Sum of Ranks of Two Complementary Matrices Suppose we have a binary matrix $A$, i.e, all the elements of $A$ are either $0$ or $1$. Let $B$ be the complementary matrix of $A$, i.e., 
$$
B_{ij}=
\begin{cases}
1,\textrm{ if }A_{ij}=0;\\
0,\textrm{ if }A_{ij}=1.
\end{cases}
$$
Then $A+B$ is a matrix with all entries being $1$. Suppose $\textrm{rank}(A)=r$, what is $\textrm{rank}(A)+\textrm{rank}(B)$?
 A: There is not enough information to tell. 
For example, let $s=\operatorname{rank} B$. If $n=2$, we can take


*

*$A=\begin{bmatrix} 1&1\\0&0\end{bmatrix}$, then $r+s=2$ 

*$A=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, then $r+s=3$ 

*$A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$, then $r+s=4$
Similarly, with $n=3$: 


*

*$A=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0\end{bmatrix}$, then $r+s=3$

*$A=\begin{bmatrix} 1&0&1\\0&1&0\\0&0&0\end{bmatrix}$, then $r+s=4$

*$A=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&0\end{bmatrix}$, then $r+s=5$

*$A=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, then $r+s=6$
A: So the question is : what is rank(B) ? 
The answer is : we cannot know just from the value of rank(A) .
For example take the following binary matrices :
\begin{pmatrix}
1 & 1 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}
And 
\begin{pmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1
\end{pmatrix}
Both these matrices have rank=1, but their complementary matrices have rank=1 and 0 respectivly.
A: You can't really say much about it, there is no formula.
Here is why 
Consider the case $A = I = \mathbb{1}_{n \times n}$ so that 
\begin{equation}
B_{i,j } = \begin{cases} 0 & \text{if } i =j  \\ 
1 & \text{if } i \neq j \\ 
\end{cases}
\end{equation}
Then $\text{rank}(A) = \text{rank}(B) =  n =$ "full rank." 
Likewise consider the case 
\begin{equation}
B_{i,j } = \begin{cases} 1 & \text{if } i =j =1  \\ 
0 & \text{o.w. } \\ 
\end{cases}
\end{equation}
then both $A$ and $B$ have a really small rank. 
