# Can an ellipsoid be moved freely inside another ellipsoid?

An origin centric ellipsoid is defined by any positive semi-definite $n$ by $n$ matrix $X$, by taking all vectors $v$ such that $v^tXv\leq1$. Call two origin centric ellipsoid equivalent if one can be obtained from the other by rotation of space.

Now suppose that $E$ and $E'$ are two equivalent centric ellipsoids that are both contained inside a third centric ellipsoid $F$. Is it always possible to continuously rotate $E$ inside $F$ until is gets to the position $E'$?

This question is a more difficult version of this one.

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You can connect your small ellipsoid to the big one by a continuous nested one-parameter family of ellipsoids, say $E_t$, $t\in [0,1]$. Nested means that $E_{t_1}\subset E_{t_2}$ if $t_1 \le t_2$. Say if $$E=\{\\, x \mid \langle x,Ax\rangle \le1 \\,\}\ \ \text{and}\ \ F=\{\\, x \mid \langle x,Bx\rangle \le1 \\,\}$$ then $$E_t=\{\\, x \mid \langle x,((1-t)\cdot A+t\cdot B)x\rangle \le1 \\,\}$$
Let $E$ and $E'$ be two ellipsoids with semiaxis correspondingly $a_1\le a_2\le\dots\le a_n$ and $a'_1\le a'_2\le \dots \le a'_n$. Note that if $E\supset E'$, then $a_i\ge a_i'$ for all $i$.
Now consider continous family of rotations of $E=E_0$ such that its axis go to the axis of $E_t$. This way we get a continuous rotation of $E_0$ inside $E_1=F$ which moves it to the standard position. One has to be bit more careful in case some of $E_t$ have equal semiaxis. The later is left as an exercise.
I think you misunderstood my question. I'm considering the case where $E$ and $E'$ are really just the same ellipsoid - one is obtained from the other simply by rotation. However I would like to now continuously rotate $E$ into the position $E'$ while maintaining that all the intermediate ellipsoid are inside a third ellipsoid $F$, which contains both $E$ and $E'$. Even if all the axes of $E$ (and $E'$) have different lengths, there are many paths in the group of rotations that can continuously rotate $E$ into $E'$. It's not clear to me which of those, if any, keeps the ellipsoid in $F$. –  puzne Sep 24 '12 at 11:29
I made my notation bit closer to yours, but still my $E'$ has nothing to do with your $E'$. –  Anton Petrunin Sep 25 '12 at 1:46