linear algebra – Direct proof of $mathfrak{so}(3)_{mathbb C}simeqmathfrak{sl}(2,mathbb C)$


It is well known that $mathfrak{su}(2)equivmathfrak{su}(2,mathbb C)$, the real Lie algebra of traceless skew-Hermitian $2times 2$ complex matrices, satisfies $mathfrak{su}(2)_{mathbb C}simeq mathfrak{sl}(2,mathbb C)$. To see this, it is sufficient to observe that any traceless matrix $A$ can be written as
$$A = ileft(frac{A+A^dagger}{2i}right) + left(frac{A-A^dagger}{2}right),$$
where both components are traceless and skew-Hermitian, and the decomposition is unique.

We also know that $mathfrak{so}(3)simeqmathfrak{su}(2)$, where $mathfrak{so}(3)$ is the real Lie algebra of traceless skew-orthogonal $3times 3$ real matrices. This follows from observing that both Lie algebras satisfy the same commutation relations, $(T_i,T_j)=epsilon_{ijk}T_k$ (or rather, we can always find bases for both spaces satisfying such relations).

This should imply that also $mathfrak{so}(3)_{mathbb C}simeqmathfrak{sl}(2,mathbb C)$ (as also mentioned in passing in this answer), but how can I show that this is the case more directly?
As far as I understand, this statement should mean that, given any traceless $2times 2$ complex matrix $A$, there is a bijection sending $A$ to two $3times 3$ real skew-orthogonal matrices. What is this decomposition?