Hermitian Operators and the Spectral Theorem
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 $W = \operatorname{span} \{ w \}^{\perp}$.
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:
- Find one eigenvector $v \in V$ of $T$ and set $E = \operatorname{span} \{ v \}$.
- Show that $E^{\perp}$ is $T$-invariant.
- 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.
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.)