Spectral Theorem for normal operators

Question

I want to prove this in the infinite dimensional Hilbert space case. What is the easiest way to go about this (What do I need to know, what theorems do I need,etc). My aim is to show every normal operator admits invariant subspaces.

Answer 1

In my view, this is the shortest way to show that a normal operator has proper invariant subspaces. It requires some familiarity with von Neumann algebras, though.

Let $N\in B(H)$ be a normal operator. The von Neumann algebra $W^*(N)$ generated by $N$ is abelian, and so it cannot be dense in $B(H)$. This means that its commutator $W*(N)'$ is non-trivial (i.e. it is not the scalar multiplies of the identity). Any non-trivial von Neumann algebra has non-trivial projections. The range of any of these projections will be invariant for $N$.

Answer 2

To rephrase what Martin Argerami said above in a more earthy way: you need to know the spectral theorem for normal operators. Once you prove that every normal operator $T$ is unitarily equivalent to a multiplication operator on $L^2(\sigma(T))$, you're done.