Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I saw in several texts, as a part of the spectral theorem for unitary operators, that given a unitary operator $U$ on a Hilbert space $H$ (say it is separable), $H$ can be decomposed as an orthogonal direct sum (finite or countable) of cyclic sub-spaces (i.e. spaces of the form $\operatorname{cls}(\operatorname{span}\{U^nx/n\in\mathbb{Z}\})$ for some vector $x$).

I couldn't find a proof for that, so if someone could give me a reference or a sketch of the proof it would be great.

share|improve this question
    
I think you can find the answer in "A Course in Functional Analysis" by Conway. I don't have it with me right now, but I'm almost positive it is in there. –  Nonliapunov Sep 25 '12 at 8:36
    
The spectral theorem for unitary operators is part of the spectral theorem for normal operators. Or if you prefer the spectral theorem for (possibly unbounded) self-adjoint operators, it's basically equivalent to that. –  Robert Israel Aug 31 '13 at 1:24

1 Answer 1

Take the following "typical" unitary operator: on the Hilbert space $L^2(\mathbb{T}, \mu)$ where $\mu$ is a finite Borel measure on the circle, define

$$ V f(z) = z f(z). $$

Then certainly the claim holds for $V$; the cyclic vector can be any trigonometric monomial and there is just one summand. (The trigonometric polynomials are dense by Stone-Weierstrass and regularity arguments.)

But the spectral theorem says an arbitrary unitary on a separable Hilbert space is unitarily equivalent to a countable direct sum of such $V$'s.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.