Over a finite field, which square matrices produce a zero quadratic form?

Question

For which matrices $A \in (\mathbb{F}_p)^{n \times n}$ do we have $x^T A x=0$ for all $x \in (\mathbb{F}_p)^n$?

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Obviously, this is the case if $A=B-B^T$ for some $B$ (which is equivalent to saying that $A$ is anti-symmetric if $p\neq 2$). Is this if an iff?

I think for $\mathbb{R}$ instead of $\mathbb{F}_p$, this is true since matrices of the form $xx^T$ span the whole space of symmetric matrices. Does that still hold for $\mathbb{F}_p$?

Answer 1

$A$ is anti-symmetric if $A^t=-A$. However

$a_{ii}=e_i^tAe_i=0$

(Here $\{e_i\}$ is the standard basis )

Then

$$(e_i-e_j)^tA(e_i-e_j)=-e_i^tAe_j-e_j^tAe_i=-a_{ij}-a_{ji}=0$$

And so $a_{ij}=-a_{ji}$

So your statement holds for any field, except for $\mathbb{F}_2$, as you can see in the comments below.

Answer 2

Oh, I see it now; it's an iff, and the argument isn't really different for $\mathbb{R}$ or $\mathbb{F}_p$. We have $$ (e_1+e_2) (e_1+e_2)^T-e_1e_1^T - e_2e_2^T = e_1e_2^T+e_2e_1^T$$ for any pair of basis vectors $e_1,e_2$, so vectors of the form $xx^T$ do span the space of symmetric matrices. Since $$x^T A x = Tr(A xx^T)\;,$$ $A$ must have zero overlap with the space of symmetric matrices and thus be anti-symmetric.