Relationship between rank of binary matrix and the NOT operator Let $A$ be a binary matrix. I'm looking for any information about the relationship between the rank of $A$ and the rank of NOT$(A)$, where NOT replaces all $0$s with $1$s, and vice-versa.
What I know


*

*These ranks can sometimes be equal. For example, applying the NOT operator to the identity matrix returns another full rank matrix.

*They can sometimes not be equal. For example, the matrix 
\begin{equation*} A=
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix}
\end{equation*}
has rank $2$, but 
\begin{equation*} \text{NOT}(A)=
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
\end{equation*}
has rank $1$.
My questions
Are there known relationships between the two ranks?
 A: If $E$ is the $n \times n$ matrix of all $1$'s, $NOT(A) = E - A$.  Now $E$ has rank $1$, and in general $$\text{rank}(A)-\text{rank}(B) \le \text{rank}(A+B) \le \text{rank}(A) + \text{rank}(B)$$
Thus the rank of $NOT(A)$ differs from that of $A$ by at most $1$.  
You gave an example where the ranks are equal, and one where $\text{rank}(NOT(A)) = \text{rank}(A) - 1$; interchange $A$ and $NOT(A)$ and you have an example where $\text{rank}(NOT(A)) = \text{rank}(A) + 1$.
A: It may help to notice that 
$$
\operatorname{not}
\begin{bmatrix} 
x_{11}       & x_{12}\\
x_{21}       & x_{22}\\
\end{bmatrix}
=
\begin{bmatrix} 
1       & 1\\
1       & 1\\
\end{bmatrix}
-
\begin{bmatrix} 
x_{11}       & x_{12}\\
x_{21}       & x_{22}\\
\end{bmatrix}
$$
and $\operatorname{rank}\begin{bmatrix}1\end{bmatrix}_{nn} = 1$.
since $\operatorname{rank}(A+ B) \le \operatorname{rank}(A) + \operatorname{rank}(B)$, you can tell that
 $$\operatorname{rank}(\operatorname{not}(A)) \le \operatorname{rank}(A) + 1$$
and also
$$\operatorname{rank}(A) = \operatorname{rank}(\operatorname{not}(\operatorname{not}(A))) \le \operatorname{rank}(\operatorname{not}(A)) + 1$$
which means $$\operatorname{abs} \left(\ \operatorname{rank}(A) - \operatorname{rank}(\operatorname{not}(A)) \ \right) \le 1$$
