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Given that $A$ is an $n\times n$, symmetric and positive semidefinite, how would you prove that the sum of the entries of $A$ is positive?

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Let $X=(1,\ldots,1)\in\mathbb{R}^n$, then since $A$ is symmetric and positive semidefinite, one has : $$XA{}^tX\geqslant 0.$$ On the other hand, one has: $$XA^tX=\sum_{i=1}^n\sum_{j=1}^na_{i,j}.$$ Whence the result.

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    So why do you need the symmetry statement? Isn't this true regardless? – 2017-02-05
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    @BenKushigian Because in the litterature, positiveness of an operator is only defined for symmetric ones. It does not make too much sense to defined positiveness for any operator, since there is no interpretation in terms of quadratic forms. – 2017-02-05