H$, 
+    \begin{align*}
+        0 &\le \angles{\begin{bmatrix}
+        P & A \\
+        A^* & P
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}}_{H^2} \\
+    &= \dpn{P\xi, \xi}{H} + \dpn{P\eta, \eta}{H} + \dpn{A\eta, \xi}{H} + \dpn{A^*\xi, \eta}{H}
+    \end{align*}
+
+    In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$, 
+    \begin{align*}
+        0 &\le \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}^2} + \dpn{P\eta, \eta}{H} - \frac{\dpn{A\eta, A\eta}{H}}{\norm{P}_{B(H)}} - \frac{\dpn{A^*A\eta, \eta}{H}}{\norm{P}_{B(H)}} \\
+        &= \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}} + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2  \\
+        &\le\norm{A\eta}_H^2 + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2 \\
+        &= \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - \dpn{A^*A\eta, \eta}{H}
+    \end{align*}
+    
+    so $A^*A \le \norm{P}_{B(H)}P$, and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. 
+    
+\end{proof}