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Suppose $A \in \mathbb{R}^{n\times n}$ (not necessarily symmetric) is weakly diagonally dominant with positive diagonals. Is it true that all eigenvalues of $A$ are non-negative (in the case of complex eigenvalues, have non-negative real parts)? Note that this holds true for strictly diagonally dominant matrices: if the matrix is symmetric with non-negative diagonal entries, the matrix is positive semi-definite.

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  • $\begingroup$ Correct! But, in the case of complex eigenvalues, we need the real part of the eigenvalues to be non-negative. $\endgroup$ – Arthur Dec 12 '18 at 4:55
  • $\begingroup$ Are you familiar with the Gershgorin circle theorem? $\endgroup$ – JimmyK4542 Dec 12 '18 at 4:59
  • $\begingroup$ Yes! I am familiar with that. $\endgroup$ – Arthur Dec 12 '18 at 5:00
  • $\begingroup$ I guess I got your point. In this case, all circles in the Gershgorin circle theorem will be on the right-half plane will possible intersection with the imaginary axis! $\endgroup$ – Arthur Dec 12 '18 at 5:05

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