Gauge fixing, invertibility and Green's functional

consider the photon in QED and the corresponding EOM of its Green’s functional in k-space: $$(k^mu k^nu-k^2g^{munu})Delta_{nurho}(k)=idelta^mu_rho.$$

Now, I understand that $U^{munu}(k):=k^mu k^nu-k^2g^{munu}$ is not injective, since $U^{munu}k_nu=0$ and thus $det U=0$. That is why $U$ is not invertible.

In the literature I read that gauge fixing solves this problem. Using the $R_xi$ gauges, one then obtains a new $U^{primemunu}=(1-xi^{-1})k^mu k^nu-k^2g^{munu}$. It follows that $U^{primemunu}k_nu=-xi^{-1}k^2k^mu$ and thus $k_nu$ does not have the eigenvalue zero anymore.

1. How can we be sure that there aren’t any other vanishing eigenvalues? Why don’t we diagonalise the operator?

Also, I remember that in scalar field theory we solved the invertibility problem by analytical continuation and then Feynman-shifting the poles away from the real axis: $p^2-m^2 mapsto p^2-m^2+iepsilon.$

We can do the same here, can’t we? If we write $U^{primemunu}=k^mu k^nu-k^2g^{munu}+iepsilon$, then we arrive at $U^{primemunu}k_nu=iepsilonneq0$ for $epsilon>0$.

2. Why do we need gauge fixing to make $U$ invertible? Why isn’t it sufficient to analytically continue the operator and then Feynman-shift its poles, as we do in scalar field theory?