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?
Best 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