non diagonally dominant = singular?

Question

There are non-strictly diagonally dominant matrices which are nonsingular [1].

However, I found no information or reference to the following question: are all non (strictly or not) diagonally dominant matrices also singular?

(The Gershgorin circle theorem does not provide a definite answer on any eigenvalue being zero in this case)

After doing a search I found that something similar (but not quite as general) was asked here before [2], 5 years ago, but received no answer. Maybe someone can answer it now.

Thanks!

Answer 1

I'm sure that $\pmatrix{0&1\\1&0}$ is not diagonally dominant.

(Diagonal dominance is a useful sufficient condition for nonsingularity, but it is not a necessary condition.)