Weakly Diagonally Dominant with Positive Diagonals

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.

• Correct! But, in the case of complex eigenvalues, we need the real part of the eigenvalues to be non-negative. – Arthur Dec 12 '18 at 4:55
• Are you familiar with the Gershgorin circle theorem? – JimmyK4542 Dec 12 '18 at 4:59
• Yes! I am familiar with that. – Arthur Dec 12 '18 at 5:00
• 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! – Arthur Dec 12 '18 at 5:05