Proof that compact self-adjoint operators have at least one non-zero eigenvector (using something analogous to min-max theorem)

Question

I checked several texts on the spectral theorem for compact self-adjoint operators (like this (PDF) and this (PDF)) on Hilbert spaces. They all mention that the cluster points of the real eigenvalues is zero. However, none of them show that at least one non-zero eigenvalue (and in turn, a legitimate eigenvector) exists for compact self-adjoint operators as claimed in this Wiki article.

As for Hermitian matrices, the key point is to prove the existence of at least one nonzero eigenvector. One cannot rely on determinants to show existence of eigenvalues, but one can use a maximization argument analogous to the variational characterization of eigenvalues.

Could someone give me or point me to a complete proof of this for compact self-adjoint operators? I'm not sure how exactly to use the min-max theorem to prove this (or something analogous to it).

Moreover, is any specific topology necessary on the space of the operators, as claimed here? And is completeness required? (To ensure that the eigenvalues/vectors don't get "lost" due to silly reasons. I'm not sure what the silly reasons could possibly be, though.)

Answer 1

Of course, topology is related to this topic since compact operators are defined in terms of compactness. Here I'll give a scheme of proof.

To prove existence of eigenvalues you need to observe several results.

1. Every operator $A$ that acts on a Banach space $E$ has a non-empty spectrum and moreover $\sup_{z \in \sigma(A)}|z| = \rho(A) = \lim||A^n||^{\frac{1}{n}}$

2. $||A^*A|| = ||A||^2$ for all operators on a Hilbert space

3. For self adjoint operator $||A^2|| = ||A||^2$ and therefore $\rho(A) = ||A||$. So for non-zero self adjoint operator you have a non-zero point in spectrum

4. There is a theorem for compact operators: if $K$ is a compact operator and $\lambda \ne 0$ is a complex number then $T = \lambda I - K$ has following properties. i) $T$ has closed image ii) $T$ is injective iff it is surjective

From all this we deduce that compact self adjoint operator has a non-zero point in spectrum and from last statement we obtain existence of eigenvectors (since operator $\lambda I - K$ is not invertible it can't be injective).

References: first statement can be found in Rudin FA (theorem 10.13); second and third statements are elementary; fourth can be found in Yosida (theorem 1, chapter 10.5)

Edit: Also from here you can deduce classical Hilbert-Schmidt theorem. You need to observe that for self adjoint operators the orthogonal complement of closed invariant subspace is invariant. Then you can construct a basis of eigenvectors by induction (compactness of operator will easily give you the convergence of eigenvalues to $0$).