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Suppose we have two ellipses in $2$-dimensions centered at the origin. It is easy to visualize that (unless one is contained in the other) they will have $4$ points of intersection. Is it known whether in three dimensions, three generic ellipsoids will have a common point of intersection (unless one is contained in the other) or is there a counterexample? Is there a similar result in $n$-dimensions?

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KReiser
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fuzzyrock1
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    You need to formulate your condition in terms of major and minor axes. – Oliver Jones Feb 08 '16 at 02:41
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    This result is known in a bit more generality as Bézout's theorem. It counts intersections with multiplicity and in projective space over an algebraically closed field, but implies that $n$ ellipsoids in $\mathbb{R}^n$ intersect in at most $2^n$ points (if the intersection is finite). – Charlie Feb 08 '16 at 03:21
  • Thanks for your tip regarding Bézout's theorem. However the real numbers are not algebraically closed. So the theorem does not seem to directly apply. Do you mind clarifying how you can claim the intersection of $n$ ellipsoids in ${\mathbb R}^n$ from this theorem? – fuzzyrock1 Feb 08 '16 at 03:47
  • @Oliver Jones: the ellipsoids can have arbitrary orientation, but are all centered at the origin. I am looking for solution to the system of equations $x^T {\mathbf Q}_i x = 1$ for $i=1,2,.., n$ and $x \in {\mathbb R}^n$ and each ${\mathbf Q}_i$ is positive semidefinite. – fuzzyrock1 Feb 08 '16 at 03:50
  • @fuzzyrock1 You need to define the condition "unless one is contained in the other". – Oliver Jones Feb 08 '16 at 04:13
  • I have added a figure to clarify the case of two dimensions and when one ellipse is contained in the other. – fuzzyrock1 Feb 08 '16 at 14:09
  • Also I was wondering if there is a negative result that says that n+1 or more (generic) ellipsoids in n dimensions will not have a single common point of intersection? The ellipsoids may mutually intersect, but there would not be a single point common to all of them. – fuzzyrock1 Feb 08 '16 at 14:35
  • @fuzzyrock1 I know what you mean but you must describe the condition mathematically. – Oliver Jones Feb 09 '16 at 07:55
  • Bezout's theorem applies because $\mathbb{R} \subset \mathbb{C}$. That's why it gives you a bound on the number of points of intersection, instead of equality. More explicitly, look at the ellipsoids in $\mathbb{C} \mathbb{P}^n$ by taking the projective closure. Apply Bezout's theorem (which gives you exactly $2^n$ points if the intersection is finite. Look in the appropriate affine chart, then intersect back down to $\mathbb{R}$. You see that you can loose points, but can't gain any. A general position argument like others suggest should guarantee equality in the way that you want. – Charlie Feb 18 '16 at 05:03

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