 Solution

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.