Currently I'm self studying functional analysis, namely the spectrum of compact operators on infinite dimensional Banach spaces. In the text, the author gives the following proposition:
Proposition A: Let $T\colon X\to X$ be a compact operator and $X$ be an infinite dimensional Banach space. Then $T_\lambda:=T-\lambda\text{Id}$ is one-to-one if and only if it is onto.
I completely understand Proposition A. The author then gives the main theorem of this chapter:
Theorem 1: For any compact operator $T\colon X\to X$ in an infinite dimensional Banach space $X$ we have
$$
\sigma(T)=\{0\}\cup\sigma_p(T).
$$
The author does not give any proof, so I assume Proposition A must, in some sense, be the proof. So my question is, does my proof below work?
Proof. It's clear that $\sigma(T)\supseteq\{0\}\cup\sigma_p(T)$. Take $\lambda\in\sigma(T)\setminus\{0\}$. By definition this means $T_{\lambda}$ is not invertible. By Proposition A it follows that $T_{\lambda}$ is neither one-to-one or onto. Since it's not one-to-one, it follows that $\text{ker}T_{\lambda}\neq\{0\}$. Thus $\lambda\in\sigma_p(A)$, and so $\sigma(T)\setminus\{0\}\subseteq\sigma_p(T)\implies\sigma(T)\subseteq\{0\}\cup\sigma_p(T)$.
Best Answer
$0 \in \sigma (T)$ because a compact operator in an infinite dimensional space cannot be invertible. By definition $\sigma_p(T) \subset \sigma(T)$. So we only have to show that $\sigma(T) \subset \{0\} \cup \sigma_p(T)$. So take $\lambda \neq 0$ is the spectrum. Suppose it does not belong to $\sigma_p(T)$. Then $\lambda $ is not an eigen value so there is no non-zero vector $x$ with $Tx=\lambda x$. Thus $(T-\lambda I)x=0$ implies $x=0$. This means $T-\lambda I$ is one-to-one. By Proposition A it follows that $T-\lambda I$ is also onto. By Open Mapping Theorem it follows that $T-\lambda I$ has a bounded inverse. But this contradicts our assumption that $\lambda \in \sigma (T)$.