Beefed up the Schur Complement Condition.

src/op/system/completely.tex

@@ -215,89 +215,194 @@
 
 \begin{lemma}[Schur Complement Condition for Positivity]
 \label{lemma:schur-complement-positive}
-    Let $H$ be a complex Hilbert space, $x, y \in B(H)$, and $z \in G(B(H))$, then
+    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
         \[
         \begin{bmatrix}
-            x &y \\
+            P &A \\
-            y^* &z
+            A^* &Q
-        \end{bmatrix} \ge 0
+        \end{bmatrix} = \begin{bmatrix}
+            P^{1/2} &0 \\
+            A^*P^{-1/2} &I
+        \end{bmatrix}\begin{bmatrix}
+            I &0 \\
+            0 & Q - AP^{-1}A^*
+        \end{bmatrix}\begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
+        \end{bmatrix}
         \]
-    if and only if $z \ge 0$ and $x - yz^{-1}y^* \ge 0$. 
+        
-\end{lemma}
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$.
-\begin{proof}
+    \end{enumerate}
-    ($\Leftarrow$): Suppose that $z \ge 0$ and $x - yz^{-1}y^* \ge 0$, then since $z$ is invertible, 
+
-    \begin{align*}
+    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
+        \[
         \begin{bmatrix}
-            x &y \\
+            P &A \\
-            y^* &z
+            A^* &Q
-    \end{bmatrix} &= \begin{bmatrix}
+        \end{bmatrix} = \begin{bmatrix}
-        I &yz^{-1} \\
+            Q^{1/2} &Q^{-1/2}A \\
             0 &I
-    \end{bmatrix}\begin{bmatrix}
-        x - yz^{-1}y^* &0 \\
-        0 &z
         \end{bmatrix}\begin{bmatrix}
             I &0 \\
-        z^{-1}y^* &I
+            0 & P - A^*Q^{-1}A
+        \end{bmatrix}\begin{bmatrix}
+            Q^{1/2} &0 \\
+            A^*Q^{-1/2} &I
+        \end{bmatrix}
+        \]
+        
+        \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}
-        I &0 \\
+            P^{1/2} &0 \\
-        z^{-1}y^* &I
+            A^*P^{-1/2} &I
-    \end{bmatrix}^*\begin{bmatrix}
-        x - yz^{-1}y^* &0 \\
-        0 &z
         \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 \\
-        z^{-1}y^* &I
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} \begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
         \end{bmatrix} 
     \end{align*}
     
-    Since the middle factor is positive, the given matrix is positive as well.
+    and
     
-    ($\Rightarrow$): Now suppose that the given matrix is positive, then for any $\xi \in H$, 
     \[
-    \dpn{z\xi, \xi}{H} =  \angles{\begin{bmatrix}
+    \begin{bmatrix}
-            x &y \\
+        P &A \\
-            y^* &z
+        A^* &Q
+    \end{bmatrix} = \begin{bmatrix}
+        P^{1/2} &0 \\
+        A^*P^{-1/2} &I
     \end{bmatrix}\begin{bmatrix}
-            0 \\
+        I &0 \\
-            \xi
+        0 & Q - A^*P^{-1}A
-        \end{bmatrix}, \begin{bmatrix}
+    \end{bmatrix}\begin{bmatrix}
-            0 \\
+        P^{1/2} &P^{-1/2}A \\
-            \xi
+        0 &I
-        \end{bmatrix}}_{H^2} \ge 0
+    \end{bmatrix}
     \]
 
-    so $z \ge 0$. As such,
+    (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. 
-    \begin{align*}
+
+    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, 
+    \[
     \begin{bmatrix}
-        x - yz^{-1}y^* &0 \\
+            I &0 \\
-        0 &z
+            0 & Q - A^*P^{-1}A
-    \end{bmatrix} &= \begin{bmatrix}
+        \end{bmatrix} = \begin{bmatrix}
-        I &-yz^{-1} \\
+            P^{-1/2} &0 \\
+            -P^{-1}A^* &I
+        \end{bmatrix}\begin{bmatrix}
+        P &A \\
+        A^* &Q
+    \end{bmatrix} \begin{bmatrix}
+            P^{-1/2} &-P^{-1}A \\
             0 &I
-    \end{bmatrix}\begin{bmatrix}
+        \end{bmatrix} 
-            x &y \\
+    \]
-            y^* &z
+
-    \end{bmatrix}\begin{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. 
-        I &0 \\
+
-        -z^{-1}y^* &I
+    (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}, 
-    \end{bmatrix} \\ 
+    \begin{align*}
-    &= \begin{bmatrix}
+        |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\
-        I &0 \\
+        &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\
-        -z^{-1}y^* &I
+        &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\
-    \end{bmatrix}^*\begin{bmatrix}
+        &= \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2}
-            x &y \\
-            y^* &z
-    \end{bmatrix}\begin{bmatrix}
-        I &0 \\
-        -z^{-1}y^* &I
-    \end{bmatrix} \\
     \end{align*}
     
-    Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^* \ge 0$ as well. 
+    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}
     \label{example:transpose-not-completely-positive}
     Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive.