Added the polar decomposition.

src/op/example/bounded.tex

@@ -3,9 +3,60 @@
 
 \begin{definition}[$B(H)$]
 \label{definition:hilbert-endomorphism}
-    Let $H$ be a Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$. 
+    Let $H$ be a complex Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$. 
 \end{definition}
 
 
 % 1. Every non-trivial ideal of B(H) contains the finite-rank operators.
 % 2. If H is separable, then the only non-trivial closed idela of B(H) are the compact operators. 
+
+
+\begin{definition}[Partial Isometry]
+\label{definition:partial-isometry}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a \textbf{partial isometry} if $T|_{\ker(T)^\perp}$ is an isometry. In which case, $\ker(T)^\perp$ is the \textbf{initial space} of $T$, and $T(H)$ is the \textbf{final space} of $T$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:partial-isometry-characterisation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. 
+\end{proposition}
+\begin{proof}
+    ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^\perp$, then $\dpn{Tx, Tx}{H} = \norm{x}_H^2$ and $\dpn{T^*Tx, x}{H} = \norm{x}_H^2$. By \hyperref[polarisation]{proposition:polarisation-complex}, for each $x, y \in \ker(T)^\perp$, 
+    \begin{align*}
+        \dpn{T^*Tx, y}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T^*T(x + i^ky), x + i^ky}{H} \\
+        &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x + i^ky, x + i^ky}{H} = \dpn{x, y}{H} 
+    \end{align*}
+
+    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection. 
+
+    ($\Leftarrow$): Suppose that $T^*T$ is a projection, then for each $x \in \ker(T)^\perp$, $\dpn{Tx, Tx}{H} = \dpn{T^*Tx, x}{H} = \norm{x}_H^2$. 
+\end{proof}
+
+\begin{theorem}[Polar Decomposition]
+\label{theorem:hilbert-polar-decomposition}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then there exists a unique pair $(P, V) \in B(H)^2$ such that:
+    \begin{enumerate}
+        \item $P$ is positive.
+        \item $V$ is a partial isometry.
+        \item $T = VP$. 
+        \item $\ker P = \ker V$.
+    \end{enumerate}
+
+    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 12.8]{Zhu}}}. ]
+    Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). For each $x \in H$, 
+    \[
+        \norm{Px}_H^2 = \dpn{Px, Px}{H} = \dpn{P^*Px, x}{H} = \dpn{T^*Tx, x}{H} = \norm{Tx}_H^2
+    \]
+
+    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry (2). Moreover, for any $x \in H$, $Tx = V_0Px = VPx$ (3).
+
+    Finally, since the initial space of $V$ is $\ol{P(H)}$, $\ker(V) = P(H)^\perp = \ker(P)$ (4). 
+    
+    It remains to show uniqueness. Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
+
+    Now, since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+\end{proof}
+
+