My question follows from another of my question: symmetric matrices and positive definiteness. Suppose the matrix $A$ is symmetric with all entries positive and each row sum is a fixed constant ($>1$). Assume that each diagonal entry is greater than the off-diagonal entries of the corresponding row (and column), by atleast 1 (that is, the difference between the diagonal entry and each corresponding off-diagonal entry is atleast more than 1).
From these conditions, can we conclude that $A$ is positive definite.
Try: I tried to write $A=sI-B$, where $I$ is the identity matrix and $s$ is the largest diagonal entry of $A$ ($s>1$, from the given conditions). Then, I am not sure how to proceed. I am not sure if the result follows by gershgorin's theorem. Any hint will be really helpful.
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For a counterexample, let $$ A = \pmatrix { 10 & 8 & 6 & 2 \cr 8 & 10 & 2 & 6 \cr 6 & 2 & 10 & 8\cr 2 & 6 & 8 & 10 } $$ and note that $\det(A)=-3120$.
