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Quadratic forms with the same roots over GF(2) for low rank problems

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 ...
Fabio Dias's user avatar
5 votes
1 answer
432 views

Number of roots of a quadratic form over GF(2)

If $Q(x) = x^T A x$ with $x \in GF(2)^n$ and $A \in GF(2)^{n \times n}$, is there a way to find how many roots $Q(x)$ has based on any properties of $A$ (e.g., rank, etc.)?
Fabio Dias's user avatar
2 votes
0 answers
100 views

Sum of binary quadratic forms over inputs of equal Hamming weight

$\DeclareMathOperator{\field}{\mathbb{F}}$ Let $n$ be a positive integer, and let $q : \field_2^{n} \rightarrow \field_2$ be a quadratic form, specified in coordinates as $$q(x)=\sum_{i =1}^n \alpha_i ...
Ben's user avatar
  • 980
3 votes
2 answers
246 views

Number of solutions of quadratic equation from a perfect pairing over $\mathbb{Z}/p^n\mathbb{Z}$

Let $p>2$ be a prime number, $V=\left(\mathbb{Z}/p^n\mathbb{Z}\right)^{2k+1}$. The bilinear form $$B:V\times V \rightarrow \mathbb{Z}/p^n\mathbb{Z}$$ is a perfect pairing. That is, mapping $x\in V$ ...
Ted Mao's user avatar
  • 453
9 votes
1 answer
360 views

Standard conjecture on u-invariants?

This is well beyond my expertise, but I just learned some of the history behind $u$-invariants of fields $F$, where ($u(F)+1$)-variable quadratic equations always have a non-trivial solution, but $u(F)...
Joseph O'Rourke's user avatar
2 votes
0 answers
111 views

Inseparable field extensions of degree p and linear independence

Let $F$ be a field of characteristic $p$; let $\alpha \in F$ such that $\alpha \neq \beta^p$ for any $\beta \in F$, and let $K := F(x)$ where $x=\sqrt[p]{\alpha}$. Is it true that the elements $1,(x-...
adam chapman's user avatar
1 vote
0 answers
111 views

Reference request: Number of elliptic and hyperbolic quadratic forms of a given rank over a finite field

My question is over the finite field $\mathbf{F}_q$ of $q$ elements. It is well known that a symmetric matrix of odd rank corresponds to a parabolic quadratic form but even rank symmetric matrices ...
Singh's user avatar
  • 179
6 votes
1 answer
1k views

orthogonal group in characteristic 2

Let $O(2,\mathbb{Z}_2)$ be the orthogonal group of order two matrices. On $\mathbb{Z}_2$ there should exist just one odd quadratic form, hence the stabilizer subgroup $O^-$ of an odd quadratic for ...
IMeasy's user avatar
  • 3,779
2 votes
1 answer
186 views

Stabilizer of a nonsingular vector in a quadratic space (char (k)=2)

suppose that $k$ is a finite field of characteristic 2 and $(V,q)$ a quadratic space, i.e., $V$ is a $k$-vector space and $q:V\to k$ quadratic form. Suppose that $\dim(V)\geq 4$ and that $q$ is non-...
César Galindo's user avatar
8 votes
0 answers
1k views

roots of quadratic forms

This may be a very silly question, but I was wondering what is known about the roots of a quadratic form over variables $x_1,\ldots,x_n,y_1,\ldots,y_m$ in the finite field $\mathbb{F}_p$. I'm not ...
Sarah's user avatar
  • 131