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For a finite-dimensional normed vector space $E$, represent a finite family by its column operator $A:{\ell }_2^m\to E$. Its operator norm is the family's weak square norm. Since $A=AQ$ for the orthogonal projection onto $(\ker A{)}^{\perp }$, summing $\Vert A{e}_j{\Vert }^2\le \Vert A{\Vert }^2\Vert Q{e}_j{\Vert }_2^2$ proves the upper bound $\sqrt{\dim E}$. In coordinates adapted to a maximal inscribed Euclidean or Hermitian ellipsoid, the John contact decomposition gives ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I$ and ${\sum }_r{c}_r=\dim E$. The vectors $\sqrt{c_r}{u}_r$ have total squared norm $\dim E$ and weak square norm one, giving equality. This holds over both real and complex scalars.

The two-state calculation is not a general formula for a ferromagnet. Exchange interactions, magnetic domains, magnetic domain-wall pinning, magnetic anisotropy, and hysteresis can determine which response is accessible. Both spin angular momentum and orbital angular momentum can contribute; a single constant gyromagnetic ratio may be only a restricted effective description. Rotation sets an internal bias, while the material and its preparation determine magnetization. The Barkhausen effect illustrates why a smooth bias can produce jumps instead of the toy smooth curve.

For a limited, reversible, isotropic linear model, suppose the same intrinsic volume magnetic susceptibility ${\chi }_v$ responds to the applied bias and to $B_{\Omega }/{\mu }_0$. For a uniformly magnetized ellipsoid along a principal axis, include its demagnetizing factor $N_d$:Solving this linear equation yieldsHere $H_{\mathrm{int}}$ and $H_{\mathrm{app}}$ are actual magnetic field intensities; the added rotational bias is an equivalent drive in the constitutive response. In SI, dimensionless ${\chi }_v$ multiplies an intensity in $A/m$, so $M_z={\chi }_v\Omega /\gamma$ would omit ${\mu }_0$ and have the wrong units. A susceptibility defined instead as $\partial M/\partial B$ has different units. State the convention before comparing formulas.

At zero applied magnetic field intensity, a stipulated ${\chi }_v=3$ gives prefactor $3/2$ in a sphere, with $N_d=1/3$, and a prefactor approaching $3$ in a long axial ellipsoid. This is a geometry check of the assumed linear model. A real cylinder does not have an exactly uniform demagnetizing field, and an irreversible ferromagnet cannot in general be represented by this single linear equation. Inferring an intrinsic gyromagnetic ratio requires an independently calibrated material response and a geometry-dependent conversion of the measured stray magnetic field.

For nonzero real $a_i$, scale the ambient coordinates by $q_i=a_i{r}_i$. On the half where $\epsilon q_i>0$, omitting $q_i$ is a manifold chart onto the open unit ball; its inverse restores $q_i=\epsilon \sqrt{1-{\sum }_{j\ne i}{q}_j^2}$. These $2(n+1)$ charts form a smooth atlas of the ellipsoid, including for negative $a_i$.

A demagnetizing factor relates a component of demagnetizing field to magnetization. The three principal factors of a uniformly magnetized ellipsoid sum to one in the International System of Units. Factors quoted in a different unit convention need conversion; state the convention and measurement geometry before applying them.

The demagnetizing field is the contribution to magnetic field intensity produced by the sample's own magnetization. Inside a uniformly magnetized ellipsoid, it is uniform and has the form ${\mathbf{H}}_d=-\mathsf{N}\mathbf{M}$; the demagnetizing factor along a principal axis gives its longitudinal component. These conditions make the scalar correction in field conventions for magnetic susceptibility useful.

