Here is a related post that I find interesting.

In linear algebra, Minkowski‘s criterion states the following.

**Theorem (Minkowski’s Criterion). **Let be an matrix with real entries such that the diagonal entries are all positive, off diagonal entries are all negative, and the row sums are all positive. Then .

This is a nice criterion and is not very difficult to prove, but for a random matrix it is asking too much. To decide whether a matrix is singular one usually looks for a row/column consisting of zeros or adding up to zero. The following result gives sufficient conditions for this to work. Unfortunately, it does not generalise Minkowski’s result.

**Theorem.** Let be a matrix with real entries such that its row sums are all , its lower diagonal entries are and its upper diagonal entries are . Then if and only if has either a row consisting entirely of zeros or *all* the row sums equal to zero.

*Proof. *Suppose that , where . Assume that . Then there exists such that and . Hence

So we must have (i) , (ii) , (iii) and (iv) either or . These boil down to having either or . Apply this argument to each row of to obtain the desired conclusion.