Let $A$ be a boolean matrix (eg with $0,1$ entries). Assume that $A$ has rank $\le r$ both over $\mathbb{F}_2$ and over $\mathbb{F}_3$. Does this imply that $A$ has low rank over the reals? This seems highly unlikely to me, but I cannot find a counterexample.Note that if we require that $A$ has low rank just over one characteristic (say $\mathbb{F}_2$) then the Hadamard matrix shows an exponential gap (i.e. rank $r$ over $\mathbb{F}_2$; rank $2^r$ over the reals), which is tight. However, this example does not seem to carry over when we require low rank in two different characteristics.Update 6/9:Li Qian gave a super-polynomial separation: a matrix $A$ with rank $r$ modulo 2 and modulo 3, but rank approximately $r^{\log r}$ over the reals. So I guess the real question is: is this the largest separation?To build it, consider first boolean functions $f:\{0,1\}^n \to \{0,1\}$. Take the $2^n \times 2^n$ matrix $A$ given by $A_{x,y} = f(x \wedge y)$, where $x \wedge y$ is the bit-wise AND. The rank of $A$ (in any field) equals the number of monomials in the polynomial representation of $f$, which is essentially controlled by the degree of $f$.Let $n=k^2$. Take $$f(x_1,\ldots,x_n)=MOD_2(MOD_3(x_1,\ldots,x_k),MOD_3(x_{k+1},\ldots,x_{2k}),\ldots,MOD_3(x_{n-k+1},\ldots,x_n))$$ where $MOD_p(\cdot)$ is the boolean function which returns 1 if its input hamming weight is divisible by $p$. Then one can check that $f$ has degree $O(\sqrt{n})$ modulo 2 and modulo 3, but linear degree over the reals. This translates to a matrix $A$ with rank $n^{O(\sqrt{n})}$ modulo 2 and modulo 3, but rank $2^{\Omega(n)}$ over the reals.
Rank mod 6 vs rank over the reals
linear algebra;matrices;boolean matrix