For matrices, why is the Lie algebra equal to the tangent space? For a matrix Lie group $G\subset GL_n(\mathbb C)$ we define the Lie algebra to be the set of matrices $X\in M_n(\mathbb C)$ such that for all $t\in \mathbb R$ we have  $\quad \exp(tX)\in G$.  
For clarity, I define the exponential map by
$$\exp(X) = \sum_{k=0}^\infty \dfrac{X^k}{k!}$$
The tangent space $T_I(G)$ at the group identity $I$ is the set of all matrices $X\in M_n(\mathbb C)$ for which there exists a differentiable path
$$\gamma : (-\epsilon,\epsilon) \to G$$
where $\gamma(0)=I$ and $\gamma'(0)=X$. 
It is proposed that Lie algebra and the tangent space are the same. 
It is clear from the differentiability of exponentiation that if $\quad \exp(tX)\in G \quad$ at every $t$ then the path $\gamma$ defined by $\gamma(t) = \exp(tX)$ shows that $X$ is in the tangent space. 
The other direction is not as obvious to me. If $X$ is in the tangent space (ie. there exists a path $\gamma$ satisfying the conditions above) then $X$ is in the Lie algebra. Can somebody please explain why this is so? 
 A: The following comes from Brian C. Hall's text (2003 edition):$\newcommand{\fg}{\mathfrak{g}}$

Corollary 2.35: suppose $G \subset GL(n;\Bbb C)$ is a matrix Lie group with Lie algebra $\fg$. Then, a matrix $X$ is in $\fg$ if and only if there exists a smooth curve $\gamma$ in $M_n(\Bbb C)$ such that
  
  
*
  
*$\gamma(t)$ lies in $G$ for all $t$
  
*$\gamma(0) = I$
  
*$\frac{d \gamma}{dt}|_{t=0} = X$
  
  
  Thus, $\fg$ is the tangent space at the identity to $G$

The proof (of your direction) is as follows:
Let $\gamma$ be a curve satisfying these conditions.
Then, we may state that $\log(\gamma(t)) \in \fg$ for all sufficiently small $t$ (see theorem 2.27: there exists a neighborhood $U$ of $0$ in $\fg$ and a neighborhood $V$ of $I$ in $G$ such that $\exp:U \to V$ is a homeomorphism). Now, define
$$
X_{\gamma} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = \lim_{h \to 0} \frac{\log(\gamma(h))}{h}
$$
since $X_{\gamma}$ is the limit of elements in $\fg$ and since $\fg$ is closed, $X_{\gamma} \in \fg$.  However, we note that (again, for sufficiently small $t$)
$$
\log(\gamma(t)) = (\gamma(t) - I) - \frac 12(\gamma(t)-I)^2
+ \frac 13(\gamma(t) - I)^3 - \cdots
$$
Differentiate term by term by applying the product rule.  All terms besides the first approach zero as $t \to 0$.  Thus, we have
$$
\frac{d}{dt} \gamma(t)|_{t=0} = \frac{d}{dt} \log(\gamma(t))|_{t=0} = 
 X_{\gamma} \in \fg
$$
which was the desired result.
