How do I obtain the 3D or 2D projection of a hyperellipsoid? The hyperellipsoid is 5D. I know the semi-axes and principal axes directions of the hyperellipsoid in 5D space. The center of the hyperellipsoid is not at origin, and the principal axes of the hyperellipsoid are not aligned with the cartesian coordinate axes.
2026-03-29 21:58:12.1774821492
Projection of a 5D hyperellipsoid
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Practice example: take the ellipsoid $$ x^2 + y^2 + z^2 + y z + z x + x y = 6, $$ find the curve on the surface (oblate spheroid) where the gradient is orthogonal to $(0,0,1),$ then find the orthogonal projection of that curve into the $xy$ plane.
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appears you have had multivariable calculus. Taking $x$ as a column vector, constant vector $b,$ and $H$ as the Hessian matrix of the defining polynomial, the object is $$ x^T H x / 2 + b^T x + c = 0 \; .$$ As a column vector, the gradient of this is $$ Hx + b. $$
For the 2-d projection, you get an ellipse. Taking $w_3 = (0,0,1,0,0),$ $w_4 = (0,0,0,1,0),$ $w_5 = (0,0,0,0,1),$ you are solving four things for an ellipse in R^5, $$ x^T H x / 2 + b^T x + c = 0 \; ,$$ $$ (Hx + b)^T w_3 = 0, $$ $$ (Hx + b)^T w_4 = 0, $$ $$ (Hx + b)^T w_5 = 0. $$ This gives an ellipse in a 2-plane in $R^5,$ and the projection to the $x_1 x_2$ plane is also an ellipse.