For $k \times n$ matrices $X,Y$ of rank $k$, find a characterization of $Y = AX$, $A \in \operatorname{Gl}(k, \mathbb{R})$ Let $F(k,n)$ (with $k < n$) denote the set of $k \times n$ matrices of of rank $k$ (entries in $\mathbb{R}$). We can give $F(k,n)$ a Euclidean topology by identifying it with a subset of $\mathbb{R}^{kn}$. 
We can define an equivalence relation $\approx$ on  $F(k,n)$ via $X \approx Y \iff Y = AX$, some $A \in \operatorname{Gl}(k, \mathbb{R})$. I would like to show that $W = \{(X,Y) \in F(k,n) \times F(k,n) : X \approx Y \}$ is a closed subset of $F(k,n) \times F(k,n)$ (with the product topology).
I have tried to work out this problem previously (For $k \times n$ matrices $X,Y$, when do we have $X = AY$ for $A \in \operatorname{Gl}(k, \mathbb{R})$?), but my ideas were not well-formed at the time, so I have made this more rigorous post. 
My attempt at a solution: show that $W$ equals a finite intersection of zero sets of continuous functions $F(k,n) \times F(k,n) \to \mathbb{R}$. Well, the continuous functions I am thinking of can be described as follows. We take $(X,Y) \in F(k,n) \times F(k,n)$ and form a $(k +1) \times (k+1)$ matrix, using $k$ rows from $X$, and then the last row we pick from $Y$ (of course, we need to delete some columns so we would be using the FULL rows from $X$ and $Y$). Then we calculate the determinant of this new matrix. This type of function is continuous since it's a continuous function of the entries in $X$ and $Y$. Let $B$ denote the collection of all functions that match this description. 
If $Y = AX$, the the rows of $Y$ are linear combinations of the rows of $X$. Hence, if we pick out a $(k+1) \times (k+1)$ submatrix as described above, it will have determinant $= 0$ since the last row will be a linear combination of the first $k$ rows.
So, up to this point I think I've shown that $W \subseteq \bigcap_{f\in B} f^{-1}(0)$. But the tricky part is showing the reverse containment. That is, if we have a pair $(X,Y) \in \bigcap_{f\in B} f^{-1}(0)$ how can I demonstrate the existence of the required $A \in Gl(k,n)$ so that $Y = AX$? Am I even on the right trick with this collection $B$ that I'm using?
Hints or solutions are greatly appreciated.    
 A: Let $X_n\in F(k,n)$, $Y_n\in F(k,n)$, and $A_n\in GL(k,\mathbb{R})$, and they are related by $Y_n= A_n X_n$. 
Our goal is to prove that $X_n\rightarrow X\in F(k,n)$, and $Y_n\rightarrow Y\in F(k,n)$ implies the existence of $A\in GL(k,\mathbb{R})$ such that $Y=AX$. 
We can pick $k$ linearly independent columns($Y^{i_1},\cdots, Y^{i_k}$) of $Y$. Combine these $k$ columns to make a $k\times k$ invertible matrix $Y'$. 
Then we have for $X_n'=(X_n^{i_1},\cdots, X_n^{i_k})$, $Y_n'=(Y_n^{i_1},\cdots, Y_n^{i_k})$ (corresponding columns for $X_n$ and $Y_n$), 
$$Y_n'=A_n X_n'.$$ 
For sufficiently large $n$, the matrices $X_n'$ and $Y_n'$ are invertible. 
We also have that $X_nX_n^T$ is invertible $k\times k$ matrix. So, we have another formula 
$$A_n=(Y_n X_n^T)(X_nX_n^T)^{-1}$$
Therefore, $A_n$ converges to some matrix $A$, but it remains to show that $A$ is invertible. For,
$$det Y_n'=det A_n det X_n'$$
We now see that any sequence of determinants above, should converge, but the LHS is nonzero. Hence, $det A_n$ converges to a nonzero value, and $A$ in invertible as claimed. 
