From the definition of tangent space to the Lie algebra of a matrix group I'm trying to understand how to go from the general definition of a tangent space to the explicit definition of the Lie algebra of a matrix group. Let $G$ be a Lie group, the Lie algebra is the tangent space at the identity
$$\mathfrak{g}=T_e G$$
We can see define the tangent space as follows: let $\gamma: (-a,a)\to G$ be a smooth curve such that $\gamma(0)=e$, we define an equivalence relation: $\gamma \sim \tau$ if for any chart $\phi:U\to\mathbb{R}^n$ such that $e\in U$ $(\phi\circ\gamma)'(0)=(\phi\circ\tau)'(0)$, we note by $[\gamma]$ the equivalence class of $\gamma$.
For a curve $\gamma$ we can define a derivation
$$\begin{aligned}\gamma'(0):& C^\infty(G)\to\mathbb{R} \\ &f\mapsto(f\circ \gamma)'(0)\end{aligned}$$
then $$T_e G =\{[\gamma]'(0): \gamma:(-a,a)\to G \textrm{ smooth, }\gamma(0)=e\}$$ where by the derivative of the equivalence class I mean the one of a representative.
First question: do I have this definition correct?
Now suppose that $G$ is a matrix group, i.e. $G\subset M_n(\mathbb{R})$, the Lie algebra is usually defined as the space of the matrices
$$\Omega =\frac{d}{dt}R(t)|_{t=0} $$
where $R(t):(-a,a)\to G$ is a smooth curve. The derivative is well defined as $R(t)$ is a matrix for any $t$. To reconcile the two definitions my idea is to consider the functions
$$\begin{aligned}x^{ij}:&G\to \mathbb{R}\\&A\mapsto A_{ij}=\langle e_i, A e_j\rangle \end{aligned}$$
for some basis $\{e_i\}$, then $R'(0)(x^{ij})=R'(0)_{ij}$ and we automatically have a matrix associated to any derivation in $T_eG$. My question is, can we reconstruct the image of the derivation applied to any smooth function on $G$ via this matrix, i.e., is it true that
$$R'(0)(f)=\sum_{ij}\alpha_{ij} R'(0)(x^{ij})$$.
For some coefficient $a_{ij}$?. This seems to rely on the fact that in some way $f(A)=\sum \alpha_{ij} x^{ij}$, hence that $f$ is linear, so this doesn't seem like the right way to do it. How can I reconcile these two definitions?
 A: Your definition of the tangent space at the identity is correct.
You seem to be trying to identify $M_n(R)$ with $\mathbb{R}^{n\times n}$, which is a valid approach. The $x^{ij}$ are the coordinate projections.
We can write the curve $$R(t)= \pmatrix{R_{11}(t) & R_{12}(t)  &\cdots & R_{1n}(t) \\ \vdots & \ddots  &\\ R_{n1}(t) & R_{n2}(t)  &\cdots & R_{nn}(t) \\} .$$
The only step you're missing is that $f: G \to \mathbb{R}$ now has to be identified as a function from a subset of $\mathbb{R}^{n\times n}$. Thus, we have
$$f\circ R: I\subset \mathbb{R} \to \mathbb{R}.$$
Applying the chain rule we have, $$\frac{d}{dt} (f\circ R(t))\lvert_0 \ = \ df\vert_{R(0) }\circ \frac{d}{dt} R\vert_0.$$
Here $df$ is the derivative (linearisation/jacobian) of $f$ at $R(0)$. It is a linear map from the space of $n\times n$ real matrices to $\mathbb{R}$ (or from $\mathbb{R}^{n\times n}$ to $\mathbb{R}$).
Careful consideration of indices will show that your $\alpha_{ij}$ are in fact the directional derivatives of $f$ at $R(0)$.
