# 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?

• 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

• Hi, I'm trying to do the same exercise (decompose a hilbert space as an orthogonal sum of cyclic subspaces for some algebra closed under adjoints) and I'm confused about one aspect of your proof. Can the following situation ever arise: you have $\bigoplus E_i \subsetneqq H$ and $\overline{\bigoplus E_i} =H$?. In this case, the orthogonal complement is $0$ so your induction wouldn't continue but you still haven't exhausted everything. – Augie March Apr 30 '18 at 20:18