Find the projection of the ellipsoid:
$ \mathcal{E}:= \left\{x \in \mathbb{R}^n, y \in \mathbb{R}^m \left| \begin{pmatrix} x^T & y^T \end{pmatrix} \underbrace{\begin{pmatrix} P_1 & P_2 \\ P_2^T & P_3 \end{pmatrix}}_{P} \begin{pmatrix} x \\ y \end{pmatrix}=1 \right. \right\} $ onto the $x$-hyperplane. The matrix $P \in \mathbb{R}^{n+m}$ is positive definite and thus $P_1 \in \mathbb{R}^{n}$ and $P_3 \in \mathbb{R}^{m}$ are nonsingular.
Professor Christian Blatter answered a similar question here: Projection of ellipsoid. Following his approach, I propose the following solution.
Note that the ellipsoid $\mathcal{E}$ can be written as the level set at the origin of the function:
$F(x,y):= \begin{pmatrix} x^T & y^T \end{pmatrix} P \begin{pmatrix} x \\ y \end{pmatrix} -1$,
that is, $ \mathcal{E}= \left\{x \in \mathbb{R}^n, y \in \mathbb{R}^m \left|F(x,y)=0 \right. \right\}$.
Let $\mathcal{E}'$ denote the projection of $\mathcal{E}$ onto the $x$-hyperplane. The points $(x',y') \in \mathcal{E}$ that generate the boundary $\partial \mathcal{E}'$ of $\mathcal{E}'$ are such that the tangent hyperplane at $(x',y')$ is parallel to the $y$-hyperplane. Thus, the normal hyperplane $n_{(x',y')}$ at $(x',y')$ is perpendicular to the $y$-hyperplane or, in other words, the $y$-component of $n_{(x',y')}$ equals the $m$-dimensional zero vector $\mathbf{0}_m$.
The gradient of $F(x,y)$ is given by
$\nabla F(x,y) = 2P\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2P_1x + 2P_2y \\ 2P_2^Tx + 2P_3y \end{pmatrix}$,
and $n_{(x',y')}$ is given by $\nabla F(x',y')$. The $y$-component of $\nabla F(x,y)$ is given by $2P_2^Tx + 2P_3y$. Therefore, the points $(x',y')$ in question lie on the hyperplane $2P_2^Tx + 2P_3y = \mathbf{0}_m$. That is, the points $(x,y) \in \mathcal{E}$ producing the projection boundary $\partial \mathcal{E}'$ are those satisfying $2P_2^Tx + 2P_3y=\mathbf{0}_m$ and $F(x,y) = 0$, i.e., the points $(x,y)$ satisfying:
$F(x,-(P_3)^{-1}P_2^Tx) = x^T\left( P_1 - P_2P_3^{-1}P_2^T \right)x - 1 = 0$.
Hence, the projection $\mathcal{E}'$ is given by
$ \mathcal{E}' = \left\{x \in \mathbb{R}^n \left| x^T\left( P_1 - P_2P_3^{-1}P_2^T \right)x = 1 \right. \right\}. $
Interestingly, the matrix $\left( P_1 - P