The first three posts in this series showed that the orthogonal projection of a (n-1)-dimensional ellipsoid from a n-dimensional Euclidean space to a (n-1)-dimensional subspace has a surface which is a (n-2)-dimensional ellipsoid. In this fourth post we will extend our insights to projections down onto a p-dimensional sub-spaces with 1 ≤ p < n. The sub-space has a (n-p)-dimensional orthogonal complement with respect to the original n-dimensional space. Studying the effect of the projection requires some preparation – and a partitioning of the original quadratic form matrix into blocks. In the end, we will find out that the matrix which defines the quadratic form of the border ellipsoid of the projection image is given by a Schur complement of the original ellipsoid’s matrix. I will first show this for a projection onto (n-1)-dimensional sub-space. Afterwards we will generalize the proof to lower dimensional sub-spaces.
Previous posts
Quadratic form
We have an ellipsoid in the ℝn. It is defined as a closed (n-1)-dim hypersurface E
\[ \text{vectors } \pmb{x}_e \text{ of (} n\text{-1)-dim ellipsoid } E\,: \quad \pmb{x}_{e} \in E = \left\{ \pmb{x}_e \,\, | \,\, (\pmb{x}_e)^T \, \pmb{\operatorname{\Sigma}}^{-1} \, \pmb{x}_{e} \,=\, 1 \, \right\} \subset \mathbb{R}^n \,. \tag{1} \]
As in previous posts we have named the matrix of the quadratic form Σ-1 to keep in touch with Multivariate Normal Distributions. The matrices Σ and Σ-1 are symmetric, invertible and positive-definite. 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 𝕊n-1 (u ∈ 𝕊n-1) :
\[ ||\pmb{u}|| = \sqrt{ u_1^2 + u_2^1 + … + u_n^2 \, } \,=\, 1 \,, \,\, \, \pmb{u} \in \mathbb{S}^{n-1} \, , \tag{3}
\]
\[ \pmb{u} \in \mathbb{S}^{n-1} \,\Rightarrow \, \pmb{\operatorname{A}} \, \pmb{u} \in E \,. \tag{4} \]
The Cartesian Coordinate System [CCS] is defined by orthogonal unit-vectors ei (1 ≤ i ≤ n). We write down our vectors x and the matrix in sectioned form – one component of vector z along en and the rest in the (n-1)-dim sub-space spanned by ei (1 ≤ i ≤ n-1):
\[ \pmb{x} \,=\, \begin{pmatrix} \pmb{y} \\ z \end{pmatrix} \,, \,\,\, \text{with} \,\,\, \pmb{x} \in \mathbb{R}^n, \,\,\, \pmb{y} \in \mathbb{R}^{n-1}, \,\,\, z \in \mathbb{R} \,. \tag{5} \]
We can always reorganize the order of unit vectors that our chosen unit vector is the last one. Remember that the change of order of unit vectors corresponds to an orthogonal transformation to a new base of the ℝn. So, without loosing generality we choose en to give us the projection’s direction below. To cover the effect upon the matrix we write it down partitioned into blocks:
\[ \pmb{\operatorname{\Sigma}}^{-1} \,\equiv \pmb{\operatorname{Q}} \,=\,
\begin{pmatrix} \pmb{\operatorname{Q}}_{yy} & \pmb{q}_{yz} \\ (\pmb{q}_{yz})^T & q_{zz} \end{pmatrix} \,. \tag{6} \]
The introduction of Q (= Σ-1) is just to make writing simpler as we can omit the {-1}. We see
\[ \pmb{\operatorname{Q}}_{yy} \,\in \, \mathbb{R}^{(n-1)\operatorname{x}\,(n-1)} \,, \,\, \pmb{q}_{yz} \in \mathbb{R}^{(n-1)\operatorname{x}1} \,, \,\, q_{zz} \in \mathbb{R} \,. \tag{7} \]
