Hermitian Operators and the Spectral Theorem

Question

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?

Answer 1

I think that this is a great question that is usually unasked and unanswered. This is quite unfortunate because it has a very simple answer:

We can use the adjoint operator $T^{*}$ to detect $T$-invariant subspaces of codimension one.

Let me show you how this works. Assume $W \subseteq V$ is a codimension one subspace (geometrically, a hyperplane). Let us choose a normal vector $0 \neq w^{\perp} \in W^{\perp}$ to $W$. If the vector $w^{\perp}$ is an eigenvector of $T^{*}$ then $W$ is a $T$-invariant subspace. To see why, let $w \in W$ and compute

$$ \left< Tw, w^{\perp} \right> = \left< w, T^{*}w^{\perp} \right> = \left< w, \lambda w^{\perp} \right> = \lambda \left< w, w^{\perp} \right> = 0 $$

which shows that $Tw \in W$.

Stated differently, this observation shows that any eigenvector $w$ of $T^{*}$ gives us a $T$-invariant codimension one subspace $W = \operatorname{span} \{ w \}^{\perp}$.

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Given the result above, what kind of condition we can impose on an operator $T$ which guarantees that if $v \in V$ is an eigenvector of $T$ then $W = \operatorname{span} \{ v \}^{\perp}$ is $T$-invariant? Well, we can try and guarantee that if $v$ is any eigenvector of $T$ then $v$ is also an eigenvector of $T^{*}$ (possibly with a different eigenvalue).

One condition which guarantees the above is $T = T^{*}$ because then $v$ is trivially an eigenvector of $T^{*}$ with the same eigenvalue. A less trivial condition which guarantees the above is $TT^{*} = T^{*}T$ because then if $v \in V$ is an eigenvector of $T$ with eigenvalue $\lambda$ then $v$ is an eigenvector of $T^{*}$ with eigenvalue $\overline{\lambda}$.

So what is the difference between the real and the complex case? The usual proof of the spectral theorem for the real and complex case goes like this:

1. Find one eigenvector $v \in V$ of $T$ and set $E = \operatorname{span} \{ v \}$.
2. Show that $E^{\perp}$ is $T$-invariant.
3. Restrict $T$ to $E^{\perp}$ and repeat the argument above for $T|_{E^{\perp}}$

When $V$ is complex, step one is trivial and doesn't use any property of $T$. To get step $(2)$, we use $T^{*}T = TT^{*}$. Step three is then again trivial because if $T$ was normal on $(V, \left< \cdot, \cdot \right>)$, then $T|_{E^{\perp}}$ will be normal on $(E^{\perp}, \left< \cdot, \cdot \right>|_{E^{\perp}})$.

When $V$ is real, step one is not trivial. The condition $T^{*}T = TT^{*}$ which is enough for step two to carry unfortunately doesn't guarantee that $T$ has even one eigenvector so we can't start the argument. However, the stronger condition $T = T^{*}$ does guarantee that $T$ has at least one eigenvector (this is a non-trivial result). This condition is also enough for step two and step three is again trivial.

Answer 2

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.)