This exercise is from "Lie groups beyond an introduction, Knapp": Identify the linear Lie algebra of
$$ G=\left\{ \begin{bmatrix} a & z \\ 0 & a^{-1} \end{bmatrix} \quad \Bigg| \quad a\gt 0, \quad z\in\mathbb{C}\right\} $$
I was thinking of using an explicit formula for the exponential of a $2\times 2$ matrix, such as the one at the end of this page:
mathworld.wolfram.com/MatrixExponential.html
And then using the fact that the linear Lie algebra is
$$ \mathcal{g}= \{ X\in M_n(\mathbb{C})\quad|\quad \exp(tX)\in G\quad\mbox{for all }t\in\mathbb{R} \}. $$ But I think this won't give a "nice" answer, is there an easier way to do this?
Two parameters, so two generators, so one commutation relation. Has to be the affine group in one dimension, where the diagonal generator is doubled and shifted by subtraction of the identity matrix.
$$ G= \begin{bmatrix} a & z \\ 0 & a^{-1} \end{bmatrix} $$ is the identity for $a=1, z=0$. At the identity, then you have $$ \partial G/\partial a \to \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \equiv x ~, $$ and $$ \partial G/\partial z \to \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \equiv y ~, $$ so, then, $$ [x,y]=2y . $$
The conventional WP normalization redefines your x to $\omega\equiv (x+1\!\!1)/2$ to yield $[\omega,y]=y$. 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 $$ G= \begin{bmatrix} a & z \\ 0 & a^{-1} \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a^{-1} \end{bmatrix} \begin{bmatrix} 1 & z/a \\ 0 & 1 \end{bmatrix} = e^{(\ln a) ~x} e^{(z/a) y} $$ and its CBH contraction.