# 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?