Question about the $\operatorname{Exp}$ map on $\operatorname{End}(V)$ and the left invariant vector field Let $V$ be a finite dimensional vector space over $\mathbb{R}$. 
We define for $T \in \operatorname{End}(V)$ with $\| T \| < \infty$
$$
\operatorname{Exp}(T) = \sum_{k=0}^{\infty} \frac{T^k}{k!}. 
$$
Then it can be shown that 
$$
\frac{d}{dt} \operatorname{Exp}(tT)= \operatorname{Exp}(tT)\cdot T. 
$$
Then the notes I am reading states : 
"Therefore, $t \rightarrow \operatorname{Exp}(tT)$ is tangential to the left-invariant vector field evaluated at $e$ is $T$."
I am struggling to understand this sentence here. I would appreciate if someone could explain me in more detail what this sentence means... In particular, which left-invariant vector field are they talking about?
Thank you.  
 A: Recall that the commutator map $$(S, T) \mapsto [S, T] := S \circ T - T \circ S$$ defines a Lie algebra structure on the vector space $\operatorname{End}(V)$ of linear maps $V \to V$. We can identify this Lie algebra with the Lie algebra $\mathfrak{gl}(V)$ of the Lie group $GL(V)$ of invertible linear transformations of $V$. (If we choose a basis of $V$, we can identify both of these with the space $M(n, \Bbb R)$ of $n \times n$ real matrices, where $n := \dim V$, endowed with the commutator $$(A, B) \mapsto A B - B A$$ of matrices.)
In particular, we can regard any linear transformation $T \in \operatorname{End}(V)$ as an element of $\mathfrak{gl}(V)$, and so it determines a left-invariant vector field $\tilde T$ on $GL(V)$ characterized by $$\tilde T_1 = T .$$
Now, the map $\gamma : \Bbb R \to GL(V)$ defined by $\gamma : t \mapsto \operatorname{Exp}(t T)$ is a smooth curve in $GL(V)$, and unwinding definitions shows that it is an integral curve of $\tilde T$, that is, that $$\gamma'(t) = \tilde T_{\gamma(t)}$$ for all $t$.
