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?

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. 

