2
$\begingroup$

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 $P=|T|$.

$\endgroup$
3
  • $\begingroup$ What do you mean by $\lvert T\rvert$? $P = \sqrt{T^\ast T}$, however. $\endgroup$ Mar 22, 2014 at 19:36
  • 1
    $\begingroup$ It means $P$ is positive semidefinite. Polar decomposition is a product $UP$ such that $U$ is unitary and $P$ is positive semidefinite. $\endgroup$
    – Git Gud
    Mar 22, 2014 at 19:36
  • $\begingroup$ @DanielFischer $|T|={\sqrt{T^*T}}$. His proof suggests he simply means $P$ is positive rather than $P=|T|$. $\endgroup$
    – user134724
    Mar 22, 2014 at 19:51

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Why does Rudin use theorem 12.32 to justify $P$ is positive? If $P=|T|$, then of course $P$ is positive. @Martin Argerami $\endgroup$
    – user134724
    Mar 22, 2014 at 19:59
  • $\begingroup$ In 12.35 b), he is proving that $UP$ is a polar decomposition. $\endgroup$ Mar 22, 2014 at 20:10
  • $\begingroup$ Yes he has to prove $T=U|T|$ and $U$ is unitary. I just don't get the point of proving $P=|T|$ is positive when you already know it. @Martin Argerami $\endgroup$
    – user134724
    Mar 22, 2014 at 20:20
  • $\begingroup$ Or is it simply that he only knows $P=|T|$ at the end of the proof. $\endgroup$
    – user134724
    Mar 22, 2014 at 20:29
  • $\begingroup$ Another thing which is odd is that he implies the polar decomposition is not unique. Yet if $U,U'$ are unitary then clearly $U|T|=U'|T|$ implies $U=U'$ $\endgroup$
    – user134724
    Mar 22, 2014 at 20:53
1
$\begingroup$

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 $|T|$.

$\endgroup$
1
  • $\begingroup$ The point is that restricting to normals guarantees a unitary. (See example above.) $\endgroup$ Jul 14, 2015 at 18:06
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.