 Solution

Keep row-vector action and the block order $(W^{\prime },U,W)$. Let $R$ be the reversal matrix of size $k$, arising from the original reversed order of $f_1,\dots ,f_k$, and let $J_U$ be the alternating bilinear form matrix on $U$. ThenFor an element of $Q$, the equation $gJ{g}^T=J$ givesThus $D$, a $k\times (2m-2k)$ matrix, is arbitrary and uniquely determines $E$. The right side of the second equation is alternating, including in characteristic two: its diagonal entries vanish because $J_U$ represents an alternating bilinear form.

For any alternating matrix $K$, the equation $T-T^T=K$ has exactly $q^{k(k+1)/2}$ solutions. For each pair $i<j$, choose one entry freely and solve for the opposite entry; each diagonal entry is free. This works in characteristic two as well as odd characteristic. Taking $T=FR$ therefore givesThis is the unipotent radical count for a symplectic parabolic subgroup. It does not incorrectly replace the alternating constraint by division by two.