I want to prove that $\{\mathrm{ad}(X) \mid X \in \mathfrak{gl}(n, \mathbb R)\}$ is the Lie algebra of $\mathrm{Ad}(\mathrm{GL}(n, \mathbb R))$.
For $g \in \mathrm{GL}(n, \mathbb R)$, $\mathrm{Ad}(g)$ is the automorphism of $\mathfrak{gl}(n, \mathbb R)$ that maps $X$ to $g X g^{-1}$. First, I succeeded in proving that $\mathrm{Ad}(\mathrm{GL}(n, \mathbb R))$ is a linear Lie group. For closure, I used the continuity of the map $\Phi : \mathrm{GL}(n, \mathbb R) \to \mathrm{Aut}(\mathfrak{gl}(n, \mathbb R))$ which maps $g$ to $\mathrm{Ad}(g)$ and the fact that an element $T$ of $\mathrm{Aut}(\mathfrak{gl}(n, \mathbb R))$ which preserves the trace can be written as $\mathrm{Ad}(P)$ with $P \in \mathrm{GL}(n, \mathbb R)$.
However, I failed to prove the result I mentioned in the introduction. If $G$ is a linear Lie group, i.e., a closed subgroup of $\mathrm{GL}(m, \mathbb R))$, then the Lie algebra of $G$ is defined as $\{X \in M(m, \mathbb R) \mid \forall t \in \mathbb R, \mathrm{exp}(tX) \in G\}$.
I feel I'm missing something obvious here...
