Is there a way using an affine transformation matrix to convert between a rectangle of zero height (i.e. effectively having 2 different end/corner points) and a rectangle of > zero height (i.e. having 4 different end/corner points)?
If not, does another kind of transformation exist that can do the same?
Thanks in advance.
The affine (linear) function
$$ f : (x,y) \mapsto \frac{r-s}{a-b}\,x + \frac{as-br}{a-b} $$
transform $[a,b]\times[c,d]$ into $[r,s]$, as it is easy to verify. Note however that $f$ is not bijective.
A bijective affine transformation can never transform $[a,b]\times[c,d]$ into $[r,s]$. In fact such a transformation $g$ would be a homeomorphism. And this is impossible: by removing an internal point from $[r,s]$ and the corresponding point (through $g$) from $[a,b]\times [c,d]$, we would again obtain, by restriction, a homeomorphisn between a connected space and a disconnected space, absurd.