= Solution
For a <Lie bracket> over $\mathbb R$ or $\mathbb C$, <antisymmetry of a Lie bracket> means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The <Jacobi identity> is
$$
\boxed{[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0.}
$$
It expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same <vector space>.
For the <matrix> <commutator>, bilinearity and antisymmetry follow directly from distributivity. Associativity of <matrix> multiplication gives
$$
[A,[B,C]]=ABC-ACB-BCA+CBA;
$$
adding the two cyclic permutations cancels every monomial. Thus the <Jacobi identity> holds in the entire <matrix> algebra. For a specified linear subspace, the only additional bracket condition is \b[closure: $AB-BA$ must belong to the subspace whenever $A,B$ do]. It is unnecessary to require closure under the separate products $AB$ and $BA$.
To determine the <special unitary Lie algebra>, let $g(t)$ be a differentiable curve in the <special unitary group> with $g(0)=I$ and $A=g'(0)$. Differentiating $g(t)^\dagger g(t)=I$ gives $A^\dagger+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\operatorname{tr}A=0$. Conversely a traceless <skew-Hermitian matrix> has $e^{tA}$ unitary and $\det e^{tA}=e^{t\operatorname{tr}A}=1$, so it really is a tangent vector. Therefore
$$
\boxed{\mathfrak{su}(n)=\{A\in M_n(\mathbb C):A^\dagger=-A,\ \operatorname{tr}A=0\},
\qquad \dim_{\mathbb R}\mathfrak{su}(n)=n^2-1.}
$$
This is a real <Lie algebra> of complex <matrices>: multiplication by $i$ generally leaves this real subspace. Its complexification is $\mathfrak{sl}_n(\mathbb C)$, not the compact algebra itself. For $A,B\in\mathfrak{su}(n)$,
$$
[A,B]^\dagger=B^\dagger A^\dagger-A^\dagger B^\dagger=BA-AB=-[A,B],
\qquad \operatorname{tr}[A,B]=0.
$$
Thus closure holds, and the already verified <commutator> identities establish all the <Lie algebra> axioms.
The <cross-product Lie algebra> on $\mathbb R^3$ is bilinear, antisymmetric and closed because the <cross product> has those properties. Its <Jacobi identity> follows from the vector triple-product identity:
$$
\mathbf x\times(\mathbf y\times\mathbf z)
=\mathbf y(\mathbf x\cdot\mathbf z)-\mathbf z(\mathbf x\cdot\mathbf y).
$$
In the cyclic sum, the coefficients cancel by symmetry of the scalar product.
For the explicit relation to the <SU(2) Lie algebra>, take the three <Pauli matrices> and define
$$
F(\mathbf x)=-\frac{i}{2}\sum_{a=1}^3x_a\sigma_a.
$$
These <matrices> are traceless and <Skew-Hermitian>, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the <Pauli matrix commutator identity>,
$$
[F(\mathbf x),F(\mathbf y)]
=-\frac14\,2i\sum_c(\mathbf x\times\mathbf y)_c\sigma_c
=F(\mathbf x\times\mathbf y).
$$
Hence \b[$F$ is a real <Lie algebra isomorphism>]. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an <Adjoint double cover from SU(2) to SO(3)>; the isomorphism of their tangent algebras does not identify the two global groups.
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