Converse of Spectral Theorem for Compact Self-Adjoint Operators

Question

If I have a bounded linear operator $A$ on a Hilbert space $H$ whose eigenvectors form an orthonormal basis for $H$ and whose corresponding eigenvalues go to $0$ then is $A$ compact and self-adjoint?

I ask because I want to prove that $A$ defined on an orthonormal basis $\{e_k\}$ as $Ae_k=e_k/(k^2+1)$ is compact and self-adjoint. I know that it is, but I'm just wondering if appealing to the spectral theorem is valid. Thanks!

Answer 1

Yes, $A $ is compact. By considering truncations of $A $ that are zero on $e_k $ for $k\geq n $, you can write $A $ as a limit of finite-rank operators.