Rational matrix eigenproblem on its pole-free domain

Let $\mathbb{C}(\lambda )$ denote the field of quotients of complex scalar polynomials. Let $R(\lambda )\in \mathbb{C}(\lambda {)}^{n\times n}$ with $n\ge 1$ be a square matrix whose entries are rational functions of $\lambda$. Write each entry as a reduced scalar fraction (common polynomial factors have been cancelled), and let $q(\lambda )$ be the monic least common multiple of their denominators. Thus the zero set $\mathcal{P}:=\{\lambda :q(\lambda )=0\}$ is exactly the set of poles of at least one entry. Define the pole-free domain $\mathrm{\Omega }:=\mathbb{C}\setminus \mathcal{P}$ and the polynomial matrix $N(\lambda ):=q(\lambda )R(\lambda )$, whose entries are polynomials as in Regular matrix polynomials and homogeneous eigenpairs. A rational eigenpair is a pair $(\lambda ,x)$ with $\lambda \in \mathrm{\Omega }$, $x\in {\mathbb{C}}^n\setminus \{0\}$, and $R(\lambda )x=0$; its $\lambda$ is a rational eigenvalue. The cleared-denominator polynomial equation is $N(\lambda )x=0$, but it represents the rational eigenproblem only on $\mathrm{\Omega }$. Any solution at a pole is an inadmissible candidate, even when the cleared polynomial matrix is singular there.