# 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

## 1 Answer

Here's how the argument goes: pick any nonzero vector v_0 \in H_0=H, and consider the A-invariant closed subspace E_0 generated by v (i.e., E_0 is the closure of the space of q(A)v_0 for q polynomial). E_0 is A-invariant and has v as a cyclic vector. Now let H_1 be the orthogonal complement of E_0; since A is self-adjoint, H_1 will also be A-invariant. We can now repeat the argument and pick v_1 in H_1 and let E_1 be the A-invariant closed subspace generated by v_1. We then let H_2 be the orthogonal complement of E_1 in H_1, or equivalently the orthogonal complement of E_0+E_1 in H. We keep repeating this by induction.

When does this process stop? Well if eventually H_n=0, we can't pick a nonzero vector v_n. We now have all of H decomposed as an orthogonal direct sum E_0+E_1+...+E_{n-1}, and each E_i is cyclic. However, there's no reason to expect H_n to be 0 for any finite n. In this case, what we can do is write H_\omega for the intersection of all the H_n (the orthogonal complement of the sum of all the E_n). If H_\omega=0, we are done since H is the sum of all the E_n. If not, we pick v_\omega \in H_\omega nonzero and get another E_\omega, and then let H_{\omega+1} be the complement of E_\omega in H_\omega, and so on.

We can keep doing this "forever" by transfinite induction, breaking off cyclic A-invariant subspaces one by one and taking intersections at infinite limits, until eventually some H_\alpha must be 0 (if nothing else, for \alpha greater than the cardinality of H H_\alpha must be 0!). We then have H as a direct sum of all the E_\beta for \beta<\alpha.

Incidentally, in the case that A is the identity, this is exactly how you get an orthonormal basis for an arbitrary Hilbert space, or a basis for any vector space (replacing orthogonal complement by quotient in that case). More generally, this shows that over a semisimple ring, every module is a direct sum of cyclic modules (which can in fact be chosen to be simple).

Also, if you assume H is separable, you can rig the construction to guarantee that it stops after only \omega steps (i.e., H_\omega=0) and thus not have to worry about tranfinite induction. Let {e_n} be a basis for H, and when you choose v_n choose it so that e_n is in the span of {v_0,...,v_n}. Every e_n will then be in the span of the E_n, so they must be all of H.

-