# Diagonal terms of the inverse of positive definite matrix

I have a matrix $$A = I + G$$ where $$G_{ii} = 0$$ for all $$i$$, and $$G_{ij} \in \{0, a\}$$ with $$a \in [0, 1[$$, $$G$$ is symmetric, and such that $$A$$ is positive definite.
We know that the inverse of $$A$$, call it $$B$$, is also symmetric positive definite, and also that $$B_{ii} > 1$$ for all $$i$$.

My question: is it true that $$B_{ii} > B_{ij}$$ for all $$i, j$$?

Any help welcome! Thanks

$$A:= \left[\begin{matrix}1 & \frac{5}{9} & 0 & 0 & 0\\\frac{5}{9} & 1 & 0 & \frac{5}{9} & \frac{5}{9}\\0 & 0 & 1 & \frac{5}{9} & \frac{5}{9}\\0 & \frac{5}{9} & \frac{5}{9} & 1 & \frac{5}{9}\\0 & \frac{5}{9} & \frac{5}{9} & \frac{5}{9} & 1\end{matrix}\right]\implies A^{-1}= \left[\begin{matrix}\frac{1053}{103} & - \frac{1710}{103} & - \frac{1125}{103} & \frac{2025}{206} & \frac{2025}{206}\\- \frac{1710}{103} & \frac{3078}{103} & \frac{2025}{103} & - \frac{3645}{206} & - \frac{3645}{206}\\- \frac{1125}{103} & \frac{2025}{103} & \frac{1503}{103} & - \frac{1260}{103} & - \frac{1260}{103}\\\frac{2025}{206} & - \frac{3645}{206} & - \frac{1260}{103} & \frac{9999}{824} & \frac{8145}{824}\\\frac{2025}{206} & - \frac{3645}{206} & - \frac{1260}{103} & \frac{8145}{824} & \frac{9999}{824}\end{matrix}\right]$$
• what's wrong with row 3? I saw $5/2$ dominates all off-diagonal entries. Jul 11, 2021 at 17:38
• Thank you very much for this counter-example. In fact my problem is even more precise but I don't think there would be a simple answer. My problem is as described above, except that $A^{-1}. 1 \geq 0$ (i.e. the sum of rows is positive). This is not verified by your counter-example where rows 2 and 3 add up to negative numbers. Jul 12, 2021 at 17:16