On Rank of matrices

Question

Let $A$ be a $\{0,1\}$ square matrix.

Let $J$ be all $1$ matrix.

Let $\bar{A}=J-A$.

Is $rk(A)\geq rk(\bar{A})-1$ and $rk(\bar{A})\geq rk({A})-1$ always true?

We are over $\Bbb R$.

Answer 1

Suppose $A \in \mathbb{R}^{n \times n}$ and let $v_1,\dots,v_k \in \mathbb{R}^n$ be a maximal set of linear independent columns of $A$, so $k = rk(A)$. Let $w \in \mathbb{R}^n$ denote the all $1$ vector. Then the span of $\{w - v_1, \dots w - v_k\}$ contains the vectors $v_2 - v_1, \dots, v_k - v_1$ and these are linear independent, so the span of $\{w - v_1, \dots w - v_k\}$ has at least dimension $k - 1$. As $w - v_1, \dots w - v_k$ appear as columns of $\overline{A}$, we have $rk(\overline{A}) \geq k - 1$. The second part follows from $\overline{\overline{A}} = A$.