In our recently finished class of functional analysis, the last theorem to be proved was the spectral theorem for compact normal operator in Hilbert space. More precisely, if $H$ is a Hilbert space and $ T \in B(H) $ is a bounded, linear, compact and normal operator. Then if $ \lambda_{n} $, $ \lambda_{m} $ are distinct eigenvalues, we have $$ V_{n} \perp V_{m}, \ \text{ where } \ V_n:=Ker(I\lambda_{n}-T), V_{m}:=Ker(I \lambda_{m} - T).$$
Moreover, $$\overline{\oplus_{n=1}^{\infty} V_n}=H$$ is an orthogonal direct sum. I have been trying to look for the extension of this theorem. It seems reasonable that we could find normalised eigenvectors $ v_{n} \in V_{n} $ with $ ||v_n|| = 1 $ such that there is an orthonormal basis $(v_n)_{n=1}^{\infty}$. Then for every $u \in H$ we would be able to form the Fourier series $$u = \sum_{n=1}^{\infty } \langle u , v_n\rangle v_{n}$$ and in particular
$$Tu=\sum_{n=1}^{\infty}\langle u, v_n \rangle \lambda_{n} v_n.$$ However, from some rough searching I have found that the reason orthonormal basis work is because they form a Schauder basis in $H$. However the convergence of Schauder basis representation needs not be unconditional, so permuting the index can get us something else. Indeed, the classical orthonormal basis are all equipped with some sort of index. Thus the question is how do we choose to index the eigenvectors $(v_n)_{n=1}^{\infty}$ to get a reasonable orthonormal basis?
Furthermore, if given arbitrary $u \in H$ we can find orthonormal basis $ (e_n)_{n=1}^{\infty} $ such that $ e_{n} $ is an eigenvector of $ T $ such that
$$Tu=\sum_{n=1}^{\infty}\langle u, e_n\rangle \mu_{n} e_{n},$$
then can we deduce anything like $ span(e_n) = span(v_m) $ for some $ n, m \in \mathbb{N} $ such that $ \mu_{n} = \lambda_{m} $ in some sense? I guess this is the same as asking if $E = E_1 \oplus E_2 $ and $F = F_1 \oplus F_2$ then can we say anything between $ E_1 $ and $F_1$, etc.
Thanks in advance!