Eigenspace for Normal Operator is Reducing Subspace

Theorem
Let $H$ be a Hilbert space over $\Bbb F \in \left\{{\R, \C}\right\}$.

Let $A \in B \left({H}\right)$ be a normal operator.

Let $\lambda \in \Bbb F$.

Then $\ker \left({A - \lambda}\right)$ is a reducing subspace for $A$.

Here $\ker$ denotes kernel.

Proof
Since $A$ is normal, we have by Kernel of Linear Transformation is Orthocomplement of Range of Adjoint: Corollary that:


 * $\ker A = \left({\operatorname{ran} A}\right)^\perp$

and in particular, that:


 * $\ker A \subseteq \left({\operatorname{ran} A}\right)^\perp$

Now, by Orthocomplement of Subset of Orthocomplement is Superset:


 * $\operatorname{ran} A \subseteq \left({\ker A}\right)^\perp$

Applying this to the normal operator $A - \lambda$, we find:


 * $\operatorname{ran} \left({A - \lambda}\right) \subseteq \left({\ker \left({A - \lambda}\right)}\right)^\perp$

We are now set up to prove that $\ker \left({A - \lambda}\right)$ is a reducing subspace for $A$.

Let $x \in \ker \left({A - \lambda}\right)$.

Then:

Therefore, $A \ker \left({A - \lambda}\right) \subseteq \ker \left({A - \lambda}\right)$; that is to say, $\ker \left({A - \lambda}\right)$ is $A$-invariant.

Now, let $x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$.

Observe that:


 * $A x = \lambda x + \left({A - \lambda}\right) x$

Now $\left({A - \lambda}\right) x \in \operatorname{ran} \left({A - \lambda}\right)$, and by our derivation above, this means that:


 * $\left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$

In conclusion, since $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ is a linear subspace of $H$, it follows that:


 * $\lambda x + \left({A - \lambda}\right) x \in \left({\ker \left({A - \lambda}\right)}\right)^\perp$

as desired.

Hence both $\ker \left({A - \lambda}\right)$ and $\left({\ker \left({A - \lambda}\right)}\right)^\perp$ have been shown to be $A$-invariant subspaces of $H$.

That is, $\ker \left({A - \lambda}\right)$ is a reducing subspace for $A$.