# Hilbert Space as direct sum of subspaces with cyclic vectors

Ok,so this should be easy, however I havent taken functional analysis for a while. But given a compact self-adjoint operator on a hilbert space H(over the complex numbers), we define v to be a cyclic vector if and only if the family q(A)v for complex polynomials is dense in H. Halmos, in the article "What does the spectral theorem say?", claims that a hilbert space (additional assumptions on its structure?) can be decomposed as a direct sum of subspaces so that the restriction of A to these spaces has a cyclic vector. He says it can be proved by "a standard transfinite argument." Well Im not up on my standards, and I was wondering if someone could break this problem down (pun intended) for me, preferably as suggesting the tools Ill need to carry out a proof?

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Of course, if you're willing to put the cart before the horse, then the spectral theorem says that $H\cong\bigoplus L^2(\mathbb R,\mu)$ where $A$ is multiplication by $x$ in each of these spaces, so $f\equiv 1$ is cyclic (by the Weierstrass approximation theorem). – Christian Remling Jun 5 '14 at 4:07