Let $n, m \in \Bbb N$. Define $$D:= \{A\in Hom(\Bbb R^n, \Bbb R^m ) |A\mathrm{ \:has \:maximum \: Rank} \}$$ Now I have to show that $D$ is an open and dense subset of $Hom(\Bbb R^n, \Bbb R^m ) $
I don't really know how to approach this, does anyone have some ideas or tipps on how to show this? Thanks in advance!
$\{A\in Hom(\Bbb R^n, \Bbb R^m ) | \mathrm{A \:has \:maximum \: Rank}\}$ dense in $Hom(\Bbb R^n, \Bbb R^m )$
To make life easier, one should reduce to the case $m=n$ as follows: To make sure that a $m\times n$-matrix has full rank you only have to find a quadratic sub-matrix (of size $m$ or $n$, depending on which number is smaller) which is invertible. So you can just pick any quadratic sub-matrix of maximal size and if you can change this sub-matrix to have full rank, the whole matrix has full rank, regardless of all other entries. This reduces to the case $m=n$.
In the $n=m$-case you can make any matrix invertible by adding $\varepsilon I$ for small $\varepsilon$, because $\det(A+\varepsilon I)$ is a polynomial in $\varepsilon$, hence has only finitely many roots.