Find the infinitesimal generator of $\theta(t,A)=\begin{pmatrix}1&t\\ 0&1\end{pmatrix}A$

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 $$\frac{d}{dt}(a+ct) \frac{\partial}{\partial a} + \frac{d}{dt}(b+dt) \frac{\partial}{\partial b} + \frac{d}{dt}(c) \frac{\partial}{\partial c} + \frac{d}{dt}(d) \frac{\partial}{\partial d} = c\frac{\partial}{\partial a} + d \frac{\partial}{\partial b} ?$$

• The infinitesimal generator of a one-parameter subgroup {g(t)} of a Lie group is found by taking the derivative with respect to t, and then evaluating at t = 0. Commented May 26, 2023 at 17:21
• Wouldn't it give the same result as we don't have any $t$'s left? @DanAsimov Commented May 26, 2023 at 17:26
• Sorry, but I have no idea what you are referring to. Commented May 26, 2023 at 17:27
• Can you help me see where I'm going south here? @DanAsimov Commented May 26, 2023 at 18:23
• A one-parameter subgroup in GL(2,ℝ) is always of the form exp(t M) where M is an arbitrary 2×2 real matrix. Its derivative at t = 0 is M. Commented May 27, 2023 at 16:36

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. $$G = \left.\frac{\mathrm{d}\theta}{\mathrm{d}t}\right|_{t=0} = \left.\frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\right|_{t=0} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$