Differentiability of $L_A(X)=AX$

Question

Let $G$ be a matrix group and $A \in G$. Define $L_A(X)= AX : G \to G$.

Show that

(1) $L_A$ is differentiable. Compute its derivative.

(2) $d(L_A)_I : T_I(G) \to T_A(G)$ is an isomorphism. [ Clearly, $L_A$ is invertible and it has a differentiable inverse by (1)]

Answer 1

The differential is given by $d(L_A)_X H = AH$.

$$\lim_{H\to 0}\frac{\|L_A(X+H) - L_A(X) - AH\|}{\|H\|} = \lim_{H\to 0}\frac{\|AX+AH - AX - AH\|}{\|H\|} = 0$$

So $d(L_A)_I$ is invertible, with the inverse being $H \mapsto A^{-1}H$.