Subspaces of matrices whose determinant is $0$

Consider matrices of size $n\times n$ over finite field $\mathbb{F}_2$. It is linear space of dimension $n^2$.

\begin{bmatrix} x_{11} & x_{12} & \dots & x_{1n} \\ x_{21} & x_{22} & \dots & x_{2n} \\ \\ x_{n1} & x_{n2} & \dots & x_{nn} \end{bmatrix}

Now consider set of matrices for which some fixed rows are linearly dependent. For example: let $L$ is set of matrices for which first two rows are the same. It is clear that $L$ is a linear subspace of dimension $n^2-n$ and for all matrices of $L$ determinant is equal to $0$.

I am interested if the opposite is true. Let $L$ is subspace of dimension $n^2-n$ and determinant is $0$ for all matrices of $L$. Can we say that we can fix some set of rows which are linearly dependant for all matrices of $L$.

One can see that there are $2^n-1$ linear combination of rows. So the question is if there are any other subspaces of dimension $n^2-n$ whose determinant is $0$ or each of such subspaces is identified by some linear combination of rows.

Solution 1:

An $n^2-n$ dimensional subspace of singular matrices in $M_n(\mathbb F)$ is either of the form $\{A\vert Ax=0\}$ for some column vector $x,$ or $\{A\vert \zeta A=0\}$ for some row vector $\zeta.$

Apparently, this was proved by Dieudonné, Sur une généralisation du groupe orthogonal à quatre variables. Arch. Math., 1 (1949). I haven't been able to access that article, but a more general result of Meshulam (with quite an easy to read proof) is available here: http://www2.math.technion.ac.il/~meshulam/eprints/maxrank.pdf - set $r=n-1$ in Theorem 3.