Matrix whose rank differs depending on the domain of the elements

Question

I am looking for a matrix in $\mathbb{F}_2$ which changes its rank if interpreted as a matrix of a different field.

Formally: I am looking for a pair of matrices $A \in \mathbb{F}_2^{m \times n}$ and $B \in \mathbb{F}_p^{m \times n}$ for some $p > 2$, such that

• $B_{ij} = \begin{cases} 1 & \text{if } A_{ij} = 1 \\ 0 & \text{else}\end{cases}$

• $\text{rank}(A) \neq \text{rank}(B)$

Do matrices like that exist?

Answer 1

Sure: choose any matrix of zeros and ones whose determinant is non-zero and even. It will be invertible over $F_p$ for any prime not dividing its determinant, but not over $F_2$.

Here's a random example: $$\left(\begin{matrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix} \right)$$