Ellipsoid

An ellipsoid is a level set $(x-c{)}^{T}A(x-c)=1$ for a real symmetric positive-definite matrix $A$. Its principal axes are the orthonormal eigenvectors of $A$, and an eigenvalue $\lambda$ gives semi-axis length $1/\sqrt{\lambda }$.

For a positive-definite real quadratic form $x^{T}Mx=c>0$, an orthonormal eigenvector of $M$ with eigenvalue $\lambda$ gives a principal semi-axis of length $\sqrt{c/\lambda }$. If an eigenvalue is repeated, any orthonormal basis of its eigenspace gives equally valid principal axes. Thus an angle attached to “the longest axis” need not be unique when the smallest eigenvalue is repeated. For example, the form with eigenvalues $6,1,1$ and short direction $(2,0,1)$ has every direction in $2x_1+x_3=0$ as a longest-axis direction. Its unit directions $(0,1,0)$ and $(1,0,-2)/\sqrt{5}$ make different angles, $0$ and $\arcsin (1/\sqrt{5})$, to the plane $x_1=0$.

Half the length of one principal diameter of an ellipsoid. For a centred quadratic form $x^{T}Sx=C$ with $S$ symmetric positive definite and $C>0$, its semi-axis lengths are $\sqrt{C/{\lambda }_i}$, where ${\lambda }_i$ are the eigenvalues of $S$. The shortest centre-to-surface distance corresponds to the largest eigenvalue.

Let $S$ be a real symmetric positive definite matrix, $C>0$ and $q\ne 0$. The Cauchy-Schwarz inequality in the $S$-weighted inner product gives $q^{T}x=(S^{-1}q{)}^{T}Sx\le \sqrt{q^T{S}^{-1}q}\sqrt{x^{T}Sx}$. Equality is attained by a positive multiple of $S^{-1}q$. Thus the displayed maximum holds. For a translated ellipsoid, add $q^{T}a$ for its centre $a$. This computes a coordinate extremum without constructing a rotation matrix.

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.

The interior bounded by an ellipsoid is a solid ellipsoid. Positive coordinate dilations multiply its symplectic capacity by the square of the dilation factor.

For the solid ellipsoid $D=\{x^2/a^2+y^2/b^2+z^2/c^2\le 1\}$ with positive semi-axes, set $x=au$, $y=bv$, $z=cw$. The Jacobian determinant of this inverse coordinate map is $abc$, and $D$ becomes the unit ball. Symmetry gives ${\int }_{B_1}{u}^{2}dV=\frac{1}{3}{\int }_{B_1}{r}^{2}dV=4\pi /15$, proving the displayed moment. The other two moments follow by interchanging the semi-axes; mixed moments vanish by reflection symmetry.

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 $\Vert x{\Vert }_E\le |x|\le \sqrt{d}\Vert x{\Vert }_E$.

For a centrally symmetric convex body in John position, there are contact points $u_j\in \partial K\cap S^{d-1}$ and positive weights $c_j$ such that ${\sum }_j{c}_j{u}_j\otimes u_j=I$. Taking the trace gives ${\sum }_j{c}_j=d$. A supporting hyperplane at $u_j$ shows $|⟨u_j,x⟩|\le \Vert x{\Vert }_K$.

Here is the maximum-volume argument for the decomposition. If $I$ were outside the convex cone generated by the contact matrices, a separating hyperplane would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the trace remains positive. The support function of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have trace $1$ and form a compact set; finite-dimensional convexity gives a finite decomposition.

A centrally symmetric convex body is in John position when its John ellipsoid is the Euclidean unit ball.

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 $a_i$.

If $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$ is invertible, $x\mapsto Dx$ sends the ellipsoid ${\sum }_i{a}_i^2{x}_i^2=1$ to the unit sphere. On $u=(v,s)\in S^n$, the two stereographic projections $v/(1-s)$ and $v/(1+s)$ omit opposite poles and have inverses $(2w/(1+|w{|}^2),\pm (|w{|}^2-1)/(1+|w{|}^2))$, with the sign chosen for the appropriate pole. Their overlap transition is $w\mapsto w/|w{|}^2$, smooth and involutive on ${\mathbb{R}}^n\setminus \{0\}$. Composing their inverses with $D^{-1}$ gives a two-chart smooth atlas of the ellipsoid. Its subspace topology is Hausdorff and has a second-countable space topology, so it is a smooth manifold. For $n=0$ the overlap is empty and each chart is a point.

A spheroid is an ellipsoid of revolution obtained by rotating an ellipse about a principal axis. Rotation about the shorter axis gives an oblate spheroid; rotation about the longer axis gives a prolate spheroid. Thus the oblate form is a member of the general family.

A prolate spheroid is an ellipsoid with two equal shorter semiaxes and a longer symmetry-axis semiaxis. Its meridional section is an ellipse rotated about its major axis. In the coordinates $R^2/(\lambda +\alpha )+z^2/(\lambda +\beta )=1$ with $\beta >\alpha$, a fixed regular $\lambda$ has equatorial radius $\sqrt{\lambda +\alpha }$ and polar radius $\sqrt{\lambda +\beta }$; its focal separation is independent of $\lambda$, giving a confocal family.

For the prolate spheroid $u^2/a^2+(v^2+w^2)/b^2=1$ with $a>b>0$, intersect the concentric sphere $u^2+v^2+w^2=R^2$. When $b<R<a$, subtraction gives two circles in the perpendicular planesTheir centers lie on the long principal axis and the line through the centers is normal to both planes. At $R=b$ the circles merge into the equator; at $R=a$ they shrink to the two polar vertices. There is no intersection for $R<b$ or $R>a$.