I understand the following construction: For a Lie group $G$, let $\widetilde{\text{Lie}}(G)$ denote the set of right-invariant vector fields on $G$ - this is a finite dimensional Lie algebra under the Lie bracket of vector fields. There is an isomorphism $\varphi\colon\widetilde{\text{Lie}}(G)\to T_eG$ defined by $X\mapsto X_e$, hence we can define a Lie bracket on $T_eG$ by $[A,B]:=[\tilde{A},\tilde{B}]_e$ for $A,B\in T_eG$ where $\tilde{A},\tilde{B}\in\widetilde{\text{Lie}}(G)$ are such that $\varphi(\tilde{A})=A$ and $\varphi(\tilde{B})=B$.
To understand this a bit more, I would like to compute the Lie bracket on $T_eG$ for an explicit Lie group $G$. In particular, I have the following question:
Question. What is the Lie bracket on $T_eG$ for the Lie group $G=\text{Gl}(n;\mathbb{R})$?
I know $T_eG=M(n;\mathbb{R})$.
For matrix groups the Lie bracket on the associated Lie algebra is generally the commutator of matrices. The justification is as follows.
In an open neighbourhood around the identity $e$ we can consider a path $\gamma(t)$ with $\gamma(0)=e$. Since we're in $GL_n$ this path has an inverse $\gamma^{-1}(t)$ which is also identity $t=0$.
Now the associated Lie algebra element is the derivative at $t=0$
$$\gamma'(0)=A\in T_eG $$
Furthermore the inverse has a derivative at the identity which we can find by taking the derivative of $\gamma(t)\gamma^{-1}(t)=e$ at $t=0$
$$\begin{align*}\gamma'(0)\gamma^{-1}(0)+\gamma(0)(\gamma^{-1})'(0)&=0\\ A+(\gamma^{-1})'(0)&=0\end{align*}$$
Which implies that $(\gamma^{-1})'(0)=-A\in T_eG$. Now let $B\in T_eG$, to find $[A,B]$ we will use the adjoint representation of $\gamma$ acting on $B$
$$\text{Ad}(\gamma(t))(B)=\gamma(t)B\gamma^{-1}(t)$$
Evaluating the derivative at $t=0$ which corresponds to the derivative at $e$, gives us the Lie Bracket $[A,B]$
$$\begin{align}\text{ad}(A)(B)=[A,B]&=\left(\text{Ad}(\gamma(t))(B)\right)'(0)\\ &=\gamma'(0)B\gamma^{-1}(0)+\gamma(0)B(\gamma^{-1})'(0)\\ &=AB-BA\end{align}$$