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

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.

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).
• Thanks, I guess what I'm missing is how this notion of "diagonal dominant" implies invertibility. Otherwise the question would be fairly obvious, as soon as we realize $A$ is not invertible when $a_{i,j} = 1/(n-1)$ for $i \neq j$. – user2566092 Sep 27 '13 at 18:02
• I also tried setting off-diagonal elements to $1/2$ for a $3 \times 3$ matrix with 1 on the diagonal, and I got that the matrix was invertible? (determinant = $1/2$) I think maybe the off-diagonal elements should be set to $-1/(n-1)$? Then the sum of the rows is $0$ so $A$ is not invertible. – user2566092 Sep 27 '13 at 18:19