I have been struggling with this problem for awhile. Given that $A$ is a strictly diagonally dominant matrix with positive diagonal entries and non-positive off-diagonal entries, show that $A$ is monotone, i.e. $A^{-1} \geq 0$, meaning $a^{-1}_{ij} \geq 0$ for all $i,j$.
I have looked extensively for some help on this problem but have not come up with anything. Any help or a link to the right resource would help me out immensely!
Best Answer
Take the definition:
A real n-by-n matrix $A=[a_{ij}]$ with $a_{ij}\le 0$ for all $i\neq j$ is an M-matrix if $A$ is nonsingular and $A^{-l}\le0$ (this mean that we don't have nonnegative elements)
If $A$ is strictly diagonal dominant, $|a_{jj}|>\sum_{j\neq i, i=1}^n |a_{ij}|$ and $a_{ij}\le0$ $i\neq j$. And now... why $a_{ij}^{-1}$ is nonnegative! Well if you do the Gauss-Jordan procedure to inverse you will note that, you inverse will have all elements positive.
Its it, make the inverse of matrix using elementary rows operations.