While playing around with certain non-negative matrices, I got stuck at the following question.


Let $A$ be a *strictly* positive-definite $n \times n$ matrix ($n \ge 3$), with ones on the diagonal, and all other entries in the range $[0,1]$. How should I go about proving a tight bound on the sum of the entries of $A^{-1}$. In symbols, how should I go about trying to compute the smallest number $\gamma(n)$ such that $$1^TA^{-1}1 \le \gamma(n).$$


Any pointers to related work, or some possible ways to attack the problem will be very useful. Of course, if you think this question is not well-formulated, please help me improve it!


**Remarks:**


a. Notice that for the special case where $A$ is the identity matrix, we have equality, and $\gamma(n)=n.$


b. If the vector of all ones, i.e., $1$, happens to be an eigenvector of $A$, then also we have an instant answer. 


**More background**


The reason I reached the above bounding problem was that I was looking at the matrix 
$$\begin{bmatrix} A & 1\\\\
1^T & n\end{bmatrix},$$
and trying to prove that it is positive semidefinite. So, either I could show that $A-11^T/n \succeq 0$, or $1^TA^{-1}1 \le n$. Now, since the bound with $n$ might not always hold, I started searching for a $\gamma(n)$ that holds. Perhaps, I need to further cleanup my question?



Tight bound for sum of entries of the inverse of a nonnegative matrix
When $n\ge3$, there does not exist such a bound. To see this, take $n=3$ and the following matrix
$$A=\begin{pmatrix}
1 & a & 0 \\\\ a & 1 & a \\\\ 0 & a & 1 \end{pmatrix}.$$
If $a< 1/\sqrt2$, it is positive definite with non-negative entries. However
$$1^TA^{-1}1=\frac{3-4a}{1-2a^2}$$
is not bounded as $a\rightarrow1/\sqrt2$.


