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I have a symmetric matrix $A$ with all entries positive and each diagonal entry is greater than other off-diagonal entries in its corresponding row and column (not necessarily strictly diagonally dominant).
My question is: Can I conclude that $A$ is positive definite (or positive semidefinite).
For a $2 \times 2$ matrix it is clearly true as we can directly take its inverse and show, but for matrices with high dimension, I am unable to conclude it. Any help in the form of hint or reference will be really helpful.

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No. For instance, since $B=\pmatrix{1&0&1\\ 0&1&1\\ 1&1&1}$ is indefinite (it has a positive trace but its determinant is $-1$), any matrix $A$ that is close to $B$, such as $A=\pmatrix{1&t&1-t\\ t&1&1-t\\ 1-t&1-t&1}$ with a small $t>0$, will be indefinite too.

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  • $\begingroup$ Can we consider only the case when the diagonal entries are greater than the corresponding off-diagonal entries in row and column by atleast 1, because numerically considering the off-diagonal entries too close to the diagonal entry is not feasible. What I mean to say is, numerically some result cannot be obtained. Please let know if I should edit the question. $\endgroup$ Oct 17, 2020 at 11:32
  • $\begingroup$ @user812951 I don't understand what you mean by "numerically some result cannot be obtained". If you pick $t=\frac1n$, then when $n$ is a sufficiently large integer, $C=2nA$ is an indefinite matrix with positive integer entries but $c_{ii}-c_{ij}\ge2$ whenever $I\ne j$. In fact, $n$ doesn't have to be very large. $C$ is already indefinite when $n=5$. $\endgroup$
    – user1551
    Oct 17, 2020 at 11:48
  • $\begingroup$ I tried out your example and it fits perfectly. I was trying out cases only when the difference between diagonal and corresponding off-diagonal entries is greater than 1. Perhaps, that's the reason why I did not get the counter-example. Thank you so much for the explanation. $\endgroup$ Oct 17, 2020 at 12:03

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