CODEWITHSUNDEEP
reigenvalue
Sara Soria
Sara Soria

Reputation: 13

How can I know the lengths of an ellipsoid without plotting them using eigen values

I have 3 parameters: A, B and C with 20000 data each of them. When I put them together in an 3D graph, they form a dot cloud with A in X axis, B in Y axis and C in Z axis. I want to fix this dots cloud in an 90% covariance error ellipsoid and determine the lengths, i.e. width, height and stretch, in > 1000 samples individually. Width will be in X axis, height in Y axis and stretch in Z axis). So, for each sample:

1) I calculate the covariation matrix with var()

covMat <- var(cbind(A,B,C)) 
covMat
           A          B          C
A 0.08269010 0.05330448 0.05329601
B 0.05330448 0.33033311 0.14017367
C 0.05329601 0.14017367 0.18732865

2) I calculate the eigenvalues with eigen()

evals<-eigen(covMat)$values
evals
[1] 0.43162433 0.10861830 0.06010923

#I obtain 3 eigenvalues ordered from largest-to-smallest

3) I determine the lengths of the 90% covariance error ellipsoid

ell.len <- 2*sqrt(6.251*evals)
ell.len
[1] 2.971533 1.666728 1.325708

The tricky thing: Since eigenvalues are always in a decreasing order, how can I know which number of the ell.len correspond to each length (width, height and stretch) since sometimes height can be larger than width and stretch, and other combinations. I want to know this because I would like to determine numerically when the ellipsoids are prolate (Height > Width > Stretch), oblate (Width > Height > Stretch) or spherical (Width = Height = Stretch) in each sample (n = 1000).

Upvotes: 0

Views: 116

Answers (1)

Edward
Edward

Reputation: 19339

You could bypass the eigen function and just use

evals <- .Internal(La_rs(covMat, TRUE))$values

since the covMat matrix is symmetric.

Upvotes: 1

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