TITLE: Hermitian Operators and the Spectral Theorem
QUESTION [6 upvotes]: I understand that in a finite-dimensional vector space $V$, a diagonalizable linear operator $T: V \to V$ decomposes $V$ into a direct sum of its invariant eigenspaces, on each of which it restricts to a scalar multiple of a projection. The Spectral Theorem says that this occurs if $T$ is Hermitian, and I'm trying to piece together a geometric intuition for the proof.
Here's my thinking: we know that the eigenspace $E$ of some eigenvector $v$ is $T$-invariant, but we need something extra (presumably involving the Hermiticity of $T$) to show that its orthogonal complement $E^{\perp}$ is also invariant: that is, we know that $Tv$ will stay inside $E$, but Hermiticity is required for $Tw$ (for some $w \notin E$) to stay outside of $E$. Once this is done, we can just restrict to $E^{\perp}$ and play the same game there, so that by induction we separate out $n = \dim{V}$ orthogonal directions to get our eigenbasis.
So what is it about Hermiticity that "stabilizes" $E^{\perp}$ under $T$? I'm not looking for a rigorous proof, but rather a natural explanation of the geometrical significance of self-adjointness. More generally, why does requiring $T$ to be normal make the Spectral Theorem biconditional?

REPLY [2 votes]: This is more of an extended comment about some hidden algebraic structure having to do with the normal operator property. This is not a direct answer because this isn't "geometric", though it fits into a larger blocks of conceptual intuition (representation theory), which is maybe close to what you mean anyway...:
If $T^* T = T T^*$,then the algebra $A = \mathbb{C}[T,T^*]$ is commutative.
From Schur's lemma we know that the irreducible finite dimensional representations of $A$ are 1 dimensional. See * below.
Then if $W$ is an $A$ invariant subspace of $V$, so is its orthogonal complement (easy computation, using that $T$ and $T^*$ fix this subspace, $<Tv, w> = <v,T^*w> = <v,w'> = 0$, and the same for $T^*v$...).
Hence, if $W$ is more than one dimensional, it isn't irreducible, so has an invariant subspace, and splits around it. By induction on dimension can prove that $V$ decomposes into orthogonal irreducibles, which are necessarily one dimensional.
These one dimensional invariant subspaces are necessarily eigenspaces of both $T$ and $T^*$, and this gives the spectral decomposition.
*Digging back into where $T^*T = TT^*$ is used here, it is in the application of Schur's lemma. Because of normality, $T$ and $T^*$ are both intertwiners for any $A$ module, and hence must act by scalars on irreducible $A$ modules (here we use the existence of an eigenvector, which requires working over $\mathbb{C}$). This implies that irreducible $A$ modules are 1 dimensional.
(About generalization to infinite dimensions : Unfortunately, the proof I know of Schur's lemma in the infinite dimensional setting requires the spectral theorem, since there isn't always going to be an eigenvector, so I don't think this argument generalizes to give intuition about that case... though if someone knows an extension I'd love to hear it.)