Minimum enclosing ellipsoid to maximal enclosed ellipsoid

Question

Given a convex body $K$ and an ellipsoid of minimal volume which contains $K$, find the maximal ellipsoid contained in $K$.

I have tried to multiply the matrix by 4 (since the eigenvalues are the reciprocals of the squares of the semi-axes, but I don't get the maximal ellipsoid.

If you need to compute the minimal containing ellipsoid, please refer to here

Thanks in advance.

Given $K$, why do you need to go from the enclosing ellipsoid to the enclosed ellipsoid? Why not simply go from $K$? There is no unique relationship between a bounding ellipsoid and a bounded ellipsoid. That is, if you're given merely the bounding ellipsoid (but not $K$), there is no unique bounded ellipsoid. — David G. Stork, Jan 15 '16 at 19:18
because i need to know the connection between the minimal enclosing and maximal enclosed ellipsoids. — user3492773, Jan 15 '16 at 19:20
That isn't an answer. If you are give $K$, then 1) find the bounding ellipsoid from $K$, 2) find the bounded ellipsoid from $K$, 3) Plot them both. — David G. Stork, Jan 15 '16 at 19:21
So there is no connection between the bounded and bounding ellipsoids? — user3492773, Jan 15 '16 at 19:21
let E be the enclosed ellipsoid, then according to John Theorem (John ellipsoid), in order to get the enclosing ellipsoid you need to do the following: c + d*(E-c) where d is the dimension of the space and c is the center of the ellipsoid E. — user3492773, Jan 15 '16 at 19:23
For a tetrahedron, minimum circumscribing ellipsoid is the Steiner ellipsoid while maximum inscribed ellipsoid is the Steiner inellipsoid. — Ng Chung Tak, Jan 31 '22 at 04:37

David G. Stork · Accepted Answer · 2016-01-18T04:42:10.240

4

Consider the yellow and orange convex figures, which have the same bounding ellipsoid but different bounded ellipsoids.

Therefore, given just the bounding ellipsoid you cannot determine the bounded ellipsoid. (Nor vice versa.) You must know the figure $K$.

A better, limiting, example: Suppose $K$ is an ellipsoid. Then the bounding ellipsoid and the bounded ellipsoid are the same!

edited Jan 18 '16 at 04:42

answered Jan 15 '16 at 19:32

David G. Stork

30,409

score 3 · Answer 2 · answered Jan 18 '16 at 14:39

David has offered the correct answer geometrically, and I'd say he deserves the checkmark :-) But I did want to address this from a computational point of view.

The computational complexity of computing either the minimum volume circumscribed ellipsoid (MinVCE) or the maximum volume inscribed ellipsoid (MaxVIE) depends greatly on the way the underlying convex volume is described. And generally, only one of the exact ellipsoids yields a tractable solution. I can't think of a non-trivial case where computing both ellipsoids is tractable.

For instance, the minimum volume circumscribing ellipsoid is tractable for the following cases:

A finite set of points $x_1,x_2,\dots, x_m$;
A convex polyhedron described by its vertices (this is really the same as the previous case);
The union of a finite number of ellipsoids.

Conversely, the maximum volume inscribing ellipsoid is tractable for the following cases:

A convex polyhedron described by its faces (e.g., a set of linear inequalities);
The intersection of a finite number of ellipsoids.

Source: Boyd & Vandenberghe, Convex Optimization, chatper 8.

score 1 · Answer 3 · answered Jan 16 '16 at 19:00

Here is a reasonable conjecture in dimension $2$:

For an equilateral triangle $T$ the smallest circumscribed ellipse is the circumcircle, and the largest inscribed ellipse is the incircle. The area ratio between these two is $4$.

Conjecture: Let $K\subset{\mathbb R}^2$ be convex and compact, and let $A_{\rm circum}(K)$, resp $A_{\rm inscr}(K)$, be the areas of the smallest circumscribed ellipse, resp., the largest inscribed ellipse. Then $${A_{\rm circum}(K)\over A_{\rm inscr}(K)}\leq4\ .$$

Minimum enclosing ellipsoid to maximal enclosed ellipsoid

3 Answers3