Let $A$ be a strictly diagonally dominant matrix and let’s assume $A$ is singular, that is, $\lambda=0\in\sigma(A)$. Then, by Gershgorin’s circle theorem, an index $i$ exists such that:

which is in contrast with strictly diagonally dominance definition. ∎

Let $A$ be a Hermitian diagonally dominant matrix with real nonnegative diagonal entries; then its eigenvalues are real and, by Gershgorin’s circle theorem, for each eigenvalue an index $i$ exists such that:

which implies, by definition of diagonally dominance,$\lambda\geq 0.$ ∎