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For $a=(a_1,a_2,a_3)$ the map has the explicit formIts diagonal entries sum to zero and its conjugate transpose is its negative, because the three coordinates are real. Hence it is traceless and a skew-Hermitian matrix.

Direct multiplication of the given basis matrices yieldsThe reverse products give the negative commutators, and equal-index commutators vanish. Bilinearity then givesThus the cross product is represented by the commutator:This realizes the cross-product model of su(2) with exactly the normalization and cyclic orientation used here.

Every traceless skew-Hermitian matrix of size two has the formso it equals $\mathcal{M}(c)$ for $c=(2p,2\mathrm{Re}w,2\mathrm{Im}w)$. If $c=0$, take $A=B=0$. Otherwise choose a unit vector $u$ perpendicular to $c$ and set $a=u$, $b=c\times u$. The vector triple product gives $a\times b=c$.

By part (iii), choosing $A=\mathcal{M}(a)$ and $B=\mathcal{M}(b)$ gives the required representationIn fact both factors can themselves be chosen as traceless skew-Hermitian matrices. The cross-product model of su(2) explains this geometric commutator construction.