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Tensor transformation. Under an orthogonal transformation $R$, the magnetic moment and magnetic field intensity have transformed components $M^{\prime }=RM$, $H^{\prime }=RH$ for a proper rotation. The physical relation $M=\chi H$ must retain its form, giving $RM={\chi }^{\prime }RH$, henceThis 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.

Parallel directions. The condition $M=\lambda H$ is the eigenvalue equation $\chi H=\lambda H$. For the matrix in the paper, expansion of $\det (\chi -\lambda I)$ givesSolving the three resulting homogeneous systems givesThe 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.

Energy extrema. The stipulated energy is $\mathcal{E}=H^T\chi H$. At fixed $H^{T}H=H_0^2>0$, a Lagrange multiplier calculation givesbecause $\chi$ is a real symmetric matrix. Thus all stationary directions are the eigenvectors, for which $M=\lambda H$. Equivalently write $H=q_0{e}_0+q_1{e}_1+q_6{e}_6$; thenThe Rayleigh quotient immediately givesThe 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.

Constant-energy surface. In the original field components,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.