Full-rank linearly independent matrices - MathOverflow most recent 30 from http://mathoverflow.net2013-05-20T19:52:50Zhttp://mathoverflow.net/feeds/question/120488http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/120488/full-rank-linearly-independent-matricesFull-rank linearly independent matricessignum2013-02-01T04:44:51Z2013-02-01T12:13:58Z
<p>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?</p>
http://mathoverflow.net/questions/120488/full-rank-linearly-independent-matrices/120510#120510Answer by Chris Godsil for Full-rank linearly independent matricesChris Godsil2013-02-01T12:13:58Z2013-02-01T12:13:58Z<p>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.</p>