Orthogonal projection and condition for tangential points
We introduce the projection P along en as a projection matrix
\[ \pmb{\operatorname{P}} \,=\, \pmb{\mathbb{I}}_n \,-\, \pmb{e}_n \, \pmb{e}_n^T \,. \tag{8} \]
What property characterizes the critical points xet,
\[ \pmb{x}_e^t \,=\, \begin{pmatrix} \pmb{y}_e^t \\ z_e^t \end{pmatrix} \,, \tag{9} \]
on the ellipsoid that give us the surface of the projection image? Answer from previous posts: en must be tangential and orthogonal to the normal vector of the ellipsoid’s surface manifold at such a tangential point. As we have seen in previous posts that the normal vector to the ellipsoid’s surface at vector xe is given by Qxe. Thus the condition is:
\[ (\pmb{e}_n)^T \, \pmb{\operatorname{Q}} \,\pmb{x}_e^t \,=\, 0 \,. \tag{10} \]
Furthermore, we know
\[ \begin{pmatrix} \pmb{y}_e^t \\ 0 \end{pmatrix} \,=\, \pmb{\operatorname{P}} \, \pmb{x}_e^t \,. \tag{11} \]
The vectors (yet , 0)T reside in a subspace Vp of the ℝn whose vectors are orthogonal to en:
\[ (\pmb{y}_e, 0) \,, \, (\pmb{y}_e^t, 0) \in S_P\, \quad \text{with} \,\,\, S_P \,=\, \left\{ \pmb{x} \,\, | \,\,\, \pmb{x} \bullet \pmb{e}_n = 0 \,, \,\,\, \pmb{x} \perp \pmb{e}_n \, \right\} \,. \]
This sub-space is a representation of the ℝ(n-1). So far, we learned nothing new compared to previous posts. But now let us draw conclusions regarding the structure of Q. A quick check of the matrix multiplication shows that condition (9) means
\[ \begin{align} (\pmb{q}_{yz})^T \, \pmb{y}_e^t + q_{zz}\,z_e^t \,=\, 0 \,, \tag{11} \\[10pt]
z_e^t \,=\, – {1 \over q_{zz} } \, (\pmb{q}_{yz})^T \, \pmb{y}_e^t \,. \tag{12}
\end{align} \]
Now we do once again the same as in previous posts. We apply condition (1) to our tangential points xet, resolve the matrix multiplication and insert eq. (12) for zet:
\[ \begin{align}
& (\pmb{x}_e^t)^T \, \pmb{\operatorname{Q}} \, (\pmb{x}_e^t) \,=\, 1 \quad \Rightarrow \\[10pt]
& (\pmb{y}_e^t)^T\, \pmb{\operatorname{Q}}_{yy} \, \pmb{y}_e^t \, +\, 2 \, \left[ (\pmb{y}_e^t)^T \, \pmb{q}_{yz}\right] * z_e^t \, +\, q_{zz}\,(z_e^t)^2 \,=\, 1 \,, \tag{13}\\[10pt]
& (\pmb{y}_e^t)^T\, \pmb{\operatorname{Q}}_{yy} \, \pmb{y}_e^t \, -\, {2 \over q_{zz} } \, \left[ (\pmb{y}_e^t)^T \, \pmb{q}_{yz} \right] \, (\pmb{q}_{yz})^T \, \pmb{y}_e^t
\,+\, { q_{zz} \over q_{zz}^2} \,\left( \, -\, (\pmb{q}_{yz})^T \, \pmb{y}_e^t \right)^2 \,, \tag{14} \\[10pt]
& (\pmb{y}_e^t)^T\, \pmb{\operatorname{Q}}_{yy} \, \pmb{y}_e^t \, -\, {2 \over q_{zz} } \, \left[ (\pmb{y}_e^t)^T \, \pmb{q}_{yz} \right] \, (\pmb{q}_{yz})^T \, \pmb{y}_e^t
\,+\, {1\over q_{zz} }\, (\pmb{y}_e^t)^T \, \pmb{q}_{yz} \, (\pmb{q}_{yz})^T \, \pmb{y}_e^t \,=\, 1 \,, \tag{15} \\[10pt]
& \Rightarrow \quad (\pmb{y}_e^t)^T\, \left[ \, \pmb{\operatorname{Q}}_{yy} \,-\, {1 \over q_{zz}} \, \pmb{q}_{yz} \, (\pmb{q}_{yz})^T \, \right] \, \pmb{y}_e^t \,=\, 1\,. \tag{16}
\end{align} \]
We found again a matrix mediating a quadratic form – this time working on the reduced vectors yet.
