Generalizing a surjection ${\bf GL}(2,\Bbb R)\cong (\bf M,\circ)$ Consider the set of all invertible $2\times 2$ matrices over $\mathbb R$ (I think we can do it over $\Bbb C$ but I didn't look at it) $${\bf GL}(2,\Bbb R)=\left\{\left(\begin{matrix}  a&b \\c &d\end{matrix}\right):a,b,c,d\in\Bbb R,ac-bd\neq 0 \right\}$$
Consider now the set $\bf M$ of all functions of the form
$$f(x)=\frac{ax+b}{cx+d}$$
again with $a,b,c,d\in\Bbb R,ad-bc\neq 0 $. If we identify each of these with a matrix 
$$\left(\begin{matrix}  a&b \\c &d\end{matrix}\right)$$ then we can define an isomoprhism between $\rm GL$ and $(\bf M,\circ)$ as groups with operations of matrix multiplications and functional composition, respectively, and identities$$\left(\begin{matrix}  1&0 \\0 &1\end{matrix}\right)=e$$
$$x=\frac{1x+0}{0x+1}=e'$$ 
since if $$\eqalign{
  & f = \frac{{ax + b}}{{cx + d}}  \cr 
  & g = \frac{{ex + f}}{{gx + h}} \cr} $$ then $$f \circ g = \frac{{\left( {ae + bg} \right)x + af + bh}}{{\left( {ce + dg} \right)x + dh + cf}}$$
which corresponds to 
$$\left(\begin{matrix}  a&b \\b &c\end{matrix}\right)\left(\begin{matrix}  e&f \\g &h\end{matrix}\right)=\left(\begin{matrix}  ae + bg&af + bh \\ce + dg &dh + cf\end{matrix}\right)$$
and similarily for inversion, $${f^{ - 1}} = \frac{1}{{ac - bd}}\frac{{dx - b}}{{ - cx + a}}$$
and $$\left(\begin{matrix}  \frac  d\Delta& \frac {-b}\Delta \\\frac{-c}\Delta &\frac a\Delta \end{matrix}\right)=e$$
I include the determinant $\Delta=ac-bd$ inside the matrix to avoid any multiplication by scalars considerations.
My question is: How can this be generalized to ${\bf GL}(n,K)$, and what is the theoretical relevance of this? I know Möbius Transformations are important in Complex Analysis, for instance, but I haven't seen any "higher dimesional" equivalent around.
ADD As users noted, for every $a\in \Bbb R$,$$\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\mapsto x$$
so the isomoprhism is actually obtained by quoting ${\bf GL}(2,\Bbb R)$ by $I=\{aI_2:a\in\Bbb R\}$
 A: This map isn't an isomorphism; all scalars are sent to the identity. The generalization goes by the name of the projective general linear group; this naturally acts on projective space, hence the name, and is the natural target for projective representations, which are important for example in quantum mechanics. 
A: You don't have an isomorphism but just a surjective map $f : \textrm{GL}_2(\Bbb{R}) \longrightarrow \textbf{M}$. The kernel of this map is all scalar multiples of the identity (as you can check) and so by the first isomorphism theorem,
$$\textbf{M} \cong \textrm{GL}_2(\Bbb{R})/ \{ aI\}$$
where $a$ ranges over all non-zero real numbers. The group on the right is called $\textrm{PGL}_2(\Bbb{R})$ (if you are working over $\Bbb{R}$). The generalisation of this is of course when you consider $\textrm{GL}_n(\Bbb{C})/\{aI\}$ that (as you may have guessed) is called $\textrm{PGL}_n(\Bbb{C})$.
I don't really know much about this group except that if you have $\textrm{GL}_n$ acting on say $\Bbb{R}^n$, then you can ask what the associated action will be on the quotient $\Bbb{R}\textrm{P}^{n-1}$ space. Now recall that real projective space $\Bbb{R}P^{n-1}$ is the space of all lines through the origin in $\Bbb{R}^n$. Suppose we let a scalar multiple of the identity act on one of these lines. Then the line just get sent to itself yes? Rather, every element on the line is sent to a scalar multiple of itself. However in terms of projective space, that line is considered as just one point and so any scalar multiple of the identity matrix becomes the "identity matrix"  on $\Bbb{R}P^{n-1}$. That's why we quotient it out to get that the "actual group" acting on $\Bbb{R}P^{n-1}$ is $\textrm{PGL}_{n}$.
