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.

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    $\begingroup$ 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. $\endgroup$ Aug 20, 2022 at 18:26
  • $\begingroup$ Indeed, the first question is false. $\endgroup$
    – Markus
    Aug 20, 2022 at 18:32
  • $\begingroup$ 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. $\endgroup$ Aug 20, 2022 at 19:26
  • $\begingroup$ @AntonPetrunin I am not sure I understand. $K$ is given and $E$ is uniquely determined by $K$. What is it to choose? $\endgroup$
    – Markus
    Aug 20, 2022 at 19:59

1 Answer 1


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.


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