The matrix of the quadratic form for the surface of the projection image is a Schur complement
Eq. (16) above gives us a certain symmetric reduced ((n-1) x (n-1)) matrix QP,red, which affects the projection image vectors (yet) ∈ SP of the vectors xet (giving the tangential points on the ellipsoid):
\[ \begin{align}
& \small{\text{Matrix of a quadratic form for vectors } \pmb{y}_e \text{ in of (} n\text{-1)-dim } } S_P \,: \\[10pt]
& \pmb{\operatorname{Q}}_{P,red} \,=\, \pmb{\operatorname{Q}}_{yy} \,-\, {1 \over q_{zz}} \, \pmb{q}_{yz} \, (\pmb{q}_{yz})^T \,. \tag{17}
\end{align} \]
It is relatively easy to show that this matrix is positive definite because Q was it. Just regard eigenvalues and the above analysis in an eigen-system. The matrix defined by eq. (17) is just the so called Schur complement of block qzz of matrix Q (= Σ-1).
Now, we could repeat the whole argumentation for our (n-2)-dimensional ellipsoid defined by eq. (16) with a further projection down to an even lower-dimensional sub-space. However, we repeat the whole process for a projection operator that eliminates (n-p) dimensions to see effects regarding the matrix structure in cases of orthogonal projections down to p-dimensional sub-space.
Generalization – orthogonal projection down to a p-dimensional sub-space
We introduce a (nxn)-projection operators Pnp and Pn(n-p) as
\[ \begin{align}
\pmb{\operatorname{P}}^n_p \,&=\,
\begin{pmatrix}
\pmb{\mathbb{I}}_p & \pmb{\mathbb{O}}_{p\operatorname{x} (n-p)} \\ \quad \pmb{\mathbb{O}}_{(n-p)\operatorname{x}p} & \quad \,\, \pmb{\mathbb{O}}_{(n-p)\operatorname{x}(n-p)} \end{pmatrix} \,=\, \pmb{\mathbb{I}}_p \,-\, \sum_{i=p+1}^n \, \pmb{e}_i \, (\pmb{e}_i)^T \,, \tag{18}\\[10pt]
\pmb{\operatorname{P}}^n_{n-p} \,&=\, \pmb{\mathbb{I}}_n \, -\, \pmb{\operatorname{P}}^n_p \,. \tag{19}
\end{align} \]
A matrix 𝕀q is the (qxq)-identiy matrix. The matrices 𝕆mxn are (mxn) zero matrices. Let us look at special sets of vectors, Vp ⊂ ℝn and Vn-p ⊂ ℝn:
\[ \begin{align} & \,\, V_p \, =\, \left\{ \, \pmb{y} \,=\, (y_1, …, y_p, \, 0, …., 0 )^T \,\, \in \mathbb{R}^n\,\, |\,\, \, \text{with} \,\,\,\, y_i = 0 , \,\,\, \text{for} \,\,\, i \gt p \right\} \,, \tag{20} \\[10pt]
& V_{n-p} \;=\, \left\{ \, \pmb{z} \,=\, (0, …, 0, z_{p+1}, …, z_n )^T \in \mathbb{R}^n\,\, |\,\, \,\text{with} \,\,\,\, z_i = 0 , \,\,\, \text{for} \,\,\, i \le p \right\} \,, \tag{21} \\[10pt]
& \, \pmb{y} \perp \pmb{z} \,\, \iff \,\, \pmb{y}^T \pmb{z} \,=\, 0 \,. \tag{22}
\end{align}\]
Vectors y ∈ Vp and z ∈ Vn-p are orthogonal to each other. The sets, Vp and Vn-p, correspond to two orthogonal sub-vector-spaces, SP and (SP)⊥, of the ℝn. Subspace SP is p-dimensional, subspace (SP)⊥ is (n-p)-dimensional. They are complementary regarding the full ℝn :
\[ \mathbb{R}^n \,=\, S_P \,\oplus (S_P)^{\perp} \,. \]
We introduce respective reduced vectors
\[ \begin{align}
\pmb{y}_{e,r} \,&= \, (y_1, ……… , y_p)^T \in S_P \sim \mathbb{R}^p\,, \\[10pt]
\pmb{z}_{e,r} \,&=\, (z_{p+1}, …, z_n)^T \in (S_P)^{\perp} \sim \mathbb{R}^{(n-p)} \,, \\[10pt]
