 Solution

The necessary and sufficient condition iswhere $\rho$ is the spectral radius of the nonnegative matrix. No irreducibility is required. We use the given nonnegative-vector characterization of its Perron eigenvalue.

For necessity, suppose $p\ne 0$ is the survival fixed point and let $S=\{i:p_i>0\}$. Since $1-e^{-z}<z$ for every $z>0$, its equation gives $(\Lambda p{)}_i>p_i$ for every $i\in S$. There are finitely many such coordinates, soOutside $S$ the inequality $\Lambda p\ge \alpha p$ still holds, because its right side is zero and its left side is nonnegative. The provided Perron–Frobenius theorem characterization therefore gives $\rho (\Lambda )\ge \alpha >1$. In particular survival is impossible at $\rho =1$ for this Poisson offspring law.

Conversely suppose $\rho >1$. The given characterization supplies a nonzero $v\ge 0$ with $\Lambda v\ge \rho v$; normalize it so ${\max }_i{v}_i=1$. For sufficiently small $\eta >0$,For example any $0<\eta \le \min (1,2(\rho -1)/{\rho }^2)$ works. Coordinates with $v_i=0$ satisfy the inequality automatically. Starting at $x_0=\eta v$, iterate $x_{t+1}=T(x_t)$. Monotonicity makes the sequence increase, and $T$ takes nonnegative vectors into $[0,1{]}^l$, so it converges to a fixed point $x_{\infty }\ge \eta v\ne 0$. Part (a) gives $p\ge x_{\infty }$, proving survival from at least one type. This proves the spectral survival criterion for multitype Poisson branching for reducible as well as irreducible reproduction matrices.