How to "find" this Lie algebra: proof that $\mathfrak{sl}$ is trace zero matrices I saw this table here on Wikipedia and it states that the Lie algebra of the special linear group $SL_n(\mathbb C)$ is the group of traceless matrices $\mathfrak{sl}_n$.
I know the definition of a Lie algebra: given a matrix group $G \subseteq GL_n(\mathbb K)$ its Lie algebra is defined to be the tangent space at $I$.
From this definition it is immediately clear what dimension the Lie algebra has to have.  What is unclear to me is: 

Assuming I don't already know the answer, how do I arrive at the insight that this tangent space is traceless matrices?

I skipped through Tapp's book in search of hints and found that the exponential map is a homomorphism from the Lie algebra into the Lie group. I also found a lemma that states that 
$$ \det e^A = e^{\mathrm{trace}(A)}$$
That is unfortunately only half-useful: If I already know what the Lie algebra is then this lets me prove that the image of the Lie algebra under the homomorpism defined by the exponential map is indeed in $SL_n$. 
So it's kind of not so useful. Surely there must be geometric insight? 
 A: 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 $e^X = I \in SL_2$.
A: 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.
