= Solution
\b[Tensor transformation.] Under an <orthogonal transformation> $R$, the magnetic moment and <magnetic field intensity> have transformed components $M'=RM$, $H'=RH$ for a proper rotation. The physical relation $M=\chi H$ must retain its form, giving $RM=\chi'RH$, hence
$$
\boxed{\chi'=R\chi R^T,\qquad\chi'_{ij}=R_{ia}R_{jb}\chi_{ab}.}
$$
This is precisely the law for a <Cartesian second-rank tensor>. Both magnetic quantities are <axial vectors>; under reflections they each acquire the same determinant sign, which cancels from the transformation of the <magnetic susceptibility tensor>.
\b[Parallel directions.] The condition $M=\lambda H$ is the <eigenvalue> equation $\chi H=\lambda H$. For the matrix in the paper, expansion of $\det(\chi-\lambda I)$ gives
$$
\det(\chi-\lambda I)=-\lambda(\lambda-1)(\lambda-6).
$$
Solving the three resulting homogeneous systems gives
$$
\boxed{\begin{array}{c|c|c}
\lambda&\text{direction of }H&\text{normalized eigenvector}\\\hline
0&(1,-1,2)&e_0=(1,-1,2)/\sqrt6\\
1&(-2,0,1)&e_1=(-2,0,1)/\sqrt5\\
6&(1,5,2)&e_6=(1,5,2)/\sqrt{30}
\end{array}.}
$$
The three directions are mutually <orthogonal>. For the zero <eigenvalue>, the induced magnetic moment is zero: the proportionality equation still holds with factor zero, though there is then no nonzero moment direction to assign.
\b[Energy extrema.] The stipulated energy is $\mathcal E=H^T\chi H$. At fixed $H^TH=H_0^2>0$, a <Lagrange multiplier> calculation gives
$$
\nabla_H[H^T\chi H-\lambda(H^TH-H_0^2)]=2\chi H-2\lambda H=0,
$$
because $\chi$ is a real <symmetric matrix>. Thus all stationary directions are the <eigenvectors>, for which $M=\lambda H$. Equivalently write $H=q_0e_0+q_1e_1+q_6e_6$; then
$$
\mathcal E=q_1^2+6q_6^2,\qquad H_0^2=q_0^2+q_1^2+q_6^2.
$$
The <Rayleigh quotient> immediately gives
$$
\boxed{\mathcal E_{\min}=0\text{ on }H=\pm H_0e_0,\qquad\mathcal E_{\max}=6H_0^2\text{ on }H=\pm H_0e_6.}
$$
The intermediate direction $e_1$ is stationary with $\mathcal E=H_0^2$ but is not an extremum on the sphere. At the minimum $M=0$, so the precise extremum condition is proportionality, including a vanishing proportionality factor.
\b[Constant-energy surface.] In the original field components,
$$
\boxed{\mathcal E=(H_1+H_2)^2+(H_3+2H_2)^2=q_1^2+6q_6^2.}
$$
For a fixed positive value $E$, $q_0$ is unrestricted and the cross-section satisfies $q_1^2/E+q_6^2/(E/6)=1$. The surface is therefore an <elliptic cylinder>, with axis along $(1,-1,2)$ and transverse semiaxes $\sqrt E$ along $e_1$ and $\sqrt{E/6}$ along $e_6$. This is an example of <level sets of a positive semidefinite quadratic form>. At $E=0$ the surface degenerates to the axis line; no negative-energy level exists.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2-susceptibility-energy-cylinder.png]
{title=The susceptibility energy has elliptic cylindrical level sets}
{description=The surface shown has E=1. Its unbounded axis is the zero-eigenvalue direction (1,-1,2); transverse semiaxes are 1 along e1 and 1/sqrt(6) along e6. Only a finite axial segment is drawn.}
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