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Let $G$ be the Lie group of affine transformations, $$\{x \mapsto Ax+b,A \in GL(n), b \in \mathbb{R}^n\}.$$ We can represent these maps as matrices $$\begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix}.$$

Now, I read that one can describe the Lie algebra $\mathfrak{g}$ as the set of matrices $\begin{pmatrix} A & b \\ 0 & 0 \end{pmatrix}$ How can I rigorously show that this is indeed the Lie algebra of $G$?

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  • $\begingroup$ $G$ is a subgroup of $GL(n+1)$, so the Lie algebra of $G$ is the subalgebra of $\mathfrak{gl}_{n+1}$ given by the tangent space of $G$ at the identity. $\endgroup$ – Alex Zorn Jul 25 '15 at 18:35
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Hint The elements of the Lie algebra $\mathfrak{g} \cong T_{\Bbb I} G$ are the tangent vectors at the identity element $\Bbb I \in G$ to curves in $G$ through that point.

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