# Show that a matrix satisfying certain conditions is non-singular

I have a square matrix $A$ satisfying the following conditions:

1. The elements on the diagonal are negative;
2. All other elements are non-negative;
3. All row sums are less than or equal to $0$;
4. There is at least one row sum that is less than $0$.

I think that $A$ is non-singular, but I cannot prove it. I was thinking to try to show that $\det{A} \neq 0$ by determining that there is no eigenvalue that is zero. Can anyone point me to a book or sketch a proof (not necessarily the one I have in mind)?

Edit 1 It seems that I am missing a condition. I will try to sketch the situation more closely. There are two more matrices $B$ and $C$ with non-negative entries and at least one element positive such that the matrix defined by

$$D := A + B + C$$

is irreducible and all rows sums of $D$ are $0$. This should immediately exclude the matrix proposed by @MooS.

Edit 2 Does the first edit imply that $A$ is actually irreducible so that $A$ is irreducibly diagonally dominant and non-singular as shown in the first sentence of this Wikipedia page?

• The base case of 1x1 matrices is trivial since it is simply a matrix containing one negative element. Since swapping two rows only makes the determinant switch signs and does not switch from non-zero to zero, we can swap the bottom row for one of the rows with sum less than $0$. We can then use induction to show that the lower-right hand (n-1)x(n-1) is non-singular since it satisfies all four properties. I'm not sure how to go from there to showing that the n by n matrix is non-singular. Commented Apr 21, 2016 at 13:37

You need all row sums to be $<0$, else you can do the following:
$$\begin{pmatrix}-1&1&0\\1&-1&0\\0&0&-1\end{pmatrix}$$
If you assume all row sums to be $<0$, the statement is true. After multiplying by $-1$, the matrix will be a Diagonally dominant matrix.
• Thanks for the answer. It seems that I am overlooking a condition, because this matrix should actually be excluded. I think the matrix $A$ is called a transient infinitesimal generator from the theory of Markov processes, if that rings any bells.