Cross-product matrix

For $a=(a_1,a_2,a_3{)}^T$, the map $b\mapsto a\times b$ has matrixIt has trace zero and annihilates $a$, so its determinant is zero.

The cross-product matrix $A\mathbf{x}=\mathbf{n}\times \mathbf{x}$ obeys $A\mathbf{n}=0$ and $A^2=\mathbf{n}{\mathbf{n}}^T-|\mathbf{n}{|}^{2}I$, by the vector triple product. On the orthogonal complement of a nonzero $\mathbf{n}$, this is $-|\mathbf{n}{|}^{2}I$; hence the two remaining complex eigenvalues are $\pm i|\mathbf{n}|$. In an orthonormal basis adapted to $\mathbf{n}$, $A$ is a planar quarter-turn multiplied by $|\mathbf{n}|$, with zero action along its axis.