Eigenspace for Normal Operator is Reducing Subspace

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Theorem

Let $\HH$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$.

Let $A \in \map B \HH$ be a normal operator.

Let $\lambda \in \Bbb F$.


Then $\map \ker {A - \lambda}$ is a reducing subspace for $A$.

Here $\ker$ denotes kernel.


Proof

We are given that $A$ is normal.

Hence by Kernel of Linear Transformation is Orthocomplement of Image of Adjoint: Corollary:

$\ker A = {\Rng A^\perp$

and in particular, that:

$\ker A \subseteq \Rng A^\perp$

Now, by Orthocomplement of Subset of Orthocomplement is Superset:

$\Rng A \subseteq \ker A^\perp$

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

$\Rng {A - \lambda} \subseteq \paren {\map \ker {A - \lambda} }^\perp$


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

Let $x \in \map \ker {A - \lambda}$.

Then:

\(\ds A x\) \(=\) \(\ds \lambda x + \paren {A - \lambda} x\)
\(\ds \) \(=\) \(\ds \lambda x\) Definition of Kernel of Linear Transformation, $x \in \map \ker {A - \lambda}$

Therefore:

$A \map \ker {A - \lambda} \subseteq \map \ker {A - \lambda}$

that is to say:

$\map \ker {A - \lambda}$ is $A$-invariant.


Now, let $x \in \paren {\map \ker {A - \lambda} }^\perp$.

Observe that:

$A x = \lambda x + \paren {A - \lambda} x$

Now $\paren {A - \lambda} x \in \Rng {A - \lambda}$, and by our derivation above, this means that:

$\paren {A - \lambda} x \in \paren {\map \ker {A - \lambda} }^\perp$

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

$\lambda x + \paren {A - \lambda} x \in \paren {\map \ker {A - \lambda} }^\perp$

as desired.


Hence both $\map \ker {A - \lambda}$ and $\paren {\map \ker {A - \lambda} }^\perp$ have been shown to be $A$-invariant subspaces of $H$.

That is, $\map \ker {A - \lambda}$ is a reducing subspace for $A$.

$\blacksquare$


Sources