Showing irreducible matrix

Question

Let $A$ be a matrix which is irreducible, so there exists no permutation matrix $P$ such that $P^{T}AP$ is upper block triangular. Let $B$ a matrix for which $A_{ij}\leq B_{ij} \leq 0$ for $i \neq j$ and $0<A_{ii}<B_{ii}$. How can I show that $B$ is irreducible as well?

Answer 1

This is not true. Consider, e.g., $$ A=\pmatrix{1&-1\\-1&1}\quad\text{and}\quad B=\pmatrix{2&0\\0&2}. $$ Both $A$ and $B$ satisfy the given conditions but $B$ is reducible.

It would be true if you replaced "$A_{ij}\leq B_{ij}\leq 0$ for $i\neq j$" by "$B_{ij}\leq A_{ij}\leq 0$ for $i\neq j$". This is easy: the only way how to make an irreducible matrix reducible is to set some off-diagonal entries to zero.