1. 2019-11-14

Here's one way to think of things:

For each element of the tangent space at $I \in G$, we can construct a flow, and thus a corresponding one-parameter subgroup $A(t)$.

We can further show that for any Lie group $G \subset GL(n,\Bbb C)$, every one-parameter subgroup has the form $A(t) = e^{tX}$ for some matrix $X = A'(0)$.

Thus, we may conclude that $X \in \mathfrak g \iff e^{tX} \in G$ for all $t \in \Bbb R$, when $G \subset GL(n, \Bbb C)$.

In this particular case, we may state that $$X \in \mathfrak{sl}_n \iff e^{tX} \in SL_n \quad \forall t \in \Bbb R \iff \det(e^{tX}) = 1 \quad \forall t \in \Bbb R \\ \iff e^{t \cdot \text{tr}(X)} = 1 \quad \forall t \in \Bbb R \iff \text{tr}(X) = 0$$ Thus, $\mathfrak{sl}_n$ consists precisely of the traceless matrices.

Note in particular the importance of considering all $t \in \Bbb R$. For example, if we take $$X = \pmatrix{2 \pi i &0 \\0&2 \pi i}$$ we find that $X \notin \mathfrak{sl}_2$, even though $e^X = I \in SL_2$.

2. 2019-11-14

Hint. Consider a smooth curve $\gamma:\mathbb R\to SL$ in your group which at O is at the identity element. What is its tangent vector at zero?

As the function $t \mapsto \det\gamma(t)$ is constant, its derivative at 0 is 0. Compute it.