Editing Spectral theorem (section)

===Cyclic vectors and simple spectrum===
A vector <math>\varphi</math> is called a [[cyclic vector]] for <math>A</math> if the vectors <math>\varphi,A\varphi,A^2\varphi,\ldots</math> span a dense subspace of the Hilbert space. Suppose <math>A</math> is a bounded self-adjoint operator for which a cyclic vector exists. In that case, there is no distinction between the direct-integral and multiplication-operator formulations of the spectral theorem. Indeed, in that case, there is a measure <math>\mu</math> on the spectrum <math>\sigma(A)</math> of <math>A</math> such that <math>A</math> is unitarily equivalent to the "multiplication by <math>\lambda</math>" operator on <math>L^2(\sigma(A),\mu)</math>.<ref>{{harvnb|Hall|2013}} Lemma 8.11</ref> This result represents <math>A</math> simultaneously as a multiplication operator ''and'' as a direct integral, since <math>L^2(\sigma(A),\mu)</math> is just a direct integral in which each Hilbert space <math>H_{\lambda}</math> is just <math>\mathbb{C}</math>.

Not every bounded self-adjoint operator admits a cyclic vector; indeed, by the uniqueness in the direct integral decomposition, this can occur only when all the <math>H_{\lambda}</math>'s have dimension one. When this happens, we say that <math>A</math> has "simple spectrum" in the sense of [[Self-adjoint operator#Spectral multiplicity theory|spectral multiplicity theory]]. That is, a bounded self-adjoint operator that admits a cyclic vector should be thought of as the infinite-dimensional generalization of a self-adjoint matrix with distinct eigenvalues (i.e., each eigenvalue has multiplicity one).

Although not every <math>A</math> admits a cyclic vector, it is easy to see that we can decompose the Hilbert space as a direct sum of invariant subspaces on which <math>A</math> has a cyclic vector. This observation is the key to the proofs of the multiplication-operator and direct-integral forms of the spectral theorem.