I'm working with the automorphism group that acts on an algebra ${\cal A}$, $Aut({\cal A}) = \{a \in End({\cal A})\ |\ a([x, y]) = [a(x), a(y)], \forall x, y \in {\cal A}\} < GL({\cal A})$. I've seen that the algebra of this group is the space of derivations, that is
$$der({\cal A}) = \{a \in End({\cal A})\ |\ a([x, y]) = [a(x), y] + [x, a(y)], \forall x, y \in {\cal A}\} $$
I can understand that if $a \in GL({\cal A})$, then $a([x, y]) = a(xy) - a(yx)$ and if $a$ is a derivation, $a(xy) - a(yx) = [a(x), y] + [x, a(y)]$. But how can I prove that this is the algebra of $Aut({\cal A})$?
I tried to apply the exponentiation of this algebra and make it act on $[x, y]$ in order to see if the result is the definition of $Aut({\cal A})$, but I didn't get it. Any help?
Since the Lie algebra of a group is the tangent space at the identity element, you can write a generic path through the group which goes through the identity, and differentiate. So what I mean is consider a path $g(t)$ in $\mathrm{Aut}(\mathcal{A})$, say parameterized with $t \in (-\varepsilon,\varepsilon)$, so that $g(0) = \mathrm{Id}$.
The condition that each $g(t)$ is in $\mathrm{Aut}(\mathcal{A})$ means that for each $t$,
$$ g(t)[x,y] = [g(t)x,g(t)y] $$
Let's make up some notation for the sake of doing concrete computations. Let's sake $e_1,\dots,e_n$ is a basis of $\mathcal{A}$, with structure constants $c^k_{ij}$, so that $[e_i,e_j] = \sum_k c^k_{ij} e_k$. Now for our chosen elements $x$ and $y$, let's define functions $a_i(t)$ and $b_j(t)$ so that $g(t)x = \sum_i a_i(t)e_i$ and $g(t)y = \sum_j b_j(t)e_j$. Then our equation above translates to
$$ g(t)[x,y] = \sum_{i,j,k} c^k_{ij} a_i(t)b_j(t) e_k $$
Now, let's take the derivative:
$$ g'(t)[x,y] = \sum_{i,j,k} c^k_{ij} \left( a_i'(t)b_j(t) + a_i(t)b_j'(t) \right) e_k $$
The right-hand side should translate to the derivation property:
$$ g'(t)[x,y] = [g'(t)x,y] + [x,g'(t)y] $$