Question
Suppose that I have an n by n square matrix $A$ whose elements are all nonnegative, and its diagonal elements are strictly less than 1. Further, for any vector $p$ with $0<p_i<1$, $p>A^Tp$, i.e. $p_i>(A^Tp)_i$. How do I show that $diag(3-4A+2A^TA)-(A+A^T-A^TA)$ is invertible?
I managed to solve the 2 by 2 case by comparing common terms in the determinant. But even a 3 by 3 matrix would make the comparison infeasible. Grateful for any comments and suggestions!