Conjecture on the 2-rank of the matrix $M_2$

Let $q=p^e$ be any odd prime power and let

$$V=M_2(\mathbb F_q).$$

In the notation of Subsection 3.2, rows are indexed by $N\in V$, and a hyperbolic-quadric column indexed by

$$B\in\mathrm{GL}_2(\mathbb F_q),\qquad C=C^{\mathsf T},$$

has incidence condition

$$NB+B^{\mathsf T}N^{\mathsf T}+C=0. \tag{1}$$

If $N_0$ satisfies (1), then $N$ satisfies it exactly when

$$(N-N_0)B$$

is skew-symmetric. Writing

$$J=\begin{pmatrix}0&1\\-1&0\end{pmatrix},$$

the complete column support is therefore the affine line

$$N_0+\mathbb F_qD, \qquad D=JB^{-1}\in\mathrm{GL}_2(\mathbb F_q). \tag{2}$$

Conversely, every invertible $D$ and every $N_0$ occur: choose

$$B=D^{-1}J,\qquad C=-(N_0B+B^{\mathsf T}N_0^{\mathsf T}).$$

Thus $M_2$ is the incidence matrix of all affine lines in $V$ with invertible matrix directions.

Choose a finite extension $E/\mathbb F_2$ containing a primitive $p$-th root of unity $\zeta$. For $A\in V$, define

$$\chi_A(N) = \zeta^{\operatorname{Tr}_{\mathbb F_q/\mathbb F_p} (\operatorname{tr}(A^{\mathsf T}N))}.$$

Because $|V|=q^4$ is odd, these additive characters form an $E$-basis of $E^V$.

Let $W\subseteq E^V$ be the span of the column incidence vectors. Every translate of a line (2) is again a column, so $W$ is translation-invariant. For the line $L_D=\mathbb F_qD$, its Fourier coefficient at $\chi_A$ is

$$\sum_{t\in\mathbb F_q} \zeta^{\operatorname{Tr}_{\mathbb F_q/\mathbb F_p} (t\operatorname{tr}(A^{\mathsf T}D))} = \begin{cases} q=1\in E,&\operatorname{tr}(A^{\mathsf T}D)=0,\\ 0,&\operatorname{tr}(A^{\mathsf T}D)\neq0. \end{cases} \tag{3}$$

Character orthogonality and translation invariance imply

$$\chi_A\in W$$

whenever there exists an invertible $D$ satisfying

$$\operatorname{tr}(A^{\mathsf T}D)=0. \tag{4}$$

Such a $D$ exists for every $A$. If $A=0$, take $D=I_2$. If $\operatorname{rank}A=1$, write

$$A=UE_{11}V,\qquad U,V\in\mathrm{GL}_2(\mathbb F_q),$$

and set

$$D=U^{-\mathsf T} \begin{pmatrix}0&1\\1&0\end{pmatrix} V^{-\mathsf T}.$$

If $\operatorname{rank}A=2$, write $A=UI_2V$ and set

$$D=U^{-\mathsf T} \begin{pmatrix}1&0\\0&-1\end{pmatrix} V^{-\mathsf T}.$$

In both cases $D$ is invertible and satisfies (4).

Therefore $W$ contains all $q^4$ additive characters, so

$$W=E^V.$$

Scalar extension preserves the rank of a matrix over $\mathbb F_2$. Consequently,

$$\boxed{\operatorname{rank}_{\mathbb F_2}(M_2)=q^4}$$

for every odd prime power $q$, as conjectured.