I was wondering if there is a proof that the space of full rank matrices $M(m\times n,R)$ with $m<n$ is a connected space using algebraic geometry.
One can construct $n \choose m$ equations by taking a the determinant of each $m \times m$ submatrix of an element of $M(m\times n,R)$.
If we allow an unknown for each entry in $M(m\times n,R)$, then we end up with $n \choose m$ equations with $m\times n$ unknowns with the zeroes being all the less than full rank matrices.
Is there somehow we can use algebraic geometry to then show the full rank matrices are connected?
If you can compute the global functions on this variety, you can tell whether it's connected or not. Recall that if $X=\bigcup_{i}X_i$ is a disconnected algebraic variety with finitely many connected components $X_i$, then $k[X]=\bigoplus_{i}k[X_i]$. So if you can't decompose $k[X]$ into a direct sum of subrings, then your variety is connected. The easiest place to demonstrate you can't do that is with constant functions.