About kernel of compact self-adjoint operators and separable Hilbert spaces

Question

I relate to this one question about corollary 4.10.2 pag. 198 of "Introduction to Hilbert Spaces - Debnath, Mikusinki" third edition, that states

Let $A$ be a compact self-adjoint operator on an infinite-dimensional Hilbert space $\mathcal{H}$. Then $\mathcal{H}$ has a complete orthonormal system $\{v_k\}$ consisting of eigenvectors of $A$

So I interpret it like "if an Hilbert space admits a compact self-adjoint operator, it will be separable" but this is not true (also this).

So what are the authors really telling, that I'm not understanding, with that proposition?

Little note:

$\ker{A}$ is sequentially closed subspace of $\mathcal{H}$, because $A$ is linear and compact (hence bounded, hence continuos), so by decomposition theorem $$\mathcal{H} = \ker{A} \oplus \ker{A}^{\perp\mathcal{H}}$$ Now, Hilbert-Schmidt theorem basically says that exists a sequence of eigenvectors $u_k$ of $A$ associated to non-null eigenvalues, such that $$ \text{cl}\,\text{Span}\{u_k\} = \ker{A}^{\perp\mathcal{H}} $$ but nothing is said about $\ker{A}$ that may also be non-separable, so that's why I cited $\ker{A}$ in the title

Answer 1

It has nothing to do with separability.

Every element of $\ker A$ is an eigenvector for $A$ with eigenvalue $0$. So you just choose an orthonormal basis of $\ker A$, countable or not, and you put it together with an orthonormal basis of $(\ker A)^\perp$ made out of eigenvectors, to get a full basis of eigenvectors.

Answer 2

It is certainly true that the orthogonal complement to the kernel decomposes as (the closure of) a countable direct sum of finite-dimensional eigenspaces, since the non-zero eigenvalues' only possible limit point is $0$. But (as alluded-to in @MartinArgerami's answer), this says nothing about the separability-or-not of the kernel itself. Yes, this does say that a compact operator on a non-separable Hilbert space must have a very large (=inseparable) kernel. Meanwhile, on separable Hilbert spaces, $0$ need not be an eigenvalue at all (though it is inevitably in the spectrum).