Sharp bound on off-diagonal entries of matrix with 1's on diagonal to make matrix invertible

Question

Suppose $A$ is an $n \times n$ matrix with all 1 on the diagonal. What is the sharp bound $\epsilon(n)$ so that $A$ is invertible if all off-diagonal entries of $A$ have absolute value less than $\epsilon(n)$? Obviously $\epsilon(n) > 0$ exists, because as all off-diagonal elements go to zero, $A$ approaches the identity matrix which is invertible.

Answer 1

Let $S$ be the set of non-negative real numbers $r$ for which if $|a_{i,j}|\lt r$ for all $i\neq j$, then $A$ is invertible. We have that $S\supset \left(0,\frac 1{n-1}\right]$ (because in this case $A$ is diagonal dominant, hence invertible). We can't hope more (take $a_{i,j}:=\frac 1{n-1}$; then $A$ is not invertible).