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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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up vote 4 down vote accepted

The answer is yes.

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.

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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
@puzne, My notation differ from yours, this is probably the reason for confusion. Moving ellipsoid in the standard position is the same as moving an ellipsoid from one position to any other position. – Anton Petrunin Sep 25 '12 at 1:38
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
Oh now I see - very nice, thanks! – puzne Oct 2 '12 at 9:36

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