Let $M = \mathrm{GL}(2, \mathbb{R})$ and define an action of $\mathbb{R}$ on $M$ by the formula $$\theta(t,A)=\begin{pmatrix}1&t\\ 0&1\end{pmatrix}A$$ for $A \in M$. Find the infinitesimal generator.
I think if we denote $A = \begin{pmatrix}a&b\\ c&d\end{pmatrix}$ we have that $$\begin{pmatrix}1&t\\ 0&1\end{pmatrix}\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\begin{pmatrix}a+ct&b+dt\\ c&d\end{pmatrix}$$ and we could identify this with a point $(a+ct, b+dt, c, d)$ in $\mathbb{R^4}$ so would the infinitesimal generator for $(a,b,c,d)\mapsto(a+ct, b+dt, c, d)$ be the vector field $$\frac{d}{dt}(a+ct) \frac{\partial}{\partial a} + \frac{d}{dt}(b+dt) \frac{\partial}{\partial b} + \frac{d}{dt}(c) \frac{\partial}{\partial c} + \frac{d}{dt}(d) \frac{\partial}{\partial d} = c\frac{\partial}{\partial a} + d \frac{\partial}{\partial b} ?$$
As Dan Asimov pointed out in the comment section, the generator $G$ of the transformation $\theta(t) = e^{tG}$ is given by its derivative at the point $t=0$, i.e. $$ G = \left.\frac{\mathrm{d}\theta}{\mathrm{d}t}\right|_{t=0} = \left.\frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\right|_{t=0} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$