# Contact points for John's ellipsoid

Suppose $$K$$ is a centrally symmetric convex body in $$\mathbb{R}^n$$ and $$E$$ is the John's ellipsoid, the ellipsoid of maximal volume inside $$K$$.

If $$E$$ and $$K$$ have exactly $$2n$$ contact points, say $$(\pm x_i)_{i=1}^{n}$$, do $$(x_i)_{i=1}^n$$ form an orthonormal basis for the Euclidean norm indiuced by $$E$$?

Naively, this statement seems true in two dimensions, but I don't know how to prove it. Or my intuition could be wrong.

Edit: removed another question (whether all points on $$\partial K$$ extreme points implies exactly $$2n$$ contact points) as it has an easy negative answer (in comments). Hopefully the remaining question is not as trivial.

• Doesn't seem right even for n=2. Let E be a circle, and K a smooth convex centrally-symmetric curve between E and some regular hexagon circumscribed about E. Aug 20, 2022 at 18:26
• Indeed, the first question is false. Aug 20, 2022 at 18:32
• You may choose a norm, so that $x_i$ forms an orthonormal basis, after that it remains to read and apply the only property of John ellipsoid mentioned in wikipedia en.wikipedia.org/wiki/John_ellipsoid. Aug 20, 2022 at 19:26
• @AntonPetrunin I am not sure I understand. $K$ is given and $E$ is uniquely determined by $K$. What is it to choose? Aug 20, 2022 at 19:59

Looks true. A necessary and sufficient condition for these points (let $$E$$ be a standard ball) is that the identity operator $$I$$ is a non-negative linear combination of projectors $$P_i$$ on lines through $$x_i$$: $$I=\sum c_i P_i.\quad\quad\quad\quad\quad(\heartsuit)$$ If $$x_i$$'s are linearly dependent, multiply $$(\heartsuit)$$ by a vector $$y$$ orthogonal to all $$x_i$$'s to get a contradiction. If not, denote by $$(z_i)_{i=1}^n$$ a biorthogonal system to $$x_i$$'s and multiply $$(\heartsuit)$$ by $$z_j$$ to get $$z_j=c_jP_jz_j$$. This means that $$z_j$$ is an eigenvector of $$P_j$$ with non-zero eigenvalue, thus $$z_j$$ is parallel to $$x_j$$. This is what you need.