In this post series we look at orthogonal projections of an ellipsoid in a n-dimensional (Euclidean) space onto a (n-1)-dimensional subspace. The ellipsoid is a closed surface in the ℝn and has a dimensionality of (n-1). Orthogonal projection means that the target subspace is orthogonal to a line of projection (defined by a vector) which is the same for all points on the ellipsoidal surface. In this post and the following one we are going to prove that the orthogonal projection of a (n-1)-dimesional multidimensional ellipsoid creates an image whose surface is a (n-2)-dimensional ellipsoid. So far the line of projection has been given by a coordinate axis of a Cartesian coordinate system. We will follow this supposition in this post, but will drop it in a forthcoming post.
Previous posts
For our objective we use insights regarding those specific points on the ellipsoid whose projection give us the surface of the ellipsoid’s image in the target subspace. Terminology and conventions are the same as in “post I”. We work in the Euclidean ℝn with a Cartesian coordinate system [CCS]. Unit vectors along the coordinate axes are written as ei. The special unit vector which gives us the line of projection onto an orthogonal subspace is named ep. The invertible matrix generating the ellipsoid from vectors of the unit sphere 𝕊n-1 is called A. The respective quadratic form is given by a symmetric, pos.-definite matrix Σ-1 whose inverse Σ fulfills Σ=AAT.
Note that our considerations do not cover optical, i.e. perspective projections onto planes in the ℝ3 via light rays coming from the edge of an ellipsoid and meeting at an eye or camera. Such projections also lead to ellipsoids, but with additional factors changing the main axes compared to orthogonal projections.
Previous results
I summarize some results o the last post. The vectors xe ∈ ℝn of the (n-1)-dim ellipsoid E fulfill the following condition given by a quadratic form
As said we have
A projection operator P (see below), which mathematically corresponds to our orthogonal projection, produces image vectors yp:
Based on geometrical considerations we have identified special points xpt on the ellipsoid which control the projection image’s surface-points: The projection of a vector xpt results in a vector ypt that has an extremal length in comparison to other projection vectors yP which have the same direction as ypt. We called these points xpt “tangential points” of the ellipsoid. The xpt are created via matrix A from special vectors un of the unit sphere 𝕊n-1 (un ∈ 𝕊n-1):
The tangent space there is orthogonal to the gradient ∇F of a function F(x1, x2, … xn)
giving us our ellipsoid as a contour hyper-surface [ F(…xi …)=0 ] and orthogonal to ep. (∇F is orthogonal to the surface.) We also saw
We symbolize the tangent space at xpt ‘s location by 𝕋(xpt). 𝕋(xpt) is defined by an orthogonal vector, which we call vT=vT(xpt). Because of (6) and the orthogonality condition, we can chose it to be:
In addition we have:
I.e.:
So far, so good. In a next step we must consider the projection. Refreshing our knowledge in linear algebra, we know that we can express a projection by a projection operator, which is nothing else than a matrix P with special properties.
Applying a projection operator
The projection operator P for projections along ep must cancel the coordinate value xp in vectors xe ∈ E. Therefore, the projection’s image EP in the (n-1)-dim subspace SP (SP ⊥ ep) in general has elements yP ∈ ℝn with the property that one component is zero
We know already that EP is a limited (n-1)-dimensional volume. The projection operator for the orthogonal projection along the direction of ep can be written as
𝕀n is the (n x n) identity matrix. Test it out! You may also check that this operator fulfills the standard condition
While SP contains all projections of points on the ellipsoid, we must analyze what properties the result of a projection of our special points xpt
have. Recalling what we did in the last post, we determined the points xpt to give us a projection vector of maximum length of yP in the direction of Pxpt.
The trick which allows for deeper analysis is to split the vector xpt into two orthogonal vectors. (I have learned this trick from studying [1] (of D. Eberly at Geometric Tools, Redmond) and from a dialog with GPT-5.)
with some factor λp specific for xpt. To get further insights, we apply condition (8). I.e.:
After some reordering we get an expression for the factor λp
The vector to the image point yPt (as soon as known) completely determines the position of xpt . Note again that this actually is a condition which requires that there is exactly one point xpt ∈ Cpt that determines the extremum of the length of yP -vectors in the direction of the gradient ∇F(xpt); see post I for the geometrical reason. Due to the curvature of the closed surface of the ellipsoid this is true.
As being a member of the ellipsoid, xpt must fulfill eq. (1):
Resolving the first bracket and reordering gives us :
Due to (8), the second term on the left side of eq. (19) is zero. Inserting λp acc. to eq. (17) results in
Let us check the properties of the middle matrix
The properties of this matrix ΣP-1 are not totally obvious. To understand what happens on a basic level, let us look at an example in 4 dimensions.
We take a symmetric matrix B and pick a unit vector e3=(0,0,1,0)T with the 3rd element being 1. We calculate Be3e3TB.
Note:
e3e3TB leaves only the 3rd row of B:
Note also that the denominator e3TBe3 just picks out the element at the crossing of the 3rd row and column of B. i.e. B[3,3]:
Therefore, Be3e3TB / e3TBe3 gives us a symmetric matrix with the 3rd row and 3rd column are identical to the respective row/column of B itself :
You see the pattern? So, B – ( Be3e3TB / e3TBe3 ) is certainly a symmetric matrix with only rank (n-1) :
It is easy to show that for a vector v = (v1, v2, 0, v4)T we have
We also have the identity
Therefore, if B were pos. definite BP would be too. The patterns shown for n=3 are structurally independent of n. You could prove this by complete induction. We omit this step and transfer our results to our matrix of concern, ΣP-1:
- ΣP-1 is symmetric, pos. definite for vectors yp in the projection’s target sub-space SP and has an appropriate rank (n-1) for an application on vectors in the sub-space. Furthermore, ΣP-1 maps elements of SP onto elements of SP. However, it gives zero when applied to vectors f * ep. And, because of eq. (20) for vectors ypt ∈ EP, we have
This mean, the projections ypt ∈ EP of our special tangential points xpt indeed fulfill the conditions of a (n-2)-dimensional ellipsoid.
The whole line of argumentation could be repeated for a changed CCS with another ep. This shows that the argument holds for any projection sub-space orthogonal to some chosen vector.
Conclusion
We have shown that the orthogonal projection of an ellipsoid in the ℝn along a given vector ep onto a (n-1) sub-space orthogonal to this vector has a surface (!) that is a (n-2)-dimensional ellipsoid. The line of argumentation was tricky. We used our geometrical intuition as a starting point. In the next post I will change perspective and follow a more direct line of proof plus a more abstract one. I will then in addition discuss what the matrix defining the quadratic form of the projection has to do with the so called Schur complement of the matrix controlling the original ellipsoid.
Links and literature
[1] D. Eberly, “Principal Curvatures of Surfaces”, Geometric Tools, Redmond WA 98052, Documentation published under the Creative Commons Attribution 4.0 International License at
https://www.geometrictools.com/Documentation/PrincipalCurvature.pdf
To view a copy of the license, visit http://creativecommons.org/licenses/by/4.0,