Given an operator $T \in B(H)$, denote its spectrum by $\sigma(T)$. We write $\Omega := \sigma(T)\setminus \{0\}$. The following result is well-known and can be assumed without proof:
Let $T \in B_0(H)$ be a self-adjoint compact operator on a Hilbert
space $H$. Given $\lambda \in \Omega$, let $P_\lambda$ be the
finite-rank projection on $\mathcal{E}_\lambda:= \ker(T-\lambda I)$.
Then $$T= \sum_{\lambda \in \Omega} \lambda P_\lambda$$ where the
unordered sum converges in the norm topology.
Given $\xi, \eta \in H$, let us define the rank-one operator $R_{\xi, \eta}: H \to H$ by $R_{\xi, \eta}(\zeta):= \langle \zeta, \eta\rangle \xi$. I want to prove the following theorem:
If $T \in B_0(H)$ is a self-adjoint operator compact operator, then
there is an orthonormal basis $\{e_s\}_{s \in S}$ for $\ker(T)^\perp$
and a collection $\{\mu_s\}_{s \in S} \in c_0(S)$ such that $$T=
\sum_{s \in S} \mu_s R_{e_s, e_s}$$ where the series converges in
norm. In particular, a self-adjoint compact operator admits an
orthonormal basis of eigenvectors.
The idea of the proof should be more or less the following:
We have an orthogonal direct sum decomposition $$\ker(T)^\perp = \bigoplus_{\lambda \in \Omega} \mathcal{E}_\lambda$$
Given $\lambda \in \Omega$, let $\mathcal{F}_\lambda$ be an orthonormal basis for $\mathcal{E}_\lambda$. Then we have $$P_\lambda = \sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}$$ and it follows that
$$T = \sum_{\lambda \in \Omega}\lambda \left(\sum_{\xi \in \mathcal{F}_\lambda} R_{\xi, \xi}\right) = \sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$
Then $\bigcup_{\lambda \in \Omega} \mathcal{E}_\lambda$ is an orthonormal basis for $\ker(T)^\perp$ and this should yield the desired decomposition. However, I feel like the equality
$$\sum_{\lambda \in \Omega}\left(\sum_{\xi \in \mathcal{F}_\lambda} \lambda R_{\xi, \xi}\right)= \sum_{\xi \in \bigcup_{\lambda \in \Omega} \mathcal{F}_\lambda} \lambda R_{\xi, \xi}$$
still requires some formal justification. We should at least show that the sum on the right converges in the norm topology! Of course, other approaches to prove my end goal are also welcome!
EDIT: It's a strong requirement to me that the convergence of all series involved is in the norm-topology. Many references only deal with the strong convergence of these series.
Best Answer
That's precisely where you get to use that your $T$ is compact.
Fix $\delta>0$, and consider $\Omega_\delta=\{\lambda\in\Omega:\ \lambda>\delta\}$. Suppose $\Omega_\delta$ is infinite. For each $\lambda\in\Omega_\delta$, fix $h_\lambda\in P_\lambda H$ with $\|h_\lambda\|=1$. The set $\{h_\lambda\}_{\lambda\in\Omega_\delta}$ is orthonormal; we can write $Th_\lambda=\lambda h_\lambda$ as $$ T(\frac1\lambda\,h_\lambda)=h_\lambda. $$ As $\lambda>\delta$, we have that $\tfrac1\lambda\,h_\lambda$ sits inside the ball of radius $1/\delta$. So $\{h_\lambda\}$ is in the image through $T$ of the ball of radius $1/\delta$. And, being orthonormal, $\{h_\lambda\}$ does not have a convergent subsequence, contradicint the compactness of $T$. We have thus shown that $\Omega_\delta$ is finite. Note that this argument also implies that each $P_\lambda$ is finite-rank.
By consider the finite sets $\Omega_{1/n}$, we prove that $\Omega$ consists of a sequence that converges to $0$, as desired.
If $\{\xi_n\}$ is a finite orthonormal set and $P_n$ denotes the orthogonal projection onto $\mathbb C\xi_n$, then $$\tag1 \Big\|\sum_n\lambda_n\,P_n\Big\|=\sup\{|\lambda_n|:\ n\}. $$ Indeed, if $\xi\in H$ with $\|\xi\|=1$, then $\xi=\eta+\sum_n c_n\xi_n$, where $\eta$ is orthogonal to all $\xi_n$. Then (since $P_n\eta=0$ for all $n$) \begin{align} \Big\|\sum_n\lambda_nP_n\xi\Big\|^2 &=\Big\|\sum_n\sum_k\lambda_nc_kP_n\xi_k\Big\|^2 =\Big\|\sum_n\lambda_nc_n\xi_n\Big\|^2 =\sum_n|\lambda_n|^2\,|c_n|^2\\[0.3cm] &\leq\sup\{|\lambda_n|:\ n\}\,\sum_n|c_n|^2\\[0.3cm] &=\sup\{|\lambda_n|:\ n\}. \end{align} Using that $\Big\|\sum_n\lambda_nP_n\xi_k\Big\|=|\lambda_k|$ it is then easy to check the equality $(1)$.
Now allow the set $\{\xi_n\}$ to be infinite. If $\lambda_n\to0$, fix $\varepsilon>0$. Choose $n_0$ such that $|\lambda_k|<\varepsilon$ for all $n\geq n_0$. Then $$ \Big\|\sum_{k=n}^m\lambda_kP_k\Big\|=\sup\{|\lambda_k|:\ k\geq n\}\leq\varepsilon $$ for all $m>n\geq n_0$. This shows that the tails of the series $\sum_n\lambda_nP_n$ go to zero, and so the series converges.