= Solution
Put $\beta=\|b\|>0$ and choose an oriented <orthonormal basis> with third vector $b/\beta$. The <Levi-Civita symbol> term has action $\epsilon_{ijk}b_kx_j=x\times b$, the negative of $b\times x$. Thus this <axisymmetric linear map> has block form
$$
B^\pm\sim\begin{pmatrix}1&\beta&0\\-\beta&1&0\\0&0&1\pm\beta^2\end{pmatrix}.
$$
The transverse eigenvalues $1\pm i\beta$ are not real. The only real <eigenvalue> and its <eigenvectors> are therefore
$$
\boxed{\lambda(B^\pm)=1\pm\beta^2,\qquad v\in\operatorname{span}\{b\}\setminus\{0\}.}
$$
The determinants are $\det B^+=(1+\beta^2)^2>0$ and $\det B^-=(1-\beta^2)(1+\beta^2)$, proving that \b[$B^+$ is always invertible, and $B^-$ is invertible exactly when $\|b\|\ne1$]. The complex transverse pair is important for this determinant even though the question asks for real eigenvalues.
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