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Given an n-dimensional ellipsoid in $\mathbb{R}^n$, is any orthogonal projection of it to a subspace also an ellipsoid? Here, an ellipsoid is defined as

$$\Delta_{A, c}=\{x\in \Bbb R^n\,:\, x^TAx\le c\}$$

where $A$ is a symmetric positive definite n by n matrix, and $c > 0$.

I'm just thinking about this because it gives a nice visual way to think about least-norm regression.

I note that SVD proves immediately that any linear image (not just an orthogonal projection) of an ellipsoid is also an ellipsoid, however there might be a more geometrically clever proof when the linear map is an orthogonal projection.

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    $\begingroup$ Yes, they do. :) $\endgroup$
    – Arnaud
    Oct 6 '20 at 8:20
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Yes they do. You can prove it by induction on the codimension of the subspace you project to. For $x\in Vect(e_1,\ldots e_{n-1})$ there exists $t \in \mathbb{R}$ such that $x+te_n$ belongs to $\Delta$ iff the discriminant of the degree $2$ equation $(x+te_n)^TA(x+te_n)\leq c$ w.r.t. the unknown $t$ is non-negative, which turns out to still be a quadratic inequality in $x$.

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    $\begingroup$ what about "regular" shadows? e.g. perspective projection shadows I guess those would not be, but I have trouble proving it. Is it trivial to show, or is this a new question? $\endgroup$
    – Jeffrey
    Oct 6 '20 at 13:33
  • $\begingroup$ I think they project to quadrics, but it could be a domain bounded by half a hyperbola for instance. $\endgroup$
    – Arnaud
    Oct 6 '20 at 15:08
  • $\begingroup$ @Jeffrey I think the projection should still be an ellipsoid if every point in the original ellipsoid is projected to some point in the plane. An ellipsoid is the convex hull of its boundary, so the projection of the ellipsoid should be the convex hull of the projection of the boundary as long as no point gets projected to “infinity”. The boundary of an ellipsoid is of degree 2, a property which is preserved when changing coordinates in projective space, and we already showed that a normal projection turns this into an ellipsoid. $\endgroup$ Oct 6 '20 at 19:13
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Yes. An ellipsoid is a linear transformation of a spherical ball, and orthogonal projection is also a linear transformation, so it suffices to show that any linear transformation whose image is a subspace sends a spherical ball to an ellipsoid in that space.

A linear transformation can be decomposed into orthogonal projection by its kernel followed by some invertible linear transformation. Orthogonal projection sends a spherical ball to a spherical ball in the subspace, so we are done.

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Indeed, ellipsoids cast ellipse shape shadows on the ground.

The intersection of any conicoid and a first degree equation plane illumination terminator between two tangential points is a conic section. It can be proved by elimination to the conic second degree equation.

enter image description here

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There are already a good answers presented, but I want to add the also one may think in a following way :

The orthogonal projection defines some subspace $\langle e_1, e_2 \ldots e_n \rangle$, and we perform an orthogonal transformation $R^{T}$, such that the matrix $A$ transforms into $R^{T} A R$, and in the rotated basis, the first $n-1$ components will correspond to that subspace. After the rotation, matrix $A$ preserves its positive definiteness, and the restriction to the $(n-1) \times (n-1)$ will be positive definite by the Sylvester's critertion. Therefore this block would define an ellipsoid in one dimension lower.

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