singularity somewhere on the spectral radius circle

55 Views Asked by At

Let $R({\lambda}) = (\lambda I -A)^{-1}$ be the resolvent operator of a matrix $A$.

Let $A$ be a non-negative matrix.

I have showed that $R({\lambda})$ is analytic except the eigenvalues. But unable to show that it must have at least one singularity somewhere on the spectral radius circle.