= Solution
Expanding the <cross product> component by component gives its <cross-product matrix>:
$$
\boxed{A=\begin{pmatrix}0&-n_3&n_2\\n_3&0&-n_1\\-n_2&n_1&0\end{pmatrix}.}
$$
It is a real <skew-symmetric matrix> and annihilates $\mathbf n$. By the <vector triple product>,
$$
A^2\mathbf x=\mathbf n\times(\mathbf n\times\mathbf x)=\mathbf n(\mathbf n\cdot\mathbf x)-\mathbf x,
$$
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> is
$$
\boxed{\lambda=0,\ i,\ -i.}
$$
Over 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 gives
$$
A=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix},\qquad \boxed{A^2=\operatorname{diag}(-1,-1,0).}
$$
Its zero <eigenspace> is $\operatorname{span}\{(0,0,1)\}$, and its $-1$ <eigenspace> is the whole plane $\operatorname{span}\{(1,0,0),(0,1,0)\}$. Hence the <eigenvalues> of $A^2$ are $-1,-1,0$, with every nonzero <vector> in the appropriate <eigenspace> an <eigenvector>.
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