How is the Lie algebra obtained from the group in a simple example on a prior post?

Question

In this answer the Lie algebra of the group

$$\left\{\begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}\middle|x,y \in \mathbb R, x\neq 0\right\}$$

is immediately given as

$$\left\{\begin{bmatrix}x&y \\ 0&0\end{bmatrix}\,\middle|\,x,y\in\mathbb{R}\right\}$$

I can't find a presentation online with a straightforward example like this including defined steps to follow.

My first inclination would be to find elements in the tangent space at the identity, $$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$

by differentiating the matrix $\begin{bmatrix} x & y \\ 0 & 1 \end {bmatrix}$ but I don't know if this is done by parameterizing a curve by $t$ in the plane $x,y$ or there is a need for two parameters. The next step may be evaluating at zero. I don't know; I have never done it. I am just curious.

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This answer should apply here (thank you Professor José Carlos Santos for your correction - any remaining mistake, mine):

Imagining the $x$ and $y$ as functions of a parameter $t$, and defining $\left.\dot x(t)\right|_{t=0} = r$ and $\left.\dot y(t)\right|_{t=0}=s,$

$$\left.\frac d{dt}\begin{bmatrix}x&y\\0&1\end{bmatrix}\right|_{t=0}=\begin{bmatrix}r&s\\0&0\end{bmatrix} \,r, s\in \mathbb R $$

Answer 1

Here is a standard Lie theory theorem:

Theorem: If $n\in\Bbb N$ and $G$ is a closed subgroup of $GL_n(\Bbb R)$, then $G$ is a Lie group whose Lie algebra is$$\mathfrak g=\{M\in GL_n(\Bbb R)\mid(\forall t\in\Bbb R):\exp(tM)\in G\}.$$

Corollary: If $n\in\Bbb N$, $G$ is a $k$-dimensional Lie subgroup of $GL_n(\Bbb R)$ and $\mathfrak g$ is a $k$-dimensional Lie algebra of $n\times n$ real matrices such that $\exp(\mathfrak g)\subset G$, then $\mathfrak g$ is the Lie algebra of $G$.

Proof: Since $\mathfrak g$ is a real vector space, if $t\in\Bbb R$ and $M\in\mathfrak g$, then $tM\in\mathfrak g$, and therefore $\exp(tM)\in G$. So, $\mathfrak g$ is a subalgebra of the Lie algebra $\mathfrak g^\star$ of $G$. But $\dim\mathfrak g=\dim G=\dim\mathfrak g^*$, and therefore $\mathfrak g=\mathfrak g^*$.

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Now, let$$G=\left\{\begin{bmatrix}x&y\\0&1\end{bmatrix}\,\middle|\, x\in\Bbb R\setminus\{0\}\wedge y\in\Bbb R\right\}.$$ Clearly, it is a closed subset of $GL_2(\Bbb R)$ and $\dim G=2$. On the other hand, let$$\mathfrak g=\left\{\begin{bmatrix}x&y\\0&0\end{bmatrix}\,\middle|\,x,y\in\Bbb R\right\}.$$Then $\dim\mathfrak g=\dim G=2$ and $\exp(\mathfrak g)\subset G$. Therefore, it follows from the corollary that $\mathfrak g$ is the Lie algebra of $G$.