TITLE: Relationship between rank of binary matrix and the NOT operator
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $$\operatorname{abs} \left(\ \operatorname{rank}(A) - \operatorname{rank}(\operatorname{not}(A)) \ \right) \le 1$$