= Solution
For $a=(a_1,a_2,a_3)$ the map has the explicit form
$$
\mathcal M(a)=\frac12\begin{pmatrix}ia_1&a_2+ia_3\\-a_2+ia_3&-ia_1\end{pmatrix}.
$$
Its 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> yields
$$
[M_1,M_2]=M_3,\qquad[M_2,M_3]=M_1,\qquad[M_3,M_1]=M_2.
$$
The reverse products give the negative <commutators>, and equal-index <commutators> vanish. Bilinearity then gives
$$
[\mathcal M(a),\mathcal M(b)]
=(a_2b_3-a_3b_2)M_1+(a_3b_1-a_1b_3)M_2+(a_1b_2-a_2b_1)M_3.
$$
Thus \b[the <cross product> is represented by the <commutator>]:
$$
\boxed{\mathcal M(a\times b)=[\mathcal M(a),\mathcal M(b)].}
$$
This realizes the <cross-product model of su(2)> with exactly the normalization and cyclic orientation used here.
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