Suppose I have a symmetric real matrix with only entries that are $\ge 0$ and I know the entries in the diagonal are larger or equal to any other entry in the row (and because of the symmetry, this also goes for the column). Note that it is not a diagonally dominant matrix, as the entries of this matrix could even all be the same. Is this enough to prove that the matrix must be positive semi-definite? If not, can somebody give me a counter-example?
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Not true for matrices of size at least $3$.
For a symmetric matrix with diagonal entries $1$ and off-diagonal elements $a$, $b$, $c$, the determinant is $$1-(a^2 + b^2 + c^2) + 2 a b c$$ This is negative for $a=b=1$, $c=0$.
orangeskid
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True, thank you. Now I'll think about why a case like this cannot happen in the problem I am concerned with. – yagod May 25 '21 at 18:59
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You are welcome. It may be true in other cases. Hope you can solve your problem. – orangeskid May 25 '21 at 19:00