I lately encountered the following problem, which I conjecture to be true, but have been unable to either give a proof or raise a counterexample.
Let $A$ be a $m \times n$ binary matrix ($m \ge 2n, n \ge 2$), i.e. each entry is either $0$ or $1$. We assume that the columns of $A$ are distinct, and each column contains at least one entry of $0$. We additionally assume that $A$ contains two non-overlapping $n \times n$ submatrices with their diagonals being all $1$'s, so without loss of generality we can have $A_{i, i} = A_{n + i, i} = 1$ for all $1 \le i \le n$ (the following is an illustration).
$$ \left(\begin{matrix} 1 & \cdot & \cdot \\ \cdot & 1 & \cdot \\ \cdot & \cdot & 1 \\ 1 & \cdot & \cdot \\ \cdot & 1 & \cdot \\ \cdot & \cdot & 1 \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \end{matrix}\right) . $$
We denote the $i$th row of $A$ by $a_i$, and denote the $(n + 1)$-dimensional extended vector $(1, a_i)$ by $\tilde{a}_i$. We let $B$ be a $m (m - 1) / 2 \times (n + 1)$ matrix, with the rows being $\tilde a_i * \tilde a_j$ for pairs of $(i, j)$ in $1 \le i < j \le m$, where $*$ denotes entrywise product of vectors.
Under all the above assumptions, is it true that this matrix $B$ has full column rank $n + 1$?