$\{\gamma'(0)\,\vert\, \gamma \in C^\infty(\Bbb R,G), \gamma(0) = I_n\}$ is a $\Bbb R$-vector space where $G$ is a closed subgroup of $GL_n(\Bbb C)$ 
Problem. Let $G$ be a closed* subgroup of $GL_n(\mathbb C)$. Define $$V:= \{\gamma'(0)\,\vert\, \gamma \in C^\infty(\Bbb R,G), \gamma(0) = I_n\}$$
Show that $V$ is a vector space over $\mathbb R$. Furthermore, determine $V$ explicitly when $G = GL_n(\Bbb C)$ and $SL_n(\Bbb C)$.
*The topology on $GL_n(\mathbb C)$ is due to the identification $M_n(\mathbb C) \cong \mathbb C^{n^2}$.

The first task is to show that $V$ is a real vector space.

*

*Consider the constant map $\gamma:\Bbb R\to G$, $t\mapsto I_n$. Then, $\gamma(0) = I_n$ and $\gamma'(0) = \mathbf{0}_n \in V$. Note that $I_n \in G$ since subgroups preserve identities. So, $\mathbf{0} := \mathbf{0}_n \in M_n(\Bbb C)$ is the zero element of the vector space.


*In addition, we must check the vector space axioms. After @KReiser's hint, I have shown that $V$ is closed under scalar multiplication. Suppose $\gamma'(0) \in V$ for some $\gamma \in C^\infty(\Bbb R,G)$ with $\gamma(0) = I_n$. Define $\gamma_a \in C^\infty(\Bbb R,G)$ for $a\in R$, by $\gamma_a(t):= \gamma(at)$. $\gamma_a$ satisfies $\gamma_a(0) = \gamma(0) = I_n$. Then, $\gamma_a'(t) = a\gamma'(at)$ implying $\gamma_a'(0) = a\gamma'(0) \in V$ for every $a\in \Bbb R$. For $\gamma_1'(0),\gamma_2'(0) \in V$, consider $\gamma := \gamma_1\gamma_2 \in C^\infty(\Bbb R,G)$ to conclude $\gamma_1'(0) + \gamma_2'(0) \in V$, i.e. $V$ is closed under addition. This was suggested by @José Carlos Santos.


*For $G = GL_n(\Bbb C)$, I can show that $V = M_n(\Bbb C)$. $V\subset M_n(\Bbb C)$ is obvious. Take $A\in M_n(\Bbb C)$, and $\gamma(t):= e^{tA}$. Then, $\gamma(0) = I_n$ and $\gamma'(0) = A \in V$. So, $V = M_n(\Bbb C)$.
Thank you!
 A: If $v_1=\gamma_1'(0)$ and $v_2=\gamma_2'(0)$, then\begin{align}(\gamma_1\gamma_2)'(0)&=\gamma_1(0)\gamma_2'(0)+\gamma_1'(0)\gamma_2(0)\\&=\gamma'_1(0)+\gamma'_2(0)\\&=v_1+v_2.\end{align}And, if $\lambda\in\Bbb R$ and $\eta(t)=\gamma_1(\lambda t)$, then$$\eta'(0)=\lambda\gamma_1'(0)=\lambda v_1.$$
A: Note that 1 is superfluous if $V \neq \emptyset$, as it follows from 2. 1 can basically be seen as the proof of non-emptyness.
And for sum check explicitly by computing $$(\gamma_1\gamma_2)'(0)= \gamma_1'(0)\gamma_2(0) + \gamma_1(0)\gamma'_2(0) = \gamma_1'(0)I_n+ I_n\gamma_2'(0) = \gamma_1'(0)+\gamma_2'(0)$$ to be really complete.
A: As I briefly said in my comment, you have to find a condition on $\gamma'(0)$
For example let $G = SL_n(\mathbb C)$ and $\gamma : \mathbb R \to G$ that is smooth.
What you have is $t \mapsto \det(\gamma(t))$ that is a constant function over $\mathbb R$. Its derivative is given by
$$t \mapsto tr\left(\gamma'(t)\cdot  {}^t Co(\gamma(t))\right)=tr\left(\gamma'(t)\cdot  \gamma^{-1}(t)\right) $$
This function as the derivative of a constant function vanishes. In particular you have : $$tr \left( \gamma'(0) \right) = 0$$
Conversely for $A$ in $M_n(\mathbb C)$ and for $t $ in $\mathbb R$ : if $tr(A) = 0$  you have $\det\left(e^{tA} \right) = e^{t \cdot tr(A)} = 1$
In the end you have $V = \{A \in M_n(\mathbb C), \quad tr (A) =0 \}$
