Let's say I map a $3 \times 1$ vector $\underline v=(x, y, z)$ by multiplying it with a $3 \times 3$ matrix of rank $2$. Would I be correct in thinking that it transforms all points in 3D space into a 2D plane in 3D space? Whereas multiplying $\underline v=(x, y, z)$ by a $2\times 3$ matrix also of rank 2 transforms all points in 3D space into all points in the 2D plane?
If my thinking is correct, would both linear maps be considered surjective? I'm quite sure $f$ is surjective in the second linear transformation I described. However, would the first transformation be considered surjective as well, since each point in the plane is "reached" by more than one point from the original 3d space? The reason I'm confused about this is because a surjective linear map is defined as $\operatorname{Im}(f)=W$, but I'm not sure whether $W$ refers to the entire 3d space or just the plane I described in 3d space; if the former is the case, then surely $\operatorname{Im}(f)$ doesn't span all of $W$, but only a subspace of it (a plane).
So in the linear map $ f \colon V \to W$, does the linear space $W$ refer to the space the vector lives in, or just to the space spanned by the vectors? In other words if a transformation is defined by $f : \mathbb{R}^n\to\mathbb{R}^m$, does the dimension $n$ and $m$ refer to the dimension of the vector space where the vector lives in (e.g. $m = 3$ for a plane in 3D space), or the just dimension that the vectors span ($m = 2$ for a plane in 3D space)?
To extend this further, can I conclude that all linear transformations by an $m \times n$ matrix $A$ is surjective if $\operatorname{rank}\le n$, or only if $m\le n$and the matrix has full rank?