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I have a hyper-ellipsoid defined as the transformation of the unit ball:

$$ \mathcal{E} = \left\{ T x + d \mid \left\Vert x \right\Vert_2 \leq 1 \right\} $$

where $T \in \mathbb{R}^{n \times n}$ is symmetrical and positive definite and $d \in \mathbb{R}^{n}$.

Now I want to plot the 2D projections of this ellipsoid ($x_1-x_2, x_1-x_3, x_2-x_3$, etc.)

How exactly do I do this?

Thanks for your help!

cedi123
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  • It uses "Schur complements" matrices as explained in a former answer of mine here – Jean Marie Feb 14 '22 at 18:32
  • @JeanMarie Could you elaborate? I'm not sure I understand how this helps me get the projections I want. – cedi123 Feb 14 '22 at 20:10
  • First of all, get rid of the d translation which can be added at the last step. Let $U=T^{-1}$ : $y=Tx \iff x=Uy$. Let us work on the sphere instead of the ball : expressing $|x|^2=1 \iff x^Tx=1 \iff (y^TU^T)(Uy)=1 \iff y^T(U^TU)y=1$. Therefore the matrix associated with the ellipsoid is $A=U^TU$. The example I have given provides the matrix associated to the ellipse which is the projection onto the $x-y$ plane. – Jean Marie Feb 14 '22 at 21:00
  • Here is a simpler method that will work for the case of a projection of a 3D ellipsoid on the 2D plane xOy. Having the equation $Y^TAY=1$ of your ellipsoid (as said in my comment earlier) with $Y=(x,y,z)$ consider it as a quadratic equation in $z$ with coefficients depending on $x,y$. Write that its discriminant is $0$ ; this will give you the equation of the projection. Say me if it works... – Jean Marie Feb 14 '22 at 22:12
  • Any comment ? ... – Jean Marie Feb 15 '22 at 06:44
  • @JeanMarie Ok so I know how to get matrix A. The matrix B in your answer in the matrix of x/y projection ellipse. In my case, the ellipsoid is in 6D. How do I have to build the schur complement to get every possible 2D projection? (x1/x2, x1/x3, x1/x4, x1/x5, x1/x6, x2/x3, x2/x3, etc.)? – cedi123 Feb 15 '22 at 06:58
  • Divide your matrix into 4 blocks with sizes $4 \times 4,4 \times 2, 2 \times 4, 2 \times 2$ resp. – Jean Marie Feb 15 '22 at 07:02
  • @JeanMarie So if I want to have the x2/x5 projection, I have to rearrange the rows and columns so that the 2nd and 5th row/column are next to each other? – cedi123 Feb 15 '22 at 07:04
  • Yes. This can be done "in-place" with indexing facilities offered for example by Matlab. – Jean Marie Feb 15 '22 at 07:07
  • Let us continue this discussion in chat. – cedi123 Feb 15 '22 at 07:10

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