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 $P=|T|$.
Asked
Active
Viewed 815 times
2
-
What do you mean by $\lvert T\rvert$? $P = \sqrt{T^\ast T}$, however. – Daniel Fischer Mar 22 '14 at 19:36
-
1It means $P$ is positive semidefinite. Polar decomposition is a product $UP$ such that $U$ is unitary and $P$ is positive semidefinite. – Git Gud Mar 22 '14 at 19:36
-
@DanielFischer $|T|={\sqrt{T^*T}}$. His proof suggests he simply means $P$ is positive rather than $P=|T|$. – Mar 22 '14 at 19:51
3 Answers
3
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.
Martin Argerami
- 205,756
-
Why does Rudin use theorem 12.32 to justify $P$ is positive? If $P=|T|$, then of course $P$ is positive. @Martin Argerami – Mar 22 '14 at 19:59
-
In 12.35 b), he is proving that $UP$ is a polar decomposition. – Martin Argerami Mar 22 '14 at 20:10
-
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 – Mar 22 '14 at 20:20
-
-
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 $U=U'$ – Mar 22 '14 at 20:53
-
The way I see the proof, he uses functional calculus to get $U$ and $P$. Since I haven't been reading the book, it is not obvious to me if at that stage of the book it is obvious that if $f(t)=|t|$, then $f(T)=(T^*T)^{1/2}$. – Martin Argerami Mar 22 '14 at 21:23
-
As for your last comment, that is not true. Let $$T=\begin{bmatrix}1&0\0&0\end{bmatrix}, \ U=\begin{bmatrix}1&0\0&1\end{bmatrix},\ U'=\begin{bmatrix}1&0\0&-1\end{bmatrix}.$$ Then $U|T|=U'|T|$, with $U\ne U'$. – Martin Argerami Mar 22 '14 at 21:24
1
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
C-star-W-star
- 16,275
1
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 $|T|$.
Disintegrating By Parts
- 87,459
- 5
- 65
- 149
-
The point is that restricting to normals guarantees a unitary. (See example above.) – C-star-W-star Jul 14 '15 at 18:06