Theorem 38.2.5 (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 $T$, and
- (5)
$V$ is a partial isometry from $\ker(T)^{\perp}$ to $\ol{T(H)}$.
- (6)
$P$ and $V$ are contained in the von Neumann algebra generated by $T$.
Proof, [Theorem 12.8, Theorem 18.9, Zhu93]. (1): Let $P = |T| = \sqrt{T^{*}T}$, then $P$ is positive (1).
(2): 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.
(3): For any $x \in H$, $Tx = V_{0}Px = VPx$.
(4), (5): Since the initial space of $V$ is $\ol{P(H)}= \ker(T)^{\perp}$, $\ker(V) = P(H)^{\perp} = \ker(P)$.
(Uniqueness): Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By Proposition 38.2.3, $W^{*}W$ 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.
Since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^{\perp}$, $V = W$ on $H$, and the polar decomposition is unique.
(6): Let $A$ be the von Neumann algebra generated by $T$. By Theorem 37.3.1, $A$ is a unital $C^{*}$-algebra, so $P = \sqrt{T^{*}T}\in A$. To see that $V \in A$, it is sufficient to apply the Bicommutant Theorem.
To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^{\perp}$, $SV|_{\ker(P)}= 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)}= 0$ as well. Therefore $V \in A'' = A$.$\square$
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