In this part, $\mathbf{K} = \mathbf{R}$. If $G$ is a closed subgroup of $\mathrm{GL}_n(\mathbf{R})$, we introduce its Lie algebra:
$$\mathcal{A}_G = \left\{ M \in \mathcal{M}_n(\mathbf{R}) \mid \forall t \in \mathbf{R} \quad e^{tM} \in G \right\}.$$
The tangent space to $G$ at $I_n$ is denoted $\mathcal{T}_{I_n}(G)$.
$\mathbf{17}$ ▷ Show that, in the particular cases $G = \mathrm{SL}_n(\mathbf{R})$ and $G = \mathrm{O}_n(\mathbf{R})$, we have $\mathcal{T}_{I_n}(G) = \mathcal{A}_G$.