\pmb{x}_e \,&=\, \begin{pmatrix} \pmb{y}_{e,r} \\ \pmb{z}_{e,r} \end{pmatrix} \,. \tag{23}
\end{align} \]
Projection Pnp produces vectors y in Vp. It is therefore natural to write the vectors xe ∈ E as a sum of a vector ye ∈ Vp and a ze ∈ Vn-p :
\[ \pmb{x}_e \,= \, \pmb{y}_e + \pmb{z}_e \,, \quad \text{with} \:\: \pmb{y}_e = \pmb{\operatorname{P}}^n_p \, \pmb{x}_e \,, \,\,\,
\pmb{z}_e = \pmb{\operatorname{P}}^n_{n-p} \, \pmb{x}_e \,. \tag{24} \]
To understand matrix-operations working on the reduced vectors, we partition the (n x n) matrix Q into respective blocks with respect to the orthogonal sub-spaces (SP ⊕ (SP)⊥):
\[ \pmb{\operatorname{\Sigma}}^{-1} \,\equiv \pmb{\operatorname{Q}} \,=\,
\begin{pmatrix} \pmb{\operatorname{Q}}_{yy} & \pmb{Q}_{yz} \\ (\pmb{Q}_{yz})^T & \pmb{Q}_{zz} \end{pmatrix} \,. \tag{25} \]
\[ \pmb{\operatorname{Q}}_{yy} \,\in \, \mathbb{R}^{p\operatorname{x}p} \,, \,\, \pmb{\operatorname{Q}}_{yz} \in \mathbb{R}^{p\operatorname{x}(n-p)} \,, \,\, \pmb{\operatorname{Q}}_{zz} \in \mathbb{R}^{(n-p)\operatorname{x}(n-p)} \,. \tag{26} \]
We use the orthogonality condition for tangent vectors along the relevant ei (p+1 ≤ i ≤ n) spanning Vn-p. You can imagine the resulting condition as a summary of searching for the the relevant points for surfaces of (n-p) subsequent projections, each projection along one of the series ep+1, ep+2, … en, eventually down to VP (or regarding the reduced vectors to SP). Note also: Due to the quadratic form, for any vector xe on the ellipsoid the y-part is maximized if the z-related part is minimized:
\[ \begin{align}
& \left(\pmb{y}_e + \pmb{z}_e \right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e \,=\, 1 \quad \Rightarrow \\[10pt]
&\left(\pmb{y}_e\right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e \,=\, 1 \,-\, \left(\pmb{z}_e\right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e \,. \tag{27}
\end{align} \]
The respective minimization condition reduces the number of relevant tangential points xet for the surface of the projection image:
\[ \left(\pmb{z}_e^t\right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e^t \, =\, 0 \,. \tag{28} \]
At these special points the surface’s normal vector Qxet (see previous posts) is perpendicular to (SP)⊥. (SP)⊥ thereby becomes a generalized tangential space at these points. This is the reason why we still can speak of an orthogonal projection down to SP.
Note that condition (28) actually means the respective original vectors uet of our xet must be vertical to all back-transformed A-1ej (p+1 ≤ j ≤ n)
\[ \begin{align}
& \pmb{u}_e^t \,=\, \pmb{\operatorname{A}}^{-1} \pmb{x}_e^t \, \\[10pt]
& \pmb{u}_e^t \,\perp\, \pmb{\operatorname{A}}^{-1} \pmb{e}_j \,, \quad \forall \,j \,\,\, \text{with} \,\,\, p+1 \,\le \, j \, \le\, n \,. \tag{29}
\end{align} \]
This can be used to find the relevant vectors uet of the unit sphere 𝕊n-1 from which we can construct the tangential points xet with the help of matrix A. In one of the next post we will use this insight for a numerical experiment.
