I am reading the following theorem from my lecture notes (English translation of German text). But I don't understand exactly what is meant from this theorem and the proof.
Theorem.
Let $H$ be a Hilbert space and $T \in K(H)$ where $K(H)$ denotes the set of compact operators from $H$ to $H$. Let $T$ be normal or self adjoint depending on whether the underlying field is $\mathbb R$ or $\mathbb C$, then there exists an orthonormal set $\{e_i \mathrel| i\in I\}$ where $I$ is either $\mathbb N$ or $\{1,2,\dotsc,k\}$ and a sequence $(\lambda_i)_{i\in I} \in \mathbb K$ which converges to $0$ such that
$$\operatorname{span}{(x_i : i ∈ I})^{⊥} = \ker(T)$$
and also $\forall x \in H$ : $Tx= \sum_{i\in I} \lambda_i \langle x, e_i \rangle e_i$ with unconditional convergence.
I am sorry here because I don't understand the theorem completely which means the translation is wrong, but can someone point out this theorem in some textbook or somewhere.
Thanks.
This is about the spectral decomposition of compact operators.
According to the theorem, a (normal / self-adjoint) compact operator can have countably many different eigenspaces, each of finite dimension, which are mutually orthogonal (due to self-adjointness / being normal). Let the eigenvectors be denoted by $e_1,e_2,e_3,...$ and the corresponding eigenvalues $\lambda_1,\lambda_2,..$ Moreover, if it has infinitely many eigenvalues, then these must tend to $0$.
Additionally, the operator must vanish outside the span of its eigenvectors.