computation with polar decomposition of bounded operator on hilbert space I am trying to prove the following homework problem: Let $T \in B(H)$ (so $T$ is a bounded operator on a Hilbert space $H$), and let $T = U|T|$ be the polar decomposition of $T$.  Prove that if $T$ is a normal operator, then $U|T| = |T|U$ and $U$ is normal.
The definition of $|T|$ is the unique positive square root of $T^*T$.  An operator $T$ is normal if $TT^* = T^*T$.
I've been messing around with calculations based on definitions (in my class, we are not assuming that $U$ is unitary, although we call it the polar part of $T$).  However, I'm stuck.  I found the following link: Polar decomposition normal operator and looked up the theorem of Rudin that is mentioned, but it made use of some things we haven't covered in my class.  Can anyone help?  I feel like it is staring me in the face and I just can't see it.
 A: Let $T = U|T|$ where $U=0$ on $\mathcal{R}(|T|)^{\perp}=\mathcal{N}(|T|)$. That uniquely defines a $U$ on the full space for which $T=U|T|$. For $\epsilon > 0$, define
$$
\begin{align}
         U_{\epsilon} & = T(|T|+\epsilon I)^{-1} \\
                  & = U|T|(|T|+\epsilon I)^{-1} \\
                  & = U\{ I-\epsilon(|T|+\epsilon I)^{-1}\}
\end{align}
$$
You can show that $U_{\epsilon}\rightarrow U$ in the strong operator topology as $\epsilon \downarrow 0$. That's enough to show that $T$ commutes with $U_{\epsilon}$ if $T$ is normal. The convergence result holds even if $T$ is not normal, and the result can be useful in other ways.
To prove $U_{\epsilon}\rightarrow U$ in the strong topology, it is enough to show that $\|U_{\epsilon}\|$ is uniformly bounded as $\epsilon\downarrow 0$, and to show that $U_{\epsilon}x\rightarrow Ux$ for all $x$ in the dense subspace $\mathcal{R}(|T|)+\mathcal{N}(|T|)$. First, a uniform bounded is
$$
       \|U_{\epsilon}\| \le 1+\|\epsilon(|T|+\epsilon I)^{-1}\| \le 2.
$$
To prove strong convergence on the dense subspace  $\mathcal{R}(|T|)+\mathcal{N}(|T|)$, it is enough to prove strong convergence separately for $x$ in each of these two subspaces. I'll leave it to you to show that $x \in \mathcal{N}(|T|)$ implies $U_{\epsilon}x=0$. Finally, suppose that $x \in \mathcal{R}(|T|)$ so that $x=|T|y$ for some $y$. Then,
$$
        U_{\epsilon}x = U\{ |T|y - \epsilon|T|y+\epsilon^{2}(|T|+\epsilon I)^{-1}y\}\rightarrow U|T|y=Ux.
$$
