Showing that the group $SL(2, R)$ acts on the upper half plane by linear fractional transformations. I need to prove that $SL(2, R)$ acts on the upper half-plane $H = \{z ∈ C\mid \text{Im}(z) > 0\}$ by linear fractional transformations. Really, I am not sure even how to begin with. I know I have to reach to the conclusion that a complex number $z$ gets mapped to $$\frac{az+b}{cz+d}$$
If I have some $z \in H$, to multiply a matrix times $z$ is just the matrix:
$$\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix} z = \begin{pmatrix}
az & bz \\
cz & dz 
\end{pmatrix}$$
How do I get from this to what I want to prove? I also tried writing $z=x +iy$ and then multiplying $$\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix} \begin{pmatrix}
x \\
y  
\end{pmatrix} = \begin{pmatrix}
ax + by \\
cx + dy 
\end{pmatrix}$$
which I guess represents the complex number $(ax+by) + i(cx + dy)$. Then I rewrote $\frac{az+b}{cz+d}$ as $\frac{a(x+iy)+b}{c(x+iy)+d}$ and tried to show they are equal. But it turned quickly into some nasty algebra which I am not even sure that it will lead me to somewhere...
Any help is indeed appreciated.
 A: As in comments: the action is more complicated than "linear" (=matrix multiplcation), though it is indeed descended from the linear action of 2-by-2 matrices on 2-vectors.
A somewhat "fancy", but (to my mind) infinitely more explanatory description of the situation is as follows. Certainly 2-by-2 complex matrices act on $\mathbb C^2$. In particular, the invertible matrices give a group action. Further, $\Omega=\mathbb C^2-\{(0,0)\}$ is stable under the action of invertible matrices.
That the action of the matrices is linear exactly means that the action commutes with scalar multiplication. Thus, that action descends to the quotient of $\mathbb C^2-\{(0,0)\}$ by non-zero scalar multiplication by $\mathbb C^\times$. The latter quotient is complex projective on-space, also known as "the Riemann sphere".
$\mathbb C$ injects to the Riemann sphere (in this model) by $z\to \pmatrix{z\cr 1}$ (or, strictly speaking, to the equivalence class of the latter in the quotient).
In this viewpoint, the linear action of the matrices on $\mathbb C^2$ descends to the "linear fractional action" on the projective space, which is the classical action on $\mathbb C\cup \infty$, where $\infty$ is the equivalence class of $\pmatrix {1\cr 0}$ in this model.
Thus,
$$
\pmatrix{a&b\cr c&d}(z) \;=\; \pmatrix{a&b\cr c&d}\pmatrix{z\cr 1} \;=\;
\pmatrix{az+b\cr cz+d} \;\tilde\; \pmatrix{(az+b)/cz+d)\cr 1}
$$
in the quotient, when $cz+d\not=0$.
