Normal Operators: Polar Decomposition (Rudin) On page 332 theorem 12.35b) of Rudin functional analysis is show that if T is normal then it has a polar decomposition $T=UP$. Does he mean that $P=|T|$? He's a bit ambiguous as to how he defines polar decomposition in whether $P$ is simply positive or whether $P=|T|$.
 A: Note that Rudin requires the $U$ in the polar decomposition to be a unitary. In that case, if $T=UP$, then 
$$
T^*T=PU^*UP=P^2,
$$
and indeed $P=|T|$. So there is no ambiguity. 
A: Let $X$ be a complex Hilbert space. For any $T\in\mathcal{L}(X)$ (normal or not), there is a unique $P\in\mathcal{L}(X)$ such that $P \ge 0$ and $P^{2}=T^{\star}T$. So your question is confusing. If you define $|T|$ to be a positive square root of $T^{\star}T$, then any $P \ge 0$ for which $P^{2}=T^{\star}T$ must be $|T|$.
A: Rudin restricts to normals as...
Definition
Given a Hilbert space $\mathcal{H}$.
Consider selfadjoint operators:
$$H:\mathcal{D}H\subseteq\mathcal{H}\to\mathcal{H}:\quad H=H^*$$

Define positivity by:*
  $$H\geq0:\iff\sigma(H)\geq0$$

*(They may be unbounded!)
Positive Operators
Given a Hilbert space $\mathcal{H}$.
Consider positive operators:
$$P^{(\prime)}:\mathcal{D}P^{(\prime)}\subseteq\mathcal{H}\to\mathcal{H}:\quad P^{(\prime)}\geq0$$
By functional calculus:
$$P^2=P'^2\implies P=P'$$
That gives uniqueness.
Closed Operators
Given two Hilbert spaces $\mathcal{H}$ and $\mathcal{K}$.
Consider a closed operator:
$$A:\mathcal{D}A\subseteq\mathcal{H}\to\mathcal{K}:\quad A=A^{**}$$
By closedness one obtains:
$$|A|:\mathcal{D}|A|\subseteq\mathcal{H}\to\mathcal{H}:\quad |A|:=\sqrt{A^*A}\geq0$$
By functional calculus:*
$$\mathcal{D}:=\mathcal{D}A=\mathcal{D}|A|$$
By square root lemma:*
$$\|A\varphi\|=\||A|\varphi\|\quad(\varphi\in\mathcal{D})$$
Thus one can find:
$$J\in\mathcal{B}(\mathcal{H},\mathcal{K}):\quad A=J|A|$$
By above and boundedness:
$$\varphi\in\overline{\mathcal{R}|A|}:\quad\|J\varphi\|_\mathcal{K}=\|\varphi\|_\mathcal{H}$$
Concluding closed operators.
Normal Operators
Given one Hilbert space $\mathcal{H}$.
Consider a normal operator:
$$N:\mathcal{D}N\subseteq\mathcal{H}\to\mathcal{H}:\quad N^*N=NN^*$$
Then it is closed:
$$N^*N=NN^*\implies N=N^{**}$$
So the above applies:
$$J\in\mathcal{B}(\mathcal{H}):\quad N=J|N|$$
And it is isometric on:
$$\varphi\in\overline{\mathcal{R}|N|}:\quad\|J\varphi\|_\mathcal{H}=\|\varphi\|_\mathcal{H}$$
By above and normality:**
$$\left(\overline{\mathcal{R}|N|}\right)^\perp=\mathcal{N}|N|=\mathcal{N}N=\mathcal{N}(N^*N)\\
=\mathcal{N}(NN^*)=\mathcal{N}N^*=\left(\overline{\mathcal{R}N}\right)^\perp$$
So it can be chosen unitary.
Partial Isometry
Given one Hilbert space $\ell^2(\mathbb{N})$.
Consider the right shift:
$$A:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A:=R$$
Its modulus is the unit:
$$A^*A=LR=1\implies|A|=1$$
Thus partial isometry is:
$$A=J|A|\implies J=R$$
But it is not unitary:
$$\mathcal{R}J=\mathcal{R}R\neq\ell^2(\mathbb{N})$$
Concluding partial isometry.
(Note it was even isometric!)
*See the thread: Square Root
**See the thread: Kernel
