Creating a matrix such that all the sub-matrices are max rank

Question

Let $A\odot B$ denote the elementwise multiplication of matrices $A$ and $B$. Given a binary matrix $B_{m \times n}=[b_{ij}]$, $b_{ij} \in \{0,1\}$, I want to find a matrix $A=[a_{ij}]$, $a_{ij}\in GF(2^q)$, such that every sub-matrix of $A\odot B$ has the maximum rank possible. Is that true that if the rows of $A$ are chosen from a Vandemonde matrix, then $A$ has the desired property?

Answer 1

Ok, I found the answer. The answer is NO. This is because for non-zero $a,b \in GF(2^q)$, $a \neq b$, we may find a $k$ such that $a^k=b^k$. Assume $a^k$ and $b^k$ are two elements of a column of a Vandermonde matrix other than the first column. Therefore, we have a sub-matrix:

$$ \left[ \begin{array}{c c} 1 & a^k \\ 1 & b^k \end{array} \right] $$

Since $a^k=b^k$ the rank of the sub-matrix is 1 while it could be 2.