 Solution

A representation of the one-arrow quiver is an $n_2\times n_1$ matrix $F$. If $r=\mathrm{rank}F$, changes of basis put it in the form with an $r\times r$ identity block and all other entries zero. Since multiplication by invertible matrices preserves matrix rank,This describes the orbit for every $0\le r\le \min (n_1,n_2)$, including the zero orbit. Its dimension isIndeed, choosing its $r$-dimensional image contributes $r(n_2-r)$ parameters in a Grassmannian, and choosing a surjective map from $k^{n_1}$ onto that image contributes $r{n}_1$. This is a rank orbit of a matrix under left-right multiplication.