Suppost that $A$ is a nonnegative matrix, and let denote the identitiy matrix with $I$ and the spectral radius of $A$ with $\rho(A)$. Note that because $A$ is nonnegative according to the Perron–Frobenius theorem $\rho(A) = \lambda_\max(A)$.
Statement. The following two are equivalent
- $I-A$ matrix is invertible and $(I-A)^{-1}$ is a nonnegative matrix
- $\lambda_\max(A) < 1$.
Note that because of the statement we also know that the Neumann series of $A$ convergent.
Question. How could we prove this statement?
Let $\lambda_{\max}<1$. Then $\rho(A)=\lambda_{\max}=1-\tau$, $\tau\in(0,1)$ and $\|A^k\|< (1-\tau/2)^k$ for $k$ sufficiently large. Thus the Neumann series $$ \sum_{k=0}^\infty A^k = (I-A)^{-1} $$ converges. As $A$ is non-negative, $A^k$ is non-negative, and by the series representation $(I-A)^{-1}$ is non-negative.
If $\lambda_{\max}=1$ then $I-A$ is not invertible.
Assume $\lambda_{\max}>1$ and that $I-A$ is invertible. If $\lambda$ is an eigenvalue of $A$ then $(1-\lambda)^{-1}$ is an eigenvalue of $(I-A)^{-1}$. By Perron-Frobenius theorem there exists a non-negative eigenvector $x$ such that $Ax = \lambda_{\max}x$. This implies $(I-A)^{-1}x = \frac1{1-\lambda_{\max}}x$. Since $\lambda_{\max}>1$ it follows $\frac1{1-\lambda_{\max}}<0$. As $x\ge0$ it follows that $(I-A)^{-1}$ cannot be non-negative.