Let $A \in \mathbb{R}^{n \times n}$ and let denote $I$ the $n \times n$ identitiy matrix.
Theorem. If $(I-A)$ is invertible and $(I-A)^{-1}$ is a nonnegative matrix and there is such a diagonal element in $(I-A)^{-1}$ which is less then $1$, then at least one of the elements in $A$ is negative.
Question. How could we prove this statement?
I've tried to use Perron–Frobenius theorem for $(I-A)^{-1}$, but after I've taken the eigenequation I don't know how to use the condition for the diagonal element.
I've also tried binomial inverse theorem and Woodbury matrix identity but they also didn't help me. Although I think using these tools lead us far away.
Let $$ B = (I-A)^{-1}, \qquad A = I-B^{-1}, $$ and let $B_{mm}<1$, $B\geq 0$, and let $e_m$ be the basis vector with $1$ in $m$-th position.
Then $Be_m = v$ has $v_m<1$ and $v\geq 0$, and $(Av)_m = v_m - 1 < 0$.