linear algebra - Symmetric diagonally dominant matrix

Suppose I have a real, symmetric, $n\times n$ matrix $A$ such that the following conditions hold:

1) All diagonal elements $a_{ii}$ are strictly positive.

2) All off-diagonal elements $a_{ij}$ are non-positive.

3) The sum of the elements in each row (and therefore also in each column since $A$ is symmetric) is nonnegative. Moreover, there exists at least one row where this sum is strictly positive.

Does it follow, then, that $A$ has full rank?

Answer

If I'm not mistaken,

$$A=\pmatrix{2&-1&0&0\\-1&2&0&0\\
0&0&1&-1\\
0&0&-1&1}$$
gives a counter-example (for $n\geq 4$, since it can be extended adding an identity matrix block of size $n-4$).
Indeed

• $a_{ii}\in\{1,2\}$ so $a_{ii}>0$.


• The extra-diagonal elements are $0$ or $-1$, hence non-positive.


• The sum of the rows are either $1$ or $0$ hence non-negative.



• The sums of the elements of the first row is $1$ which is positive.


• $A$ is symmetric.