Let $A\_{n}(F) $ denote the $n \times n$ skew symmetric matrices over a finite field $F$. Suppose $n$ be even and $N$ be a subspace of $A\_{n}(F) $. Now if all the non-zero matrices in $N$ are invertible, then the maximum the dimension of $N$ will be $n/2$. The upper bound of this maximum dimension follows from Chevalley warning theorem .
Also I know that there is a such type of subspace of dim $n/2$. But the way to get this result is tricky. Does there any simplest process to find this type subspace $N$ of $\dim n/2$? Provide me some example of such type subspace.



Problem concerning about an $n$-subspace of $ A_{n}(F) $
If you take $n/2 \times n/2$ matrix $A$, with an irreducible minimal polynomial (as can be constructed with a [companion matrix](https://en.wikipedia.org/wiki/Companion_matrix)), the subspace $W = \text{span}\left(I, A, \ldots, A^{n/2-1}\right)$ will be a subspace of matrices of dimension $n/2$ where all non-zero matrices are invertible.


You can create the subspace of $n \times n$ skew symmetric matrices by taking matrices of the form


$$\left[\begin{array}{cc}0 & X \\ -X & 0 \end{array}\right]$$


for $X \in W$.


