 Solution

Every eigenvector with eigenvalue different from one lies in $E=\mathrm{span}\{a,b\}$, because $(\lambda -1)y=Ky\in E$. The elementary overlaps are $⟨a,a⟩=1/2$, $⟨a,b⟩=1/3$, $⟨b,b⟩=1/4$. ThusIts characteristic polynomial is $\lambda (\lambda +4)$, so the finite-multiplicity eigenpairs areEach eigenspace has dimension one. On $E^{\perp }$, both integrals defining $K$ vanish, so $Ly=y$. Thus $\boxed{\lambda =1}$ is also an eigenvalue, of infinite multiplicity, and there are no others.