TITLE: Polar decomposition of Bounded Normal Operator on Hilbert Space
QUESTION [9 upvotes]: It is well known that if $T$ is a bounded linear operator on a infinite dimensional Hilbert space $H$ then there exists unique partial isometry $U$ such that $T=U \vert T \vert$,where $\vert T \vert =(T^*T)^{1/2}$.Such a decomposition is called a polar decomposition of $T$.I am trying to solve the following problem:

Suppose $T$ is Bounded Normal operator on $H$ ,then there exists unique unitary operator $U$ such that $T=U \vert T \vert$

I don't have any clean way to show the existence of such a unitary operator.Can someone please give me some idea to prove this?

REPLY [7 votes]: For any bounded operator $T$, the operator $T^{\star}T$ is a positive selfadjoint operator with unique positive square root $|T|=(T^{\star}T)^{1/2}$. And,
$$
           \|Tx\|^{2}=(T^{\star}Tx,x)=(|T|^2x,x)=(|T|x,|T|x)=\||T|x\|^{2},\;\;\; x \in H.
$$
The operators $T$ and $|T|$ have the same null spaces, and both maps induce injective linear maps from $H/\mathcal{N}(T)$ into $X$. Therefore, there is a linear map
$$
                     U : \mathcal{R}(|T|) \rightarrow \mathcal{R}(T)
$$
such that $U|T|=T$. $U$ is isometric because $\|U|T|x\|=\|Tx\|=\||T|x\|$. So it extends uniquely by continuity to an isometric map from $\overline{\mathcal{R}(|T|)}$ onto $\overline{\mathcal{R}(T)}$.
Now assume that $T$ is normal. Then $\|Tx\|^{2}=(T^{\star}Tx,x)=(TT^{\star}x,x)=\|T^{\star}x\|^2$, which gives $\mathcal{N}(T)=\mathcal{N}(T^{\star})$. Hence,
$$
  \mathcal{R}(|T|)^{\perp} 
       =\mathcal{N}(|T|)
       =\mathcal{N}(T)
       =\mathcal{N}(T^{\star}) 
       = \mathcal{R}(T)^{\perp}.
$$
So $U$ has a unitary extension to all of $H$ obtained by setting $U=I$ on $\mathcal{N}(T)$. $U$ is not unique because $U$ may be chosen to be any unitary map on $\mathcal{N}(T)$. For example, in the most extreme case where $T=0$, the operator $U$ can be any unitary map of $H$.