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For a real symmetric matrix $A$ with nonnegative eigenvalues, the finite-dimensional spectral theorem changes the equation to ${\sum }_i{\lambda }_i{q}_i^2=E$ in an orthonormal eigenbasis. If $E>0$ and $A$ has positive rank, the positive-eigenvalue coordinates form an ellipsoid, while the zero-eigenvalue coordinates are arbitrary. The level set is therefore a cylinder over that ellipsoid, with axis directions given by the kernel of a linear map of $A$. For rank two in three dimensions, it is an elliptic cylinder. At $E=0$ the level set is exactly the kernel of a linear map; for $E<0$ it is empty. This distinguishes a positive semidefinite matrix from a positive-definite matrix, whose positive level sets are bounded ellipsoids.

Choose the outward normal vector on the side of the elliptic cylinder. Parametrize it bySince ${\mathbf{r}}_{\theta }\times {\mathbf{r}}_z=(\frac{1}{2}\cos \theta ,\sin \theta ,0)$ points outward, the parameter rectangle has its bottom edge traversed with increasing $\theta$, its right cut edge with increasing $z$, its top edge with decreasing $\theta$, and its left cut edge with decreasing $z$. After identification, the two cut edges are the same physical curve with opposite directions. Their line integrals cancel for a single-valued vector field. Viewed from above, the bottom rim is counterclockwise and the top rim clockwise.

To apply Stokes theorem to a flux rather than a curl integral, first find a vector potential:This vector potential is smooth everywhere. On either horizontal rim, $d\mathbf{r}$ is horizontal whereas $\mathbf{A}$ is vertical, giving $\mathbf{A}\cdot d\mathbf{r}=0$. For a cut at ${\theta }_0=\pi /4$, $xy=1/4$, so the upward cut contribution is ${\int }_{-1}^1{z}^2/4dz=1/6$ and the downward one is $-1/6$. ConsequentlyA direct check uses the oriented surface element above: $\mathbf{F}\cdot ({\mathbf{r}}_{\theta }\times {\mathbf{r}}_z)=\frac{1}{2}{z}^2\cos 2\theta$, whose integral over a full turn vanishes. Reversing the chosen orientation still gives zero.

Choose the rotated Cartesian coordinatesTheir inverse is $x=(X-Y)/\sqrt{2}$, $y=(X+Y)/\sqrt{2}$. Consequently $x^2+y^2=X^2+Y^2$ and $xy=(X^2-Y^2)/2$, giving the canonical form of a quadricThe orthogonal diagonalization identifies the eigenvalues $1+a/2,1-a/2,1$ of the symmetric matrix of the quadratic form.

For $|a|<2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $(1+a/2{)}^{-1/2},(1-a/2{)}^{-1/2},1$. At $a=0$ it is a sphere.

For $a=2$, the equation is $2X^2+Z^2=1$, an elliptic cylinder parallel to the $Y$ axis. For $a=-2$, it is $2Y^2+Z^2=1$, an elliptic cylinder parallel to the $X$ axis. The absent coordinate is unrestricted; its disappearance does not turn the surface into an ellipse.

For $a>2$, the negative coefficient is that of $Y^2$: this is a one-sheet hyperboloid with axis $Y$. For $a<-2$, the negative coefficient is that of $X^2$: it is a one-sheet hyperboloid with axis $X$. This gives the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid, including both transition values.

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.

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

The coordinate $x_1^{\prime }$ does not occur. The perpendicular cross-section isThus the surface is an elliptic cylinder extending in both directions along the zero-eigenvalue axis. Its cross-sectional semiaxes are $\sqrt{k/{\lambda }_2}$ and $\sqrt{k/{\lambda }_3}$. It becomes a circular cylinder when the two positive eigenvalues coincide.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

For $u>0$, eliminating $v$ givesThus constant $u$ surfaces are confocal elliptic cylinders parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\le a$.

For generic $v$, eliminating $u$ givesThese are hyperbolic cylinders with the same focal lines. A specified angular value selects the half-branch with $\mathrm{sgn}x=\mathrm{sgn}(\cos v)$ and $\mathrm{sgn}y=\mathrm{sgn}(\sin v)$, because $u\ge 0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\ge a$ or $x\le -a$; at $v=\pi /2,3\pi /2$ they are $x=0$, $y\ge 0$ or $y\le 0$.

Finally, constant $z$ gives horizontal planes. The three surface families express the orthogonality of elliptic cylindrical coordinates.

Use the principal coordinatesPart (c) puts the equation in the canonical formIf $c>2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $1/\sqrt{6}$, $1/\sqrt{3}$ and $1/\sqrt{2c-4}$ along the three eigenvectors listed in part (c).

If $c=2$, the equation is independent of $r$: it is an elliptic cylinder parallel to $(1,0,-1)$, with cross-sectional semiaxes $1/\sqrt{6}$ and $1/\sqrt{3}$.

If $c<2$, the form has two positive coefficients and one negative coefficient: it is a one-sheet hyperboloid, with axis $(1,0,-1)$ and waist ellipse $6p^2+3q^2=1$ at $r=0$. Its axial scale is $1/\sqrt{4-2c}$.

These are the three cases of the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid. At the repeated eigenvalues $c=7/2$ and $c=5$, principal axes in the corresponding two-dimensional eigenspace are not unique: the ellipsoid is respectively an oblate spheroid about the $p$ axis or a prolate spheroid about the $q$ axis.