Full-rank linearly independent matrices

Can we find $n^2$ full-rank matrices in $\mathbb{F}^{n \times n}$ which are linearly independent (i.e. when vectorized are linearly independent)? If not, how many such matrices can be found?

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Yes--the set of full-rank matrices is Zariski-open (and non-empty) in $\mathbb{F}^{n\times n}$, so if $k$ is less than $n^2$, the set of full-rank matrices not in the span of $k$ chosen matrices is also Zariski-open and non-empty. (Strictly speaking, this argument only works over finite fields, but it can easily be fixed). This question is more appropriate for math.stackexchange.com. –  Daniel Litt Feb 1 '13 at 5:18

If the characteristic of $\mathbb{F}$ is not two and $E_{i,j}$ $(1\le i,j\le n)$ is the "standard basis", then the matrices $I+E_{i,j}$ are invertible. They are linearly independent if $n+1\ne0$ in $\mathbb{F}$. If the characteristic of $\mathbb{F}$ is greater than 2, we can use $I-E_{i,j}$ instead and these are linearly independent if $n-1\ne0$. So we have explicit examples except in characteristic two.

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If n > 1, any order n matrix is the sum of exactly two invertible matrices, even if the field has characteristic two. So I would expect an invertible basis for characteristic 2 also. Gerhard "Ask Me About Binary Matrices" Paseman, 2013.02.01 –  Gerhard Paseman Feb 1 '13 at 23:06
I agree that there should be an invertible basis in characteristic 2. (Exercise for the reader?) –  Chris Godsil Feb 1 '13 at 23:38
Consider the cycle q=(1 2 3 ...n) on the set of columns of an order n matrix over F_2 with n > 1. Let Q=q applied to the order n identity matrix, so the main diagonal is shifted "out of the way". In addition to the n(n-1) matrices I + E_ij for i distinct from j, take also the n matrices Q + E_ii. I think this or a slight modification to resolve parity issues should work as a basis in characteristic 2. Gerhard "Ask Me About System Design" Paseman, 2013.02.03 –  Gerhard Paseman Feb 4 '13 at 6:18