$a_{ii}=0, a_{ij}+a_{ji}=1, 1\leq i<j\leq n$. This matrix has rank $\geq n-1$?

Question

Let $A$ be an $n\times n$ matrix with entries $a_{ij}$. Assume that $a_{ii}=0, a_{ij}+a_{ji}=1, 1\leq i<j\leq n$. This matrix has rank $\geq n-1$?

How to prove then? What is known is that $A+A'$ is invertible.

Answer 1

supposing this is in reals:

compute the same thing 2 different ways -- if rank is $\leq n-2$, apply rank-nullity-- that means $\dim\ker\big(A\big)\geq 2$, so you can find a unit length vector $\mathbf v$ in the kernel of $A$ that is orthogonal to the ones vector, $\mathbf 1$. Then

$0= \mathbf v^T \big(A+A^T\big)\mathbf v =\mathbf v^T \big(\mathbf {11}^T-I_n \big)\mathbf v =\mathbf v^T \big(-I_n \big)\mathbf v=-1$
which is a contradiction.

the technique is nicely illustrated in miniature 8
https://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf