Left Invariant Vector field of matrix Lie group In Hamilton's "Mathematical Gauge Theory" I am trying to understand this proof for the left invariant vector fields of the group $G = GL(d, \mathbb{R})$. I understand how we can assume $X\in Mat(n\times n, \mathbb{R}) = T_eG$, since an isomorphism between $T_eG$ and $Mat(n\times n, \mathbb{R})$ can be established.

what I don't quite understand how exactly we pushforward a matrix instead of a tangent vector. In addition, at the last step, I'm not sure how the derivative is supposed to be applied. As far as I understand, the symbol $\dot{\gamma}(0)$ isn't a derivative in the traditional sense, its just
notation for the tangent vector at the curve since we cant apply real analysis to a curve $\gamma : \mathbb{R}\rightarrow G$, so applying this derivative like how they want you to at the end doesn't make sense.
 A: Okay here is an attempt at a more full answer:
You aren't pushing forward a matrix, you are pushing forward the map $L_A : G \to G; B \mapsto AB $. This is clearly a diffeomorphism of $G$ (by definition of a Lie group) and differentiating (i.e. pushforward) it at $e \in G$ gives a map $D_e L_A: T_e G \to T_AG$ which is defined by $D_e L_A (X) = \left.\frac{d}{dt}\right|_{t=0}L_A(\gamma(t))$ where $\gamma$ is any curve in $G$ with $\gamma(0) = e$ and $\gamma'(0) = X$ for $X \in T_eG$. So far none of this depends on matrices anywhere, I only care how $A$ defines a diffeomorphism on G.
If we consider $G = \operatorname{GL}(n,\mathbb{R}) \subset M_{n\times n}$ then we can clearly identify $T_eG \cong M_{n\times n}$ but also every $T_AG \cong M_{n\times n}$ as well. Note everything is now inside a vector space so we do indeed now have things like $\gamma'(0)$ defined as the traditional derivative rather than as an abstract tangent vector to a manifold.
Now the last line is just computing the derivative of a product of curves of matrices: $A\cdot \gamma(t)$. This follows product rule as you might hope:
$$\left.\frac{d}{dt}\right|_{t=0} (A\cdot \gamma(t)) = \left.\frac{d}{dt}\right|_{t=0}A \cdot \gamma(t) +A\cdot \left.\frac{d}{dt}\right|_{t=0}\gamma(t)$$
Since $A$ is just a constant this last is just $ A \cdot \gamma'(0) = A \cdot X $
