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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

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The answer is no. For a random counter-example, consider

$A:= \left[\begin{matrix}1 & \frac{5}{9} & 0 & 0 & 0\\\frac{5}{9} & 1 & 0 & \frac{5}{9} & \frac{5}{9}\\0 & 0 & 1 & \frac{5}{9} & \frac{5}{9}\\0 & \frac{5}{9} & \frac{5}{9} & 1 & \frac{5}{9}\\0 & \frac{5}{9} & \frac{5}{9} & \frac{5}{9} & 1\end{matrix}\right]\implies A^{-1}= \left[\begin{matrix}\frac{1053}{103} & - \frac{1710}{103} & - \frac{1125}{103} & \frac{2025}{206} & \frac{2025}{206}\\- \frac{1710}{103} & \frac{3078}{103} & \frac{2025}{103} & - \frac{3645}{206} & - \frac{3645}{206}\\- \frac{1125}{103} & \frac{2025}{103} & \frac{1503}{103} & - \frac{1260}{103} & - \frac{1260}{103}\\\frac{2025}{206} & - \frac{3645}{206} & - \frac{1260}{103} & \frac{9999}{824} & \frac{8145}{824}\\\frac{2025}{206} & - \frac{3645}{206} & - \frac{1260}{103} & \frac{8145}{824} & \frac{9999}{824}\end{matrix}\right]$
(check row 3)

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  • $\begingroup$ what's wrong with row 3? I saw $5/2$ dominates all off-diagonal entries. $\endgroup$
    – Zhanxiong
    Jul 11, 2021 at 17:38
  • $\begingroup$ @Zhanxiong last minute bug. Thanks. Now fixed. $\endgroup$
    – user8675309
    Jul 11, 2021 at 17:42
  • $\begingroup$ 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. $\endgroup$
    – user20638
    Jul 12, 2021 at 17:16

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