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Consider the following ellipse, generated by the bounding region of the following points

ps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};
rec = N@BoundingRegion[ps, "FastEllipse"];
Graphics[{rec, Red, Point@ps}]

enter image description here

We have that the ellipse 'rec' is given in the form

Ellipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]

How can I retrieve the lengths of the two main axes of such ellipsoid? Following this representation and Mathematica's general definition of Ellipsoid

enter image description here

I tried the following, using the eigenvalues of rec[[2]]

eigs = Eigenvectors[Inverse[rec[[2]]]]
eigv = Eigenvalues[Inverse[rec[[2]]]];
lens = 2/Sqrt[eigv]

Out[]= {2.06559, 3.57771}

where the 2 factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get

Graphics[{rec, Red, Point@ps,
  Blue, Line[
   RegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/
      2, (lens[[1]] eigs[[1]])/2}],
  Line[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/
      2, (lens[[2]] eigs[[2]])/2}]}]

enter image description here

Is this correct? Is there a quicker way of doing this?

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3 Answers 3

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Stolen from @J.M.'s answer, https://mathematica.stackexchange.com/a/239797/4999, with one correction (is that enough to make it not a duplicate?):

Nodes = ps;
ellipsoidBR = BoundingRegion[Nodes, "FastEllipse"]; (* not "FastEllipsoid" *)
center = ellipsoidBR[[1]];
{vals, vecs} = Eigensystem[ellipsoidBR[[2]]];
{a, b} = Sqrt[vals];
major = N@{center - a vecs[[1]], center + a vecs[[1]]}
minor = N@{center - b vecs[[2]], center + b vecs[[2]]}
Graphics[{ellipsoidBR, Red, Point@ps, Green, Point@center, 
  Line@{major, minor}}, Frame -> True]
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Here's the alternate way using SingularValueDecomposition:

pt = rec[[1]]; mat = rec[[2]];
{u, s, v} = SingularValueDecomposition[(mat + Transpose[mat])/2]; 
func = Composition[AffineTransform[{u, pt}], ScalingTransform[Sqrt[Diagonal[s]]]];

and the length:

EuclideanDistance @@@ {func@{{0, -1}, {0, 1}}, func@{{-1, 0}, {1, 0}}}

{2.06559, 3.57771}

Graphics[{rec, Blue, Line[func@{{-1, 0}, {1, 0}}], 
  Line[ func@{{0, -1}, {0, 1}}]}]

enter image description here

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I think the help of Ellipsoid is incomplete because it does not explain the input: Ellipsoid[p,[CapitalSigma]], where the second argument is called the "weight matrix".

You will remember that an ellipse (for simplicity I am explaining the 2D case, nD is similar and assume the ellipse is centered at the origin) can be written by:

x^2/rx^2 + y^2/ry^2 == 1

We may write this as:

{x,y}.{{1/rx^2,0},{0,1/ry^2}}.{x,y} == r.mat.r == 1

Note that the inverse Sqrt of the eigenvalues of m0 are the half axes of the ellipse.

If we now rotate the coordinate system by a rotation matrix: rot (r'=rot.r where r' are the new coordinates) the ellipse will be rotated (in the inverse sense) in the new coordinates:

r'. Transpose[rot].mat. rot .r' == r' . mat' . r' == 1

Therefore a rotated ellipse may be represented by a an symmetric (positive definite) matrix: mat'. This is called "weigh matrix" in the help.

Note that the eigenvalues of the matrix are not changed by a rotation. The eigenvectors point in the directions of the half axes.

Here is an example: Let

rx=2;
ry=1;
m0=DiagonalMatrix[{1/rx^1,1/ry^2}]
{x,y}.m0.{x,y}==1

This represents an axis aligned ellipse with half axes rx and ry:

Region[ImplicitRegion[{x, y} . m0 . {x, y} == 1, {x, y}], 
 Axes -> True]

enter image description here

If we now rotate the matrix m0:

rot = RotationMatrix[-Pi/4];
m= Transpose[rot].m0.rot;

we get a rotated ellipse:

Region[ImplicitRegion[{x, y} . m . {x, y} == 1, {x, y}], Axes -> True]

enter image description here

Threrfore, the half axes are obtained by the inverse Sqrt of the eigenvalues of the weight matrix. And the directions of the axes are given by the eigenvectors.

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