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Let $A=(a_{ij})$ be a positive semi-definite, symmetric matrix, of order $3\times 3$ satisfying: $$ \Sigma_{j=1}^{3} a_{ij}=0 $$ for $i=1,2,3$ (i.e.- the sum of each row is zero).

Prove: $a_{ij}\leq 0, \quad \forall i\neq j$.

I know that $det(A)\geq 0$, and that the determinant of every principal submatrix is also nonnegative. I also know that the trace of every principal submatrix is nonnegative, which implies that the diagonal elements are nonnegative. But I wasn't able to use these properties in order to deduce the required claim.

Will you please help me with this?

Thank you very much

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Are you sure?

Define the matrix $A$ as $$A=\begin{pmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{pmatrix}.$$ Clearly, the matrix $A$ is symmetric and the sum of the coefficients of each of its rows is nil. Moreover, the eigenvalues of $A$ are $0$ (double eigenvalue), and $6$, hence $A$ is positive semi-definite.

Yet, $a_{13}=1>0$.

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  • $\begingroup$ You are right. I was absolutely certain that my claim is right, but obviously it was wrong... Thank you very much !!! $\endgroup$ – georgia Feb 11 '16 at 18:46

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