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Show that every normal operator on a separable Hilbert space has a square root that commutes with it. Uniqueness?
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My attempt: Let $T$ be a normal operator. By polar decomposition $T=U|T|$ where $U$ is a partial isometry and $|T|$ is positive. Now using functional calculus $\phi: C(\sigma(|T|))\to C^∗ (|T|,1)$ , there is a sequence $\{f_n \}$ of continuous functions in $C(\sigma(|T|))$ such that $x^{ 1/2} =\lim f_n (x)$ so $|T|^ {1/2}$ is the unique square root of $ |T|$ .

But what about square root of $T$ ? Is it $|T|^{ 1/2}?

I do not use of separablity of Hilbert space $H$ in my attempt. Also I think the exercise is always true, not just for separable Hilbert space. And $|T|^{1/2}$ is always unique. Where is my mistake?

Also if I have mistaken, Please give me an example of a normal operator on a non-separable Hilbert space that its square root is not unique. Thanks.

niki
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  • For uniqueness, think about the identity. – Phillip Andreae Jan 08 '15 at 20:24
  • Take $T ={\rm id}$. This has the square roots $T$ and $-T$ that both commute with $T$ (this works as long as your definition of a square root $S$ is just that $S^2=T$). – PhoemueX Jan 08 '15 at 20:24
  • @PhoemueX : Does it just satisfy for separable Hilbert spaces? – niki Jan 08 '15 at 20:27
  • @PhillipAndreae: Does it just satisfy for separable Hilbert spaces? – niki Jan 08 '15 at 20:52
  • I don't see how your argument gives a square root for $T$. What is your square root? For example, support $T$ is unitary. – Disintegrating By Parts Jan 08 '15 at 23:29
  • @T.A.E. : I added something to the proof which clears my idea. – niki Jan 09 '15 at 06:41
  • @niki : What is your expression for $A$ such that $A^{2}=T$? – Disintegrating By Parts Jan 09 '15 at 06:57
  • @T.A.E.: I think I'm in a wrong way. See $T$ as a continuous function on a subset of $\Bbb C$. We say, a positive linear functional has a unique square. So by polar decomposition $T=U|T|$ where $U$ is a partial isometry and $|T|$ is positive. Now using functional calculus $\phi:C(\sigma(|T|)) \to C^*(|T|,1)$, there is a sequence ${f_n}$ of continuous functions in $C(\sigma(|T|))$ such that $x^{1/2}=\lim f_n(x)$ so $|T|^{1/2}$ is the unique square root of $|T|$. But what about square root of $T$. Is it $|T|^{1/2}$, too? – niki Jan 09 '15 at 07:30
  • @niki: A necessary condition missing to admit any square root is that your operator is positive: $T\geq0$ – C-star-W-star Jan 09 '15 at 15:10
  • @T.A.E. : So do you mean that nonpositive normal operators do not have square root? – niki Jan 09 '15 at 15:47
  • @niki: I'm sorry I really don't get this exercise. Maybe someone else can help? – C-star-W-star Jan 09 '15 at 16:23

2 Answers2

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A square root for an operator is not unique because square roots of complex numbers are not unique. If $A$ is a diagonal matrix on $\mathbb{C}^{n}$, then there are $2^{n}$ possible square roots for $A$ that you can spot right away. It's worse for a general Hilbert space.

If $N$ is bounded and normal on a Hilbert space, then the Spectral Theorem for $N$ gives a Borel spectral measure $E$ for which $$ N = \int \lambda dE(\lambda). $$ If $\sqrt{\lambda}$ is some branch of the square root, then you can define $\sqrt{N}$ by $$ \sqrt{N} = \int \sqrt{\lambda} dE(\lambda). $$ By the Borel functional calculus, $\sqrt{N}^{2}=\int\lambda dE(\lambda) = N$. By the way, you can't get this type of thing using the $C^{\star}$ algebra continuous functional calculus because $\sqrt{\lambda}$ cannot be assumed to be continuous on the spectrum of $N$. It's not that the result fails; it's that the technique of continuous functional calculus fails.

Check what assumptions you have for your version of the spectral theorem, especially concerning separability.

Disintegrating By Parts
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  • https://books.google.com/books?id=gtnMW_uH7DcC&pg=PA47&lpg=PA47&dq=Spectral+theorem+%2B+Conway%27s+a+course+in+operator+theory&source=bl&ots=TAL6CceRfG&sig=vRZQ9C5fs4OcfaIZV2nuqRRiUtA&hl=en&sa=X&ei=OxuwVJPoL5Dtar6VgOAH&ved=0CDoQ6AEwBA#v=onepage&q=Spectral%20theorem%20%2B%20Conway's%20a%20course%20in%20operator%20theory&f=false is the link of Conway's book, about spectral theorem. I can not find any think related to this question so I put it's link. – niki Jan 09 '15 at 18:21
  • I think janko's answer is ok. Is not it? – niki Jan 09 '15 at 18:23
  • @niki I think that there is a mistake in my reasoning. If the unitary has the whole circle in the spectrum, then the function which I proposed for the square root is not continuous on the spectrum. I guess that in this exceptional case one has to use $L^\infty$ calculus for normal operators. Thus, the answer given by T.A.E. is more complete. "My solution" is ok if the spectrum of $U$ is not the whole circle. – Janko Bracic Jan 09 '15 at 18:27
  • @niki : As I mentioned in the problem, the continuous functional calculus does not apply in the case of a general normal $N$. The Borel functional calculus does apply. If the space is separable, then there is a Borel $\sqrt{N}$ which is a strong limit of continuous functions of $N$ because there exists a ray $\mathcal{R}={ re^{i\theta} : 0 < r < \infty}$ which does not intersect the point spectrum of $N$. Then $\int_{\sigma}fdE = \int_{\sigma\setminus \mathcal{R}}fdE$ can be approximated in the strong operator topology by a sequence of continuous function of $N$. – Disintegrating By Parts Jan 09 '15 at 23:08
  • Where does separable Hilbert space use? – niki Jan 15 '15 at 08:36
  • @niki : I don't know where separable is needed in this problem. I don't see it. – Disintegrating By Parts Jan 15 '15 at 13:35
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By Theorem 12.35(b) of Walter Rudin's Functional Analysis, if $T\in B(H)$ is normal, then it has a polar decomposition $T=UP$, where $U$ and $P$ commute with each other and with $T$ and $U$ is unitary and $P≥0$. The spectrum of $P$ is a bounded subset of $[0,\infty)$, hence $\sqrt{z}$ can defined as a nonnegative square root of a nonnegative number. In this way, via the cont. functional calculus for $P$, we have unique positive $\sqrt{P}$. The spectrum of $U$ is a subset of the unit circle $C=\{ e^{it}; -\pi<t\leq \pi\}$. Define square root on $C$ by $\sqrt{e^{it}}=e^{it/2}$. This is well defined Borel function on $\sigma(U)$. Hence there is $\sqrt{U}$ in the von Neumann algebra generated by $U, U^*$ and $I$. By Fuglede theorem $\sqrt{U}$ and $\sqrt{P}$ commute. Hence $(\sqrt{U}\sqrt{P})^2=UP=T$. Note that square roots are not unique. Already the $2\times 2$ identity matrix has a lot of square roots (for each idempotent $P$ one has $(2P-I)^2=I$.

Janko Bracic
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