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Is a symmetric real matrix with diagonal entries strictly greater than $1$ and off-diagonal entries positive but strictly less than $1$ necessarily positive-semidefinite?

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up vote 6 down vote accepted

Nope. Just playing around with my computer, I found the matrix

[$\frac{11}{10}$ $\frac{1}{100}$ $\frac{99}{100}$]

[$\frac{1}{100}$ $\frac{11}{10}$ $\frac{99}{100}$]

[$\frac{99}{100}$ $\frac{99}{100}$ $\frac{11}{10}$]

with determinant $\frac{-25179}{31250}$.

Is this perhaps a misremembering of the definition of a diagonally dominant matrix?

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Not quite misremembering as much as failing to find a counterexample and being unable to prove the necessity but still not believing in it's veracity. Thanks for your example! – daegan Feb 7 '11 at 21:48
@daegan: no problem. To construct a counterexample, it of course helps to be aware of the theorem about diagonally dominant matrices, so as to stay away from that case. I did remember that theorem (luckily for me, I learned about Gershgorin's Theorem in a talk I heard last year) so the above was only the third matrix I tried (or technically the fourth: the third one also worked but the coefficients were a little more complicated: it appears in a previous version of this answer). – Pete L. Clark Feb 8 '11 at 3:31

If the sum of each row is positive, then no eigenvalue can be non-negative: take any eigenvector, and pick the largest (in magnitude) coordinate. After applying the matrix, even if the signs are all perfectly against us, still $0$ will not be crossed (or reached to).

So the correct formulation is $\geq 1$ on the diagonal, and the off-diagonals in each row sum to less than $1$.

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By Gershgorin's Theorem we can conclude that the absolute sum of all non-diagonal entries in each row must be less than 1. Then the matrix will have positive eigenvalues for diagonal entries > 1.

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