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:
- $\operatorname{ran} \left({A - \lambda}\right) \subseteq \left({\ker \left({A - \lambda}\right)}\right)^\perp$
In particular: Link to result proving $A - \lambda$ normal (not up presently).
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this in more detail, feel free to use the talk page.
Once you have done so, you can remove this instance of {{MissingLinks}} from the code.
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:
| \(\displaystyle A x\) | \(=\) | \(\displaystyle \lambda x + \left({A - \lambda}\right) x\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \lambda x\) | $\quad$ Definition of kernel, $x \in \ker \left({A - \lambda}\right)$ | $\quad$ |
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 $A$.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis ... (previous) ... (next): $II.5.6$
