A strictly diagonally dominant matrix is non-singular. In other words, let be a matrix satisfying the property
Proof: Let ; then a non-zero vector exists such that ; let be the index such that , so that ; we have
in contrast with strictly diagonally dominance definition.
Remark: the Levy-Desplanques theorem is equivalent to the well-known Gerschgorin circle theorem. In fact, let’s assume Levy-Desplanques theorem holds true, and let a complex-valued matrix, with an eigenvalue ; let’s apply Levy-Desplanques theorem to the matrix , which is singular by definition of eigenvalue: an index must exist for which , which is Gerschgorin circle theorem. On the other hand, let’s assume Gerschgorin circle theorem holds true, and let be a strictly diagonally dominant complex matrix. Then, since the absolute value of each disc center is strictly greater than the same disc radius , the point can’t belong to any circle, so it doesn’t belong to the spectrum of , which therefore can’t be singular.