Show that every normal operator on a separable Hilbert space has a square root that commutes with it. Uniqueness?
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.
Best Answer
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.