Let $Q_1(x)=x^TA_1x$ and $Q_2(x)=x^TA_2x$ with $x\in GF(2)^n$, $A_i\in GF(2)^{n\times n}, i \in \{1, 2\}$. If $rank(A_1)=rank(A_2)=2$, is it possible that $Q_1(x)$ and $Q_2(x)$ can have the same roots in $GF(2)$ with $A_1 \neq A_2$ (excluding the trivial case when $A_2$ is just a permutation of $A_1$)?
I imagine there must be some easy way to write the roots when the rank of $A_1$ and $A_2$ is such a low number? Is there any literature covering this problem or a similar problem? Thank you very much for any answers!
EDIT: Not quite the answer to the question but I did make some progress. If $rank(A_i)=2$ then from this answer we know that $Q_i(x)$ has either $2^{n-2}$ roots (case $\tau_i=-1$) or $2^{n-1}+2^{n-2}$ roots (case $\tau_i=1$).
So, for case $\tau_i=-1$, the roots of $Q_i(x)$ will all be in the null space of $A_i$, as the null space of $A_i$ has dimension $n-2$ because $rank(A_i)=2$. Therefore, $Q_1(x)$ will have the same roots of $Q_2(x)$ if $A_1$ and $A_2$ have the same null space and are both of type $\tau_i=-1$.
I still need to work out the case when they are both of type $\tau_i=1$. I imagine I will need to do some kind of enumeration over the row space of $A_i$, but that shouldn’t be too hard given the low rank.
And, obviously, if they aren’t of the same $\tau$ type then they cannot possibly have the same roots as they will have a different number of roots each.
