Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to understand the constraints on exactly fitting an ellipse/ellipsoid to a small set of points as the number of dimensions increases. The ellipse/ellipsoid is always centred at the origin.

I conjecture that in $\mathbb{R}^n$, any $n$ points can have an infinite number of ellipsoids centred at the origin fitted through them so long as the $n$ points satisfy some constraint to avoid colinearity/coplanarity, which would lead to degenerate solutions (circles/spheres/spheroids) or no solution at all. I'm not clear on the precise nature of this constraint as $n$ increases.

Suppose we identify and satisfy these constraints so that special cases are excluded. Can this infinite number of ellipsoids be reduced to a unique ellipsoid by adding 1 more point, subject to additional constraints? What are those additional constraints? And is 1 more point always enough?

For example, if you have three points (plus origin) in $\mathbb{R}^2$, you can select the two points that make the widest angle with respect to the origin, and draw a line between them. The remaining point can only be fitted to an ellipse through the other two points if it lies "outside" the line with respect to the origin. I imagine that some similar constraint should exist in $\mathbb{R}^n$ - what is this constraint?

share|cite|improve this question

This is a partial answer.

Can this infinite number of ellipsoids be reduced to a unique ellipsoid by adding 1 more point, subject to additional constraints?

An general ellipsoid (as well as hyperboloids and other quadrics) can be described by a quadratic form. You can think of this as a symmetric $r\times r$ matrix, with $r=n+1$. It has $\frac{r(r+1)}2$ different entries. But you actually have one less degree of freedom due to homogenity. The centering at the origin eats another $n$ degrees of freedom, since you can achieve all other quadrics using translation in $n$ dimensions. So you have

$$\frac{r(r+1)}2-r = \frac{r(r-1)}2 = \frac{n(n+1)}2$$

real degrees of freedom. Each defining point contributes one real degree of freedom: it can be freely moved in $n$ dimensions, but even for a given quadric, it can still move insie the cuadric, i.e. a $(n-1)$-dimensional manifold, without changing its definition.

So in 2D, you have 3 defining points, in 3D you have 6, and so on. So no, $n+1$ points won't be enough to uniquely describe a quadric.

share|cite|improve this answer
Thanks - this has corrected my thinking on the question - hopefully I'll find the rest of the answer with a bit of perseverance :-) – omatai Apr 1 '13 at 20:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.