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.



Contact points for John's ellipsoid
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.


