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Proofreading.
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Bokuan Li
2026-10-14 19:52:42 -04:00
parent df08e24cd4
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src/op/system/two.tex
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\section{On the Subject of 2×2 Matrices} \section{On 2×2 Matrices}
\label{section:op-sys-2by2} \label{section:op-sys-2by2}
\begin{lemma} \begin{lemma}
@@ -194,7 +194,7 @@
\item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. \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 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 $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^*(P^{-1/2}A) \ge 0$.
\item $Q \ge 0$, $A^*(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A)^*(Q^{-1/2}A) \ge 0$. \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{enumerate}
\end{proposition} \end{proposition}
\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ] \begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
@@ -267,7 +267,7 @@
so so
\[ \[
0 \le Q - (P^{-1/2}A)^*(P^{1/2}A) \le Q - A^*(P+\eps)^{-1}A 0 \le Q - (P^{-1/2}A)^*(P^{-1/2}A) \le Q - A^*(P+\eps)^{-1}A
\] \]
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 $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. 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 $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
@@ -310,8 +310,8 @@
and and
\begin{align*} \begin{align*}
\dpn{A + A^*\xi, \xi}{H} &= 2\text{Re}\dpn{A + A^*\xi, \xi}{H} \\ \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{(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 + 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 \\ &= (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\
\end{align*} \end{align*}
@@ -339,7 +339,7 @@
In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$, In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$,
\begin{align*} \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)}} \\ 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 \\ 0 &\le \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 \\ &\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} &= \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - \dpn{A^*A\eta, \eta}{H}
\end{align*} \end{align*}