35.2 $B(H)$
Definition 35.2.1 ($B(H)$).label Let $H$ be a complex Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$.
Definition 35.2.2 (Partial Isometry).label Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if $T|_{\ker(T)^\perp}$ is an isometry. In which case, $\ker(T)^{\perp}$ is the initial space of $T$, and $T(H)$ is the final space of $T$.
Proposition 35.2.3.label 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.
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 polarisation, for each $x, y \in \ker(T)^{\perp}$,
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 $x \in \ker(T)^{\perp}$, $\dpn{Tx, Tx}{H}= \dpn{T^*Tx, x}{H}= \norm{x}_{H}^{2}$.$\square$
Theorem 35.2.4 (Polar Decomposition).label 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:
- (1)
$P$ is positive.
- (2)
$V$ is a partial isometry.
- (3)
$T = VP$.
- (4)
$\ker P = \ker V$.
The pair $(P, V)$ is the polar decomposition of $T$.
Proof, [Theorem 12.8, Zhu93]. Let $P = |T| = \sqrt{T^{*}T}$, then $P$ is positive (1). For each $x \in H$,
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 $x \in H$, $Tx = V_{0}Px = VPx$ (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 Proposition 35.2.3, $W^{*}W$ 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.$\square$
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