# symmetric matrices and positive definiteness

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.

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.
• @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 $n=5$. Oct 17, 2020 at 11:48