A cylinder generally has spatially varying magnetization and demagnetizing field. A volume-sensitive instrument and a short central pickup coil consequently need different effective corrections: one averages over the sample, while the other samples central flux. A shape factor copied from an ellipsoid is not an exact correction for every cylinder, especially with nonlinear response or hysteresis. ^[ref]

The electric field produced by the bound charge of a polarized body is its depolarization field. For a uniformly polarized ellipsoid in vacuum, the internal depolarization field is uniform and opposes the relevant component of the electric polarization: $E_{\mathrm{dep},i}=-N_i{P}_i/{ϵ}_0$. Its dimensionless geometric factors obey $N_x+N_y+N_z=1$.

Let $A$ be a real positive-definite matrix and let the columns of $U$ be an orthonormal basis of a two-dimensional linear subspace in three-dimensional Euclidean space. Points in that central plane are $x=Uq$, with $|x{|}^2=|q{|}^2$. Restricting the ellipsoid $x^{T}Ax=1$ gives $q^T(U^{T}AU)q=1$. The compressed matrix $U^{T}AU$ is a positive-definite matrix, and its extreme eigenvalues give the distance formulas in the title by the Rayleigh quotient. The two extremal axes are the images under $U$ of the corresponding eigenvectors. This restriction reduces a problem with two Lagrange multipliers to a two-dimensional quadratic form.

For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

In 2006, Yilong Han and colleagues dispersed plastic ellipsoids in water inside a thin glass cell, using a very dilute suspension to limit particle interactions. A camera on an optical microscope recorded 30 frames per second. Image analysis extracted centre positions and long-axis angles, with reported angular resolution about $1^{\circ }$. Their shape supplied the observable axis. Reflected interference colours calibrated cell thickness; nearly constant particle interference colour checked vertical wandering. One tracked particle had dimensions $2.4\times 0.3\times 0.3\mu \mathrm{m}$ in an $846\mathrm{nm}$ gap. Angular displacement variance grew linearly with lag; displacement resolved along the particle's axes and in the laboratory showed how turning washes out translational anisotropy. These are observations of confined particles: walls alter drag, and slight out-of-plane tumbling affected positional resolution. They do not establish the bulk sphere formula. The paper's Figure 1A–B shows simulations; Figure 1F shows a measured trajectory.^[ref]

For interpreting such images, the director distinction is essential: identical ends give an angle modulo $\pi$. A continuously unwrapped trace can follow small successive turns, but it does not make the ends physically distinguishable. The invariant planar axis autocorrelation is $⟨\cos (2\Delta \varphi )⟩$, and its decay determines $D_r$ through $e^{-4D_{r}t}$. Separately measuring position and rigid body orientation is what makes the connection between translational anisotropy and loss of angular memory experimentally accessible.

In vacuum, magnetization is zero and magnetic flux density obeys $\mathbf{B}={\mu }_0\mathbf{H}$. The applied quantities ${\mathbf{H}}_{\mathrm{app}}$ and ${\mathbf{B}}_0={\mu }_0{\mathbf{H}}_{\mathrm{app}}$ mean the field of the external sources at the sample position with the sample absent. The internal magnetic field intensity includes the sample's demagnetizing field; its internal magnetic flux density additionally includes ${\mu }_0\mathbf{M}$. A field quoted in the tesla is a $B$ quantity, even when an instrument informally calls it “$H$”.

For a uniformly magnetized ellipsoid, along a principal axis with demagnetizing factor $N_d$ in the International System of Units,For a sphere, $N_d=1/3$. The axial value approaches zero in the long slender ellipsoid limit and the normal value approaches one in the thin flat ellipsoid limit. Thus a stipulated intrinsic volume magnetic susceptibility ${\chi }_v=3$ would yield apparent values $1.5$ in a sphere and approximately $3$ in a long axial sample. These are calculated geometry effects, not measured material values. For fixed $N_d$, implicit differentiation gives the same conversion for local differential magnetic susceptibility, $dM/d{H}_{\mathrm{app}}={\chi }_{\mathrm{diff}}/(1+N_d{\chi }_{\mathrm{diff}})$.

For isotropic turning in three dimensions, the tip of a marked unit vector $\mathbf{u}$ wanders over the unit sphere. Its probability density function $f$ obeys the orientational Fokker-Planck equationHere ${\nabla }_{\mathbf{u}}^2$ is the Laplace-Beltrami operator on the unit sphere. The factor two follows from the sphere surface derivative identities: ${\nabla }_{\mathbf{u}}^2{u}_i=-2u_i$, so each mean vector component decays at rate $2D_r$. This is a different orientation space from the circle of planar rotational diffusion, whose corresponding rate is $D_r$. For the same illustrative $D_r=0.25{\mathrm{s}}^{-1}$ and $t=2\mathrm{s}$, this three-dimensional autocorrelation is $e^{-1}\simeq 0.368$.

One unit vector records a pointing direction, not every aspect of a rigid body's rigid body orientation: rotation about that vector remains unspecified. Also, $C_1$ requires an identifiable arrow. An unmarked ellipsoid's long axis alone does not supply one, and projecting a three-dimensional trajectory onto a camera plane does not automatically produce the planar model.

The John ellipsoid of a convex body is its inscribed ellipsoid of largest volume. For the unit ball $K$ of a $d$-dimensional normed vector space, change coordinates so that this ellipsoid is the Euclidean unit ball. The resulting coordinates are called John position. In these coordinates the John contact decomposition implies $B_2^d\subseteq K\subseteq \sqrt{d}{B}_2^d$, or $\Vert x{\Vert }_E\le |x|\le \sqrt{d}\Vert x{\Vert }_E$.

Every finite-dimensional vector subspace $E$ of a Banach space $X$ is the range of a bounded linear projection with the displayed norm bound. In maximal inscribed Euclidean or Hermitian ellipsoid coordinates, write the John contact decomposition as ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I_E$. Its contact functionals ${\varphi }_r(x)=u_r^{\ast }x$ have norm one on $E$. Extend them to $X$ by the Hahn-Banach theorem and set $Px={\sum }_r{c}_r{\tilde{\varphi }}_r(x)u_r$. This restricts to the identity on $E$. The contact synthesis operator has Euclidean norm one, the weights sum to $\dim E$, and the Euclidean unit ball is contained in the norm unit ball; these facts give $\Vert Px{\Vert }_E\le \sqrt{\dim E}\Vert x{\Vert }_X$.

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.

The mass second-moment tensor about a centre of mass is $I_{jk}=\int y_j{y}_{k}dm$. It measures the shape of a mass distribution and appears in tidal torque theory. It differs from the mechanical inertia tensor, which is ${\mathcal{I}}_{jk}=\mathrm{tr}(I){\delta }_{jk}-I_{jk}$. For a uniform ellipsoid of mass $M$ and semiaxes $a_1,a_2,a_3$, its principal values are $M{a}_j^2/5$.

For a positive-definite symmetric matrix $A$, the ellipsoid $x^{T}Ax=1$ has nearest points to the origin exactly at $x={\lambda }_{\max }^{-1/2}u$, where $u$ ranges over unit vectors in the largest-eigenvalue eigenspace. Indeed, the spectral theorem for real symmetric matrices gives $x^{T}Ax\le {\lambda }_{\max }|x{|}^2$, with equality precisely in that eigenspace. A simple largest eigenvalue gives two opposite nearest points; multiplicity $m>1$ gives a scaled $(m-1)$-sphere of nearest points.

Parametrize the ellipsoid by $\mathbf{x}=(b\sin \theta \cos \varphi ,b\sin \theta \sin \varphi ,a\cos \theta )$, with positive semiaxes $a,b$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. The cross product ${\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi }$ gives the displayed outward vector area element. For $0<\theta <\pi$, it is the positive multiple $a{b}^2\sin \theta$ of the outward normal direction $(x/b^2,y/b^2,z/a^2)$, fixing its orientation. A flux integral can therefore be evaluated directly as $\int \int \mathbf{F}(\mathbf{x})\cdot ({\mathbf{x}}_{\theta }\times {\mathbf{x}}_{\varphi })d\theta d\varphi$, without computing the length of the normal separately. The scalar surface area element is the Euclidean norm of this vector element.

The symmetric part of the displayed matrix isIt equals the matrix of 8 plus $3I$, so the same orthonormal eigenbasis gives eigenvalues $6,1,1$. In coordinates $u,v,w$ along ${\mathbf{v}}_3,{\mathbf{v}}_{-2},{\mathbf{e}}_2$, respectively, the level set isAll three eigenvalues are positive, so this is an ellipsoid with semi-axis lengths $1,\sqrt{6},\sqrt{6}$. The short axis is along $(2,0,1)$, and the plane of the longest axes is $2x_1+x_3=0$.

A natural longest-axis direction in the $x_1{x}_3$ plane is ${\mathbf{v}}_{-2}=(1,0,-2)/\sqrt{5}$. The angle $\eta$ of a unit vector to the plane $x_1=0$ has $\sin \eta =|v_1|$, givingfor that choice. Degeneracy qualification: the two longest semi-axes are equal, so “the longest axis” does not specify a unique direction. The equally valid longest axis ${\mathbf{e}}_2$ lies in $x_1=0$ and has angle zero. Generally $\mathbf{v}=\cos \psi {\mathbf{v}}_{-2}+\sin \psi {\mathbf{e}}_2$ is a longest-axis unit vector, withThis is the principal-axis freedom in a repeated ellipsoid eigenspace; a unique answer needs an extra axis-selection convention.

An orthogonal transformation to the eigenvectors preserves distance from the centre and writes the surface as $u^2+v^2+3w^2=3$. It is an ellipsoid with semi-axes $\sqrt{3},\sqrt{3},1$. ThereforeThe original $y$ axis is one of the eigen-directions with eigenvalue one, and the centre has $a_y=0$. Consequently $y^2\le 3$, attained when the other centred coordinates vanish, so $\boxed{y_{\max }=\sqrt{3}}$.

For $A^{\prime }$, the centre and constant remain $\mathbf{a}=(-1,0,0)$ and $C=3$. Its eigenvalues are $3,1,-1$, so rotated coordinates put the surface in the form $3u^2+w^2-y^2=3$. This is a one-sheet hyperboloid. Since $3\Vert \mathbf{x}{\Vert }^2\ge {\mathbf{x}}^{\mathsf{T}}{A}^{\prime }\mathbf{x}=3$, the nearest centre-to-surface distance is at least one, attained in the eigen-direction with eigenvalue three. The coordinate $y$ is unbounded, because arbitrarily large $|y|$ can be balanced by $u$ or $w$. Thus the nearest distance is one, and there is no maximum value of $y$.

For the printed nonsymmetric $A^{\prime \prime }$, use the symmetric part of a matrix:The skew-symmetric part contributes zero to this scalar. The same centre solves $2S\mathbf{a}+\mathbf{b}=0$, and again $C=3$. Sylvester's positive principal minors are $2,2,5/2$, so this surface is an ellipsoid.

Let ${\lambda }_{\ast }$ be the largest eigenvalue of $S$. Its characteristic equation is $4{\lambda }^3-20{\lambda }^2+27\lambda -10=0$, and the Rayleigh quotient givesFor the maximum $y$, the support function of a centred ellipsoid gives the bound below. Equivalently, apply the Cauchy-Schwarz inequality in the $S$-weighted inner product: $y^2\le 3{\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y$. The inverse entry is ${\mathbf{e}}_y^{\mathsf{T}}{S}^{-1}{\mathbf{e}}_y=6/5$, and equality is attained for a centred vector proportional to $S^{-1}{\mathbf{e}}_y$. Thus

Using the eigenvalues of the raw nonsymmetric $A^{\prime \prime }$ would be incorrect: it is its symmetric part that determines the surface and its Euclidean geometry.