Why we still diagonalize compact operators even tho we lack invertibility. We know that any compact symmetric operator on a Hilbert space, has a orthogonal base of eigenvectors. But we also know that $0$ is in the spectrum if $X$ is infinite dimensional, which makes the operator non-invertible(in particular they are never surjective).
Question;
Why we would care to diagonalise in infinite dimensional at all since never get as good info as in finite dimension (non-zero eigenvalues on the diagonal).
Added later ;
Plausibel answer or claim ;
Diagonalisation of a compact operator $C$ in infinite dimensions seems more related to the invertabiity(or solvability) of Iλ−C (a fredholm equation of 2nd kind) for compact operator C rather then properties of C itself as a linear map which often is the object of intrest in finite dimension i.e linear algebra. That might be a reason for my confusion. From a historcial point of view we seem to care about solvability of Iλ−C and qualitive properties of solutions rather then C itself as a map from one vector space into another,in some sense. So when we "diagonalize" we really just cheack solvabilty of Iλ−C, we are not looking for a nice bijection(the full diagonal matrix) of the vectorspaces as we do in linear algebra.
Or simply the MAIN reason we started to study the spetrum of an operator $C$ was to determine if $I \lambda - C$ is solvable. 
Am I on to something or just way off?
 A: For any bounded selfadjoint operator $T$ on a Hilbert space $X$, you can reduce to the case where $T$ is injective because $X=\mathcal{N}(T)\oplus\overline{\mathcal{R}(T)}$ and both of these closed subspaces are invariant under $T$. That allows you to reduce to looking at $T$ on the Hilbert space $Y=\overline{\mathcal{R}(T)}$, and $T$ remains selfadjoint on this invariant subspace.
The operator $T : Y\rightarrow Y$ has dense range and is injective. If the range of $T$ is closed, then $T$ is a topological isomorphism, which means that the unit ball of $Y$ is compact, leading to the conclusion that $Y$ is finite-dimensional. Otherwise, the range of $T$ cannot be closed, which means that $0 \in\sigma(T)$.
A: You asked for intuition so I am going to allow myself to play a little loose...
Let $U$ and $V$ be  infinite vector spaces with a countable bases $\{e_i\}$ and $\{f_i\}$. Suppose that a linear map $T:U\rightarrow V$ is diagonalisable such that
$$T\left(\sum_i\alpha_i e_i\right)=\sum_i\alpha_i \lambda_i f_i.$$
If none of the $\lambda_i$ are zero, and so $T$ invertible, then the range of $T$ will be in a closed ball of infinite dimension and so $T$ cannot be compact.
