Re: Elipsoid fitting question

Hello Robert,

Thank you very much for your response. I forgot to mention that the
observation set is indeed sampled from an elipsoid (perhaps each
measurement is a bit noisy, but still the overall structure looks like
an elipsoid). I used H for denoting the least squares case and Q for
the covariance method. Anyway, H not being positive semidifinte does
not happen due to the more or less elliptical distribution of the
points as well as enforcing H to be pos. def. throughout the
optimization.

I want to know if there would be a difference in inv(Q) and H. I ask
this because I have seen papers on the complicated optimization
technique on H, subject to H being positive definite, but I do not
understand why they do not simply use the covariance idea like inv(Q).
Maybe the first one has advantages that I have not figured out?

Thanks

Golabi

On Feb 24, 12:19 am, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Golabi Doon <golabid...@xxxxxxxxx> writes:
Hello Everyone,

I am interested in estimating an elipsoid from sample points
X_1,...X_n where each X_i is a vector in R^d. For simlicity let's
assume the center of the elipsoid coincides with the origin. I know
two methods for doing this:

1. Computing covariance matrix Q=(1/n)*sum_i ( X_i*X_i' ), where X_i'
means transpose of X_i. Then I can describe the ellipsoid by X'*inv(Q)
*X=c where c is some appropriate constant.

2. H = ArgMin sum_i (X_i'*h*X_i-1)^2, where "h" is a nxn variable
matrix, subject to "h" being positive definite. This is simply a
constrained least squares method Then the ellipsoid would be X'*H*X=b
for some appropriate constant b.

My question is, other than difference in a constant factor, are inv(Q)
and H different in general? If so, under what conditions one
outperforms the other?

Your help would be highly appreciated.

Linear least squares for the equations
  (X_i)' Q X_i = 1
where Q is an unknown d x d symmetric matrix.  If the resulting Q is not
positive semidefinite, it means the sample points are a better fit to a
hyperboloid than to an ellipsoid.
--
Robert Israel              isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada- Hide quoted text -

- Show quoted text -

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  • Re: Elipsoid fitting question
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