Determining the Lie algebra of a matrix Lie group I have the following Lie group:
$$G:=\bigg\{\begin{pmatrix} x&0\\y &1/x\end{pmatrix}\bigg\vert x, y\in\mathbb{R} \land y\neq0\bigg\} $$
Now I should calculate the corresponding Lie algebra.... 
Obviously, $G$ is a subgrouo of the general linear group $\mathrm{GL}_{2}(\mathbb{R})$.....But i have no idea how to construct the Lie algebra, which is the set of all left invariant vector fields. 
 A: Two parameters, so two generators, so one commutation relation. Has to be the affine group in one dimension, here transposed.
$$
G= 
 \begin{bmatrix}
    x  & 0  \\
    y  & 1/x 
\end{bmatrix}  
$$
goes to the identity for $x=1, y=0$. At the identity, then, you have $$
\partial G/\partial x  \to  \begin{bmatrix}
    1  & 0  \\
    0  &  -1 
\end{bmatrix} \equiv a ~, $$
and 
$$
\partial G/\partial y  \to  \begin{bmatrix}
    0  & 0  \\
    1  &  0 
\end{bmatrix} \equiv b ~, $$
so, then,
$$
[b,a]=2b .
$$
Exponentiating an arbitrary linear combination of these two Lie algebra elements yields the generic form G you started with, since the relevant CBH expansion sums elegantly to a simple closed form: your starting point.
Consider the wisecrack
$$
G= 
 \begin{bmatrix}
    x  & 0  \\
    y  & 1/x 
\end{bmatrix} = 
 \begin{bmatrix}
    x  & 0  \\
    0  & 1/x 
\end{bmatrix}  
 \begin{bmatrix}
    1  & 0  \\
    y x  & 1 
\end{bmatrix} = e^{(\ln x) ~a}  e^{xy~ b}  ,
$$
and its CBH contraction,
$$
=\exp \left ( \ln x ~a -2xy\frac {\ln x}{1-x^2}   b   \right ) = e^{\omega(a + b y/\!\sinh \omega)},
$$
upon the definition $x\equiv e^\omega$.
