Consider an $n\times n$ matrix $A$, and it's perturbation matrix $dA$.
Let for simplicity $A=I$ be a matrix with ones on the diagonal and zeros off-diagonal.
Let $dA$ have zeros on the diagonal and off-diagonal elements in the range $(0,\frac{1}{n})$.
Question: Can I argue that the inverse matrix has the off-diagonal elements less than $\frac{1}{n}$ as well? Normalize them by the diagonal ones that get perturbed up a little bit.
I ran a bunch of simulations on my computer and it seems like this is indeed true. For n=30, the off-diagonal elements hardly reach 30% of my target bound. For n=100, it hardly exceeds 25% of it.
The best theory I could find tells the following. Pick a 1- or $\infty$ norm of the matrix, then $$\frac{||(A+dA)^{-1}-A^{-1}||}{||A^{-1}||} \leq K(A) \cdot \frac{||dA||}{||A||}, \quad K(A)=||A||\cdot ||A^{-1}||$$
What this means is that if the original matrix $A+dA$ was diagonally dominant, then the inverse $(A+dA)^{-1}$ is diagonally dominant as well, because the conditioning number $K(A)$ is equal to 1.
The inequality holds for any sub-multiplicative norm and the problem is that the max-norm is not sub-multiplicative, or I would have got my result immediately.
I would really appreciate if somebody knows a way to show this or at least some clue. Thanks!