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Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.