polar decomposition in von Neumann algebras
 Let $\mathcal{M}$ be a von Neumann algebra^{} acting on a Hilbert space^{} $H$ and $T\in \mathcal{M}$. If $T=VR$ is the polar decomposition^{} for $T$ with $KerV=KerR$, then both $V$ and $R$ belong to $\mathcal{M}$.
Proof :

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As $\mathcal{M}$ is a ${C}^{*}$algebra^{} (http://planetmath.org/CAlgebra), it is known that $R=\sqrt{{T}^{*}T}$ belongs to $\mathcal{M}$. (proof will be added later)

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To see that $V$ also belongs to $\mathcal{M}$, by the double commutant theorem, it suffices to show that $V$ belongs to ${\mathcal{M}}^{\prime \prime}$ (the double commutant of $\mathcal{M}$).
Suppose $S\in {\mathcal{M}}^{\prime}$. We intend to prove that $V$ commutes with $S$.
For $x\in H$ we have that
$$TSx=STx=SVRx$$ and
$$TSx=VRSx=VSRx$$ So $SV$ and $VS$ agree on $\overline{RanR}$.
As $R$ is selfadjoint^{}, ${\overline{RanR}}^{\u27c2}=KerR$, and so it remains to show that $SV$ and $VS$ agree on $KerR$. Recall that, by hypothesis, $KerR=KerV$.
Let $x\in KerR$. We have that $RSx=SRx=0$ and therefore
$$S(KerR)\subseteq KerR=KerV$$ and so we can conclude that $VS$ is identically zero in $KerR$.
Clearly $SV$ is also identically zero on $KerR=KerV$.
Thus $VS$ and $SV$ agree on $KerR$. Therefore $SV=VS$ and so $V\in {\mathcal{M}}^{\prime \prime}=\mathcal{M}\mathrm{\square}$
Title  polar decomposition in von Neumann algebras 

Canonical name  PolarDecompositionInVonNeumannAlgebras 
Date of creation  20130322 17:28:54 
Last modified on  20130322 17:28:54 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  7 
Author  asteroid (17536) 
Entry type  Result 
Classification  msc 47A05 
Classification  msc 46L10 