Why doesn't the inverse of this matrix always exist?

Question

Take $A$ square, non-negative, irreducible matrix, with Perron root $\rho(A)$. Let $Q = \rho(A)I - A$. Then it is argued (e.g. in the Handbook of Linear Algebra, Chapter 9) that the group inverse of $Q$ always exists. But I fail to see why the actual inverse doesn't always exist (and must then be the group inverse) using the following argument: By the Perron-Frobenius theorem, $\rho(A)$ distinct from the other eigenvalues, $0$ cannot be in the spectrum of $Q$, and so $Q$ must be invertible.

Answer 1

$Q$ must be singular. By Perron-Frobenius theorem, $\rho(A)$ is an eigenvalue of $A$. Therefore $0=\rho(A)-\rho(A)$ is an eigenvalue of $Q=\rho(A)I-A$ and the usual inverse of $Q$ doesn't exist.

However, since $A$ is irreducible, $\rho(A)$ is a simple eigenvalue of $A$. Therefore $0$ is a simple eigenvalue of $Q$ and $\operatorname{rank}(Q^2)=\operatorname{rank}(Q)$. Hence the group inverse (here I suppose that the author means Drazin inverse) of $Q$ exists.