Spectral theory for compact normal operators.

Question

Let $T$ be a compact normal operator on an infinite dimensional Hilbert space $\mathcal H.$ Then I came across the fact that the spectrum $\sigma (T)$ of $T$ is either finite or countably infinite containing $0.$ I have seen the following claims in the lecture note I am following.

$(1)$ Let $\sigma (T) = \{\lambda_1,\lambda_2,\cdots, \lambda_k\} \cup \{0\}.$ Let $\mathfrak h_j = \text {ker} (T - \lambda_jI),\ $ $1 \leq j \leq k.$ Let $P_j$ be the projection onto $\mathfrak h_j.$ Let $T_k = \sum\limits_{i=1}^{k} \lambda_i P_i.$

$$\color {red} {\text {Then}\ \sigma(T-T_k) = \{0\}.}$$

$(2)$ If $\sigma (T)$ is countably infinite then write $$\sigma(T) = \{\lambda_1,\lambda_2, \cdots \} \cup \{0\}.$$

$$\color{red} {\text {Then}\ \sigma (T - T_n) = \{\lambda_{n+1}, \lambda_{n+2}, \cdots \} \cup \{0\},\ \text {for all}\ n \geq 1.}$$

Can anybody make the statements in red clear to me? Thanks for your time.

Answer 1

The statements are immediate consequences of what is known as the spectral theorem in its compact, normal version. Conway's book is the place to look for this theorem.

Theorem (spectral theorem, normal, compact version)

Let $T$ be a compact, normal operator in $\mathbb{B}(H)$. Then $T$ has at most countably many distinct eigenvalues $\{\lambda_n\}$ and if they are countably many then $\lambda_n\to0$. If $P_n$ denotes the projection onto the eigenspace $\ker(T-\lambda_n I)$, then the projections $\{P_n\}$ are pairwise orthogonal and $$T=\sum_n\lambda_nP_n$$ in the sense that $$\|T-\sum_{k=1}^n\lambda_nP_n\|_{\mathbb{B}(H)}\xrightarrow[n\to\infty]{}0. $$

The claims follow directly from this theorem. (1) follows trivially and for (2) note that $T-T_n=\sum_{k\geq n+1}\lambda_kP_k$, so $T-T_n$ is a compact, normal operator and its only eigenvalues are $\{\lambda_k\}_{k=n+1}^\infty$, thus $\sigma(T-T_n)=\{\lambda_k\}_{k=n+1}^\infty\cup\{0\}$.