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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.

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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

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.

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