Let us return to the partitioned matrix Q. Expanding condition (28) and checking for relevant sub-matrices gives us:
\[ \begin{align}
& \left(\pmb{z}_e^t\right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e^t \,= \left(\pmb{z}_e^t\right)^T \pmb{\operatorname{Q}} \, ( \pmb{y}_e^t + \pmb{z}_e^t) \,= \\[10pt]
& \left(\pmb{z}_{e,r}^t\right)^T [\pmb{\operatorname{Q}}_{yz}] ^T \, \pmb{y}_{e,r}^t \,+\, \left(\pmb{z}_{e,r}^t\right)^T \pmb{\operatorname{Q}}_{zz} \, \pmb{z}_{e,r}^t \,= \\[10pt]
& \left(\pmb{z}_{e,r}^t\right)^T \left( \, [ \pmb{\operatorname{Q}}_{yz}]^T \, \pmb{y}_{e,r}^t \,+\, \pmb{\operatorname{Q}}_{zz} \, \pmb{z}_e^t \, \right) \,= \, 0 \quad \Rightarrow \\[10pt]
& \pmb{z}_{e,r}^t \,=\, – \, [ \pmb{\operatorname{Q}}_{zz} ]^{-1} \, \left(\, [ \pmb{\operatorname{Q}}_{yz}]^T \, \, \pmb{y}_{e,r}^t\,\right) \,. \tag{30}
\end{align} \]
Now we evaluate the quadratic form for our tangential points on the ellipsoid, which give us the surface points of the projection image,
\[ \begin{align}
& (\pmb{x}_e^t)^T \pmb{\operatorname{Q}} \, \pmb{x}_e^t\,=\, 1 \,, \tag{31} \\[10pt]
& \left( (\pmb{y}_{e,r}^t)^T,\, (\pmb{z}_{e,r}^t)^T \right) \, \begin{pmatrix} \pmb{\operatorname{Q}}_{yy} & \pmb{Q}_{yz} \\ (\pmb{Q}_{yz})^T & \pmb{Q}_{zz} \end{pmatrix} \,\begin{pmatrix} \pmb{y}_{e,r}^t \\ \pmb{z}_{e,r}^t \end{pmatrix} \,=\, 1\,. \tag{32}
\end{align} \]
Due to our solution of condition (28), the right side of eq. (27) reduces to 1 and eq.(28) can be simplified for our tangent vectors:
\[ \left(\pmb{y}_e^t \right)^T \pmb{\operatorname{Q}} \, \pmb{x}_e^t \,=\, 1 \,. \tag{33} \]
This leads to
\[ \left( (\pmb{y}_{e,r}^t)^T,\, (\pmb{\mathbb{O}}_{n-p,1})^T \right) \, \begin{pmatrix} \pmb{\operatorname{Q}}_{yy} & \pmb{Q}_{yz} \\ (\pmb{Q}_{yz})^T & \pmb{Q}_{zz} \end{pmatrix} \,\begin{pmatrix} \pmb{y}_{e,r}^t \\ \pmb{z}_{e,r}^t \end{pmatrix} \,=\, 1\,. \tag{34} \]
The zero sub-vector on the left side leads to the following expressions on the left side to
\[ ( \pmb{y}_{e,r}^t )^T \pmb{\operatorname{Q}}_{yy} \, \pmb{y}_{e,r}^t \,+\, ( \pmb{y}_{e,r}^t )^T \pmb{\operatorname{Q}}_{yz} \, \pmb{z}_{e,r}^t \,=\, 1 \,. \tag{35} \]
We insert our result of eq. (30) to get
\[ \begin{align}
(\pmb{y}_{e,r}^t )^T \pmb{\operatorname{Q}}_{yy} \, \pmb{y}_{e,r}^t \,-\, (\pmb{y}_{e,r}^t )^T \pmb{\operatorname{Q}}_{yz} \pmb{\operatorname{Q}}_{zz}^{-1} \pmb{\operatorname{Q}}_{yz}^T \, \pmb{y}_{e,r}^t \,=\, 1 \,, \\[10pt]
(\pmb{y}_{e,r}^t )^T \left[ \pmb{\operatorname{Q}}_{yy} \,-\, \pmb{\operatorname{Q}}_{yz} \pmb{\operatorname{Q}}_{zz}^{-1} \pmb{\operatorname{Q}}_{yz}^T \right] \, \pmb{y}_{e,r}^t \,=\, 1 \,. \tag{36}
\end{align} \]
Again, we find a reduced matrix QP,red which provides a quadratic form for the reduced vectors in sub-space SP (corresponding to the target space of projection Pnp) :
\[ \pmb{\operatorname{Q}}_{P,red} \,:=\, \pmb{\operatorname{Q}}_{yy} \,-\, \pmb{\operatorname{Q}}_{yz} \pmb{\operatorname{Q}}_{zz}^{-1} \pmb{\operatorname{Q}}_{yz}^T \,\, \in \,\, \mathbb{R}^{p\,x\, p} \,. \tag{37} \]
