Let $K$ be an algebraically closed field, and $V\subseteq K^n$ a vector subspace of dimension $k$. For a basis of $V$, suppose the map from $V$ to $K^k$ which takes elements of $V$ to their coordinates is onto. Now let $S_V\subseteq M_{n\times k} $ be the set of matrices $A\in M_{n\times k}$ such that the image $Im(A)=A(K^k)=V$.
I need to show that $\dim \overline{S_V}=k^2$ (Here by $\overline{S_V}$ I mean the closure in the Zariski topology in the variety $M_{n\times k}$).
The exercise recommends as a hint using the map $\phi:S_v\to M_{k\times k}$ which takes matrices of $S_V$ to their top $k\times k$ blocks.
I know the dimension of $M_{k\times k}$ is $k^2$, which surely must help to prove the equality.
Also, I think is clear that $\phi$ the restriction to $S_V$ of the same map from $M_{n\times k}$ to $M_{k\times k}$, which is a morphism of irreducible varieties which is onto and also dominant. However, I'm stuck at this point, since I don't exactly know how to use the definition of $S_V$ or the fact that the coordinate map $V\to K^k$ is onto to move on.
Any hint about how to get in the right direction would be very helpful.
New answer (there was a confusion on the problem statement; see below): $S_V$ consists of matrices $A\in M_{n\times k}$ (i.e., linear transformations $K^k\to K^n$) such that $A(K^k)=V$. In particular, it is a subset of the Zariski closed subset $T_V\subset M_{n\times k}$ consisting of linear transformations $A\colon K^k\to K^n$ such that $A(K^k)\subset V$, and we may instead take the closure of $S_V$ inside $T_V$.
But $T_V$ can simply be identified with the set of linear transformations $A\colon K^k\to V$, i.e., $M_{k\times k}$, upon choosing a basis for $V$. Now, $S_V\subset M_{k\times k}$ consists of linear transformations $A\colon K^k\to K^k$ which are surjective, i.e., such that $\det(A)\ne0$. But this is a Zariski open subset of $T_V$, hence $\overline S_V=T_V$. We already know that $\dim(T_V)=k^2$, since $T_V\simeq\mathbb A^{k^2}$ as an algebraic variety.
Old answer: Assume without loss that $V=K e_1\oplus \cdots\oplus Ke_k\subset K^n=Ke_1\oplus\cdots\oplus Ke_n$. Then your variety $S_V$ consists of matrices $A\in M_{n\times k}$ such that $Ae_1,\dots,Ae_k$ spans $K^k$, i.e., $\det(Ae_1,\dots,Ae_k)\ne0$. Thus $S_V$ is a Zariski open subset of $M_{n\times k}$, i.e., the closure is $M_{n\times k}$, of dimension $nk$.