Polar decomposition normal operator Let $T \in \mathcal{B}(\mathcal{H})$ be normal.
I have to show that there exists a unitary operator $U \in \mathcal{B}(\mathcal{H})$
such that $T^*=UT$ and give necessary and sufficient conditions on $T$ in terms of the norm
of $T$, the range of $T$, the kernel of $T$, or something similar for $U$ to be unique.
For the first part of the question I use Theorem 12.35(b) of Walter Rudin's Functional Analysis, which states:

If $T \in \mathcal{B}(\mathcal{H})$ is normal, the T has a polar decomposition $T=UP$ in which $U$ and $P$ commute with each other and with $T$.
Here $U$ is unitary and $P\geq 0$.

Thus there exists unitary $V$ and a $P \geq 0$ such that $T=VP$.
Then
$$
T^* = (VP)^* = (PV)^* = V^*P^* = V^*P,
$$
since $VP$ commute and $P=P^*$.
Also
$$
T=VP \implies V^*T=V^*VP = P,
$$
since $V$ is unitary. Combining the last two equalities gives
$$
T^*=V^*P=V^*V^*T=UT,
$$
where $U=(V^*)2$.
For the second part of the question im still puzzling. Since $T$ is normal I have
$||Tx||=||T^*x||$ and from theorem 12.34 that the positive square root of $T^*T$ is the only operator
such that $||\sqrt{T^*T}x||=||Tx||$. Combining with the first part gives
$$
||Tx||=||T^*x||=||UTx||=||\sqrt{T^*T}x||.
$$
I don't know how to proceed from here. Any help would be appreciated.
 A: I'm assuming you are asking about the uniqueness of $U$ (at least that's what Rudin's exercise asks). 

The result is that $U$ is unique if and only if $T$ has dense range. 

Note that, since $U$ is a unitary, $$\tag{1}(\mbox{Im}\,T^*)^\perp=\ker T=\ker T^*=(\mbox{Im}\,T)^\perp.$$ 
Assume first that $U$ is unique. Let $P$ be the orthogonal projection onto $(\mbox{Im}\,T)^\perp$. Note that $U$ maps $\overline{\mbox{Im}\,T}$ into $\overline{\mbox{Im}\,T^*}=\overline{\mbox{Im}\,T}$. This means that $U(I-P)=(I-P)U(I-P)$. Multiplying by $P$ on the left, $PU(I-P)=0$. We can repeat the same reasoning for $T^*$, to get $PU^*(I-P)=0$. So $PU=PUP$, $PU^*=PU^*P$, and taking adjoint we get $UP=PUP=PU$. Now consider, for $\lambda\in\mathbb C$ with $|\lambda|=1$, $V=\lambda P+U(I-P)$; this is a unitary, since 
$$
V^*V=(\overline\lambda P+(I-P)U^*)(\lambda P+U(I-P))=I+2\mbox{Re}\,\overline\lambda PU(I-P)=I,
$$
and 
$$
VV^*=(\lambda P+U(I-P))(\overline\lambda P+(I-P)U^*)=P+U(I-P)U^*=P+(I-P)=I.
$$
Also, $VT=\lambda PT+U(I-P)T=0+(I-P)UT=(I-P)T^*=T^*$. By the uniqueness $U=V$, that is $\lambda P+U(I-P)=U=UP+U(I-P)$. So $\lambda P=UP$. As $\lambda$ was arbitrary, this shows that $P=0$. That is, $T$ has dense range.
Conversely, if $T$ has dense range, then $T$ and $T^*$ are injective, by $(1)$. Given $y\in\mbox{Im}\,T$, there exists $x$ such that $y=Tx$. This $x$ is unique by the injectivity of $T$. Then $Uy=UTx=T^*x$, so $U$ is defined uniquely on a dense subspace, and so it is defined uniquely. 