\[ (\pmb{y}_{e,r}^t)^T \, \pmb{\operatorname{Q}}_{P,red} \, \pmb{y}_{e,r}^t \,=\, 1 \,. \tag{38} \]
As we projected the “tangential” vectors, this quadratic form controls the (p-1)-dimensional surface of the projection image of the original ellipsoid. Note that as soon as matrix QP,red is given, it picks the right vectors of sub-space SP which represent the ellipsoidal hull of the projection image. This means that we can safely omit the “t” and just refer to vectors yp ∈ SP :
\[ (\pmb{y}_p)^T \, \pmb{\operatorname{Q}}_{P,red} \, \pmb{y}_p \,=\, 1 \,, \quad \text{for} \,\,\, \pmb{y}_p \,\in\, S_P \,. \tag{39} \]
The matrix QP,red actually is the so called Schur complement of the original matrix Q with respect to block Qzz (based on a block partitioning with respect to the complementary sub-spaces SP and (SP)⊥). Let us write our Schur complement of Q as “Q/Qzz“. Then QP,red defines a (p-1)-dimensional ellipsoid ESp and we can summarize
Summary
\[ \begin{align}
& \text{For } \,\, E = \left\{ \pmb{x}_e \,\, | \,\, (\pmb{x}_e)^T \, \pmb{\operatorname{Q}} \, \pmb{x}_{e} \,=\, 1 \, \right\} \subset \mathbb{R}^n \, \\[10pt]
& \text{ and compl. orthogonal sub-spaces } S_P\,,\, (S_P)^{\perp} \,: \,\, \mathbb{R}^n \,=\, S_P \,\oplus (S_P)^{\perp}, \,\,\, S_P \sim \mathbb{R}^p \,, \\[10pt]
& \text{with } \, \pmb{\operatorname{Q}} \,=\,
\begin{pmatrix} \pmb{\operatorname{Q}}_{yy} & \pmb{Q}_{yz} \\ (\pmb{Q}_{yz})^T & \pmb{Q}_{zz} \end{pmatrix} \,, \\[10pt]
& \quad \quad \: \pmb{\operatorname{Q}}_{yy} \,\in \, \mathbb{R}^{p\operatorname{x}p} \,, \,\, \pmb{\operatorname{Q}}_{yz} \in \mathbb{R}^{p\operatorname{x}(n-p)} \,, \,\, \pmb{\operatorname{Q}}_{zz} \in \mathbb{R}^{(n-p)\operatorname{x}(n-p)} \, \\[10pt]
& \Rightarrow \\[10pt]
& \text{(}p\text{-1)-dim ellipsoid } E_{Sp} \text{ resulting from orth. projection of } \, E \subset \mathbb{R}^n \,\, \text{onto } S_P \,\, : \\[10pt]
& E_{Sp} \:=\, \left\{ \pmb{y}_{p} \,\, | \,\, (\pmb{y}_{p})^T \, \pmb{\operatorname{Q}}_{P,red} \, \pmb{y}_{p} \,=\, 1 \,\, \land \,\, \pmb{y}_{p} \in S_P \,\, \land \,\, \pmb{\operatorname{Q}}_{P,red} \,=\, \pmb{\operatorname{Q}}/\pmb{\operatorname{Q}}_{zz} \, \right\} \subset S_P \,. \tag{40}
\end{align} \]
Conclusion
An orthogonal projection of an (n-1)-dimensional ellipsoid in the ℝn down to a p-dimensional sub-space results in a p-dimensional volume figure. The (p-1)-dimensional surface of this figure is an ellipsoid, too. The matrix controlling the quadratic form in the projection’s target sub-space is a Schur complement of the original matrix Q (= Σ-1). The Schur complement relates to matrix blocks which reflect complementary orthogonal p– and (n-p)-dimensional sub-spaces of the ℝn.
In the next post
we are going to use this result to find out, how the respective inverse matrices Q-1= Σ and [QP,red]-1=ΣP,red are related. We shall see that this relation is rather simple. Stay tuned …