I am reading Invariant Subspaces by H. Radjavi and P. Rosenthal. I need help understanding the proof of the spectral theorem for compact normal operators, as stated below.
Theorem $1.4.$ (Spectral Theorem for Compact Normal Operators). If $A$ is a compact normal operator, then there exists an orthonormal basis $\{e_\alpha\}_{\alpha\in I}$ such that each $e_\alpha$ is an eigenvector of $A$.
I will reproduce the proof below, and ask questions simultaneously. Firstly, we are assuming $A\ne 0$, right?
By Zorn's lemma there is a maximal orthonormal set $\{e_\alpha\}_{\alpha\in I}$ consisting of eigenvectors of $A$.
- How is Zorn's lemma being applied here? I know that if $A \ne 0$ is compact and normal, then $\varnothing \ne \sigma(A)\setminus\{0\} = \sigma_p(A)\setminus\{0\}$ where $\sigma_p(A)$ is the point spectrum of $A$ (eigenvalues of $A$). So, there exists $\lambda\ne 0$ such that $\lambda\in \sigma_p(A)$. I believe we might have to do something with $E(\lambda):= \{x\in \mathcal H: Tx = \lambda x\}$, the eigenspace corresponding to $\lambda$.
We need only show that $\bigvee_{\alpha\in I} e_\alpha = \mathcal H$. Let $M = \bigvee_{\alpha\in I} e_\alpha$. By Theorem $1.2$, the one-dimensional space spanned by $e_\alpha$ reduces $A$ for each $\alpha$, and therefore $M$ reduces $A$. If $M^\perp\ne \{0\}$, consider $A\vert_{M^\perp}$. The operator $A\vert_{M^\perp}$ is compact and normal.
- I know that if $A$ is compact and $A = \left[\begin{matrix}W & X\\ Y& Z \end{matrix}\right]$, then each of $W,X,Y,Z$ are compact (see this post for details on matrix notation). So, $A\vert_{M^\perp}$ is compact. Why is $A\vert_{M^\perp}$ normal?
If $A\vert_{M^\perp}$ were $0$, the set $\{e_\alpha\}$ would not be maximal.
- I wish to verify my understanding of why $A\vert_{M^\perp} = 0$ is not permitted. We have $M^\perp \ne \{0\}$, and $A\vert_{M^\perp} = 0$ would mean that there exists $0\ne z\in M^\perp$ with $A\vert_{M^\perp} z = 0 = 0z$, i.e. $z$ is an eigenvector for $A\vert_{M^\perp}$. If $Az = 0$, then $Az = 0z$, i.e. $z\in M^\perp$ is an eigenvector for $A$, but this contradicts the maximality of $M$. The other possibility is that $Az \in M$. Why does this contradict the maximality of $M$?
Thus, $\sigma\left(A\vert_{M^\perp} \right)$ contains a non-zero complex number, by Theorem $1.1$, and hence, by the Fredholm alternative, $\sigma_p\left(A\vert_{M^\perp} \right)$ is not empty. Thus, $A$ has an eigenvector in $M^\perp$, contradicting the maximality of $\{e_\alpha\}$.
- Perhaps this is the same question as (3) above, but $\sigma_p\left(A\vert_{M^\perp} \right) \ne \varnothing$ means that $A\vert_{M^\perp}$ has an eigenvector in $M^\perp$. How does this imply that $A$ has an eigenvector in $M^\perp$?
Notation.
- $\bigvee_{\alpha\in I} e_\alpha$ denotes the closed linear span of $\{e_\alpha\}_{\alpha\in I}$, i.e. $\overline{\operatorname{span} \{e_\alpha\}_{\alpha\in I}}$.
- $A\vert_{M^\perp}$ should be interpreted as the compression of $A$ to $M^\perp$, given by $M^\perp \xrightarrow{A} \mathcal H \xrightarrow{P_{M^\perp}} M^\perp$ where $P_{M^\perp}$ is the projection map onto $M^\perp$.
Other Theorems.
Theorem $1.1.$ If $A$ is normal, then $\|A\| = r(A)$ (the spectral radius of $A$).
Theorem $1.2.$ If $A$ is normal and $Ax=\lambda x$, then $A^*x = \overline\lambda x$.
Best Answer
You don't need to assume $A\neq 0$. If $A=0$, then any ONB of $\mathcal H$ consists of eigenvectors of $A$.
You don't need to know much about compact operators for this step. Let $\mathcal F$ be the set of all orthonormal sets of eigenvectors of $A$, ordered by inclusion. Every chain in $\mathcal F$ has an upper bound, namely the union. Thus $\mathcal F$ has a maximal element by Zorn's lemma.
A reducing subspace for $A$ is also reducing for $A^\ast$. Thus $A^\ast(M^\perp)\subset M^\perp$ and you an check that $(A|_{M^\perp})^\ast=A^\ast|_{M^\perp}$, which clearly commutes with $A|_{M^\perp}$ by normality of $A$. Hence $A|_{M^\perp}$ is normal.
Since $z\in M^\perp$, you have $A|_{M^\perp}z=Az$. That is just the definition of the restriction of a map.
See 3. If $z\in M^\perp$ and $A|_{M^\perp}z=\lambda z$, then $Az=A|_{M^\perp}z=\lambda z$ by the definition of the restriction.