Orthogonal projections of multidimensional ellipsoids – III – arbitrary vectors for the projection direction and cuts of the unit sphere

The first two posts of this series showed that the orthogonal projection of an ellipsoid in the ℝⁿ onto a (n-1) dimensional sub-space has a (n-2)-dimensional ellipsoidal surface. The tangential points on the original ellipsoid which control the surface of the projection could be derived by linear transformation A of well defined, but special vectors of the (n-1)-dimensional unit sphere. A just the generates the vectors of ellipsoid from vectors of the unit sphere. In this post we shall invert our perspective – just for fun:
We pick an arbitrary vector v_P (= Av) of the ellipsoid to give us the direction of the projection. Then we take the (n-1)-dimensional space orthogonal to the related vector v of the unit sphere. We look at points in the intersection of this space with the unit sphere. We show that the linear transformation A of vectors in the intersection give us tangential points on the ellipsoid’s surface with tangent vectors having the same direction as v_P . We afterwards derive the matrix Σ_P describing the ellipsoidal border of the projection’s image and the respective quadratic form. This allows us to drop previous assumptions of the projection subspace being orthogonal to some chosen coordinate axis. In addition we will find a nice relation of v_P with a factor determining the matrix Σ_P.

Previous posts

Generating matrix of the ellipsoid

The ellipsoid E in the ℝⁿ is given by the following quadratic form:

\[ \text{vectors } \pmb{x}_e \text{ of (} n\text{-1)-dim ellipsoid } E\,: \quad \pmb{x}_{e} \in E = \left\{ \pmb{x}_e \,\, | \,\, \pmb{x}_e \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{x}_{e} \,=\, 1 \, \right\} \,. \tag{1} \]

The matrix Σ has a Cholesky decomposition

\[ \pmb{\operatorname{\Sigma}} \,=\, \pmb{\operatorname{A}} \, \pmb{\operatorname{A}}^T \,. \tag{2} \]

Matrix A generates the ellipsoid when applied to vectors u of the unit sphere 𝕊^n-1 (u ∈ 𝕊^n-1) :

\[ ||\pmb{u}|| = \sqrt{ u_1^2 + u_2^1 + … + u_n^2 \, } \,=\, 1 \,, \,\, \, \pmb{u} \in \mathbb{S}^{n-1} \, , \tag{3} \]

\[ \pmb{u} \in \mathbb{S}^{n-1} \,\Rightarrow \, \pmb{\operatorname{A}} \, \pmb{u} \in E \,. \tag{4} \]

Orthogonal projection

Let us pick a vector v_P ∈ E. This vector shall give us the direction of the orthogonal projection P onto a subspace S_P ⊥ v_P:

\[ \begin{align} & \pmb{v} = \left[ \pmb{\operatorname{A}}^{-1}\, \pmb{v}_P\right] \, \in \, \mathbb{S}^{n-1} \,, \tag{5} \\[10pt] & \pmb{\operatorname{P}} \,=\, \pmb{\mathbb{I}}_n – {1 \over ||\pmb{v}_P||^2 } \, \pmb{v}_P \, ( \pmb{v}_P)^{\operatorname{T}} \,:=\, \pmb{\mathbb{I}}_n – \, \pmb{e}_{\pmb{v}_P} \, ( \pmb{e}_{\pmb{v}_P} )^{\operatorname{T}} \,, \tag{6} \\[14pt] & \pmb{\operatorname{P}} \, \pmb{x_e} \in S_P \subset \mathbb{R}^n \,, \quad \text{with} \,\,\,\, S_P \perp \pmb{v}_P \,. \tag{7} \end{align} \]

The unit vectors e_vp are aligned v_P. We could have started with an arbitrary v, but picking a vector v_P of the ellipsoid to define the direction of the projection is somewhat more natural. Choosing v_P ∈ E has a nice side effect of giving us an important factor in a rather simple form; see below.

Cutting the unit sphere

Vectors s to points in the (n-1)-dim subspace S_v orthogonal to v have the property:

\[ S_{\pmb{v}} \,=\, \left\{ \pmb{s} \,\, | \,\, \pmb{s} \bullet \pmb{v} \,= 0\, \right\} \,. \tag{8} \]

Cutting with the unit sphere gives us special vectors u_v, which fulfill

\[ \pmb{u}_{\pmb{v}} \bullet \pmb{v} \,=\, 0 \,\, \land \,\, \pmb{u}_{\pmb{v}} \in S_{\pmb{v}} \,\, \land \,\, \pmb{u}_{\pmb{v}} \in \mathbb{S}^{n-1} \, . \tag{9} \]

Transformation to tangential points on the ellipsoid (with respect to v_P)

We apply A to our special vectors u_v.

\[ \pmb{x}_{\pmb{v}}^t \,=\, \pmb{\operatorname{A}} \, \pmb{u_v} \,\,\, , \quad \pmb{x}_{\pmb{v}}^t \in E \,. \tag{10} \]

The first thing we want to check in comparison to the results of the previous two blocks is that eq. (8) of post II is fulfilled. This eq. states that the tangent vectors in the direction of v_P are perpendicular to vectors orthogonal to the ellipsoidal surface manifold (see post II for details). This is obviously true:

\[ (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \pmb{x}_{\pmb{v}}^t \,=\, \pmb{v}^{\operatorname{T}} \pmb{\operatorname{A}}^{\operatorname{T}} (\pmb{\operatorname{A}}^{\operatorname{T}})^{-1} \, \pmb{\operatorname{A}}^{-1} \pmb{\operatorname{A}} \pmb{u_v} \,=\, \pmb{v}^{\operatorname{T}} \, \pmb{u_v} \,=\, 0\,. \tag{11} \]

As in post II we split the vectors to the tangential points :

\[ \pmb{x}^t_{\pmb{v}} \,=\, \pmb{y}_{\pmb{v}}^t + \lambda \, \pmb{v}_P, \,\,\,\, \text{with} \,\, \pmb{y}_{\pmb{v}}^t \,=\, \pmb{\operatorname{P}} \, \pmb{x}^t_{\pmb{v}} \,\, \land \,\, \pmb{y}_{\pmb{v}}^t \perp \pmb{v}_P \,. \tag{12} \]

λ is specific for x_v^t. Acc. to eq. (11), we have

\[ (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \left( \pmb{y}_P^t + \lambda \, \pmb{v}_P \right) \,=\, 0 \,. \tag{13} \]

After some reordering we find

\[ \lambda_p \,=\, -\, { (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{y}_{\pmb{v}}^t \over (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{v}_P } \,=\, -\, (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{y}_{\pmb{v}}^t\,. \tag{14} \]

The denominator disappears due to the quadratic form condition (1). This condition for vectors of the ellipsoid also means

\[ \left[ \pmb{y}_{\pmb{v}}^t + \lambda \, \pmb{v}_P \right]^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, ( \pmb{y}_{\pmb{v}}^t + \lambda \, \pmb{v}_P ) \,=\, 1 \,. \tag{15} \]

Expanding and using eq. (11) once again gives

\[ \begin{align} \left[ \pmb{y}_{\pmb{v}}^t \right]^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \left( \pmb{y}_{\pmb{v}}^t + \pmb{v}_P \, \lambda \,\right) \,=\, 1 \,, \tag{16} \\[10pt] \left[ \pmb{y}_{\pmb{v}}^t \right]^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \left( \pmb{y}_{\pmb{v}}^t – \pmb{v}_P \, (\pmb{v}_P)^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{y}_{\pmb{v}}^t \, \right) \,=\, 1 \,, \tag{17} \\[10pt] \left[ \pmb{y}_{\pmb{v}}^t \right]^{\operatorname{T}} \, \left[\, \pmb{\operatorname{\Sigma}}^{-1} \, -\, \| (\pmb{v}_P) \|^2 \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{e}_{\pmb{v}_P} \, ( \pmb{e}_{\pmb{v}_P} )^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \right] \pmb{y}_{\pmb{v}}^t \, \,=\, 1 \,. \tag{18} \\[10pt] \end{align} \]

Eq. (18) shows again that the projected vectors of our tangential points on the ellipsoid fulfill a quadratic form condition. Note that the matrix product of the unit vector with its transposed defines a matrix!

Note that the vectors y_v^t give us the points on the surface of the projection image (see post I for the reasoning).

Comparing eq. (18) with eq. (21) in post II gives us the nice relation

\[ \left[ \, ( \pmb{e}_{\pmb{v}_P} )^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{e}_{\pmb{v}_P} \, \right]^{-1} =\, \|\pmb{v}_P\|^2 \,. \tag{19} \]

This relation is sometimes convenient when we need to numerically calculate the elements of the matrix

\[ \pmb{\operatorname{\Sigma}}_P^{-1} \,=\, \pmb{\operatorname{\Sigma}}_{\pmb{v}_P}^{-1} \,=\, \left[\, \pmb{\operatorname{\Sigma}}^{-1} \, -\, \| (\pmb{v}_P) \|^2 \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{e}_{\pmb{v}_P} \, ( \pmb{e}_{\pmb{v}_P} )^{\operatorname{T}} \, \pmb{\operatorname{\Sigma}}^{-1} \, \right] \,. \tag{20} \]

According to eq. (18) this matrix [Σ_P]^-1 obviously determines the (n-2)-dimensional ellipsoidal surface of the projection’s image via a quadratic form condition.

Conclusion

In this post we have used an arbitrary vector of the ellipsoid to define the direction along which we project the ellipsoid to an orthogonal subspace. We again saw that the vectors to the (n-2)-dimensional surface of the projection’s image fulfill the condition for a lower-dimensional ellipsoid. We could derive the related matrix at the center of the respective quadratic form. We could write it down in a somewhat simpler form than in the previous posts.

In the next post, I will show what the matrix [Σ_P]^-1 for the quadratic form for the projection’s ellipsoidal surface has to do with the Schur complement of the matrix Σ^-1 which controlled the original ellipsoid. Stay tuned …