Assume $A$ is a finite dimensional associative $\mathbb{R}-$algebra, with identity $1( \neq 0)$, let $A^{\times}$ be the set of all invertible elements of $A$, then it's easy to see that $A^{\times}$ is a Lie group and open in $A$. My question is: what is its Lie algebra?
The answer seems to be $A$, with Lie bracket $[a, b]=ab-ba$, but I don't know why is it so.
I suggest thinking of the canonical embedding of $A$ as a subalgebra of $\operatorname{End} A$, $x \mapsto (y \mapsto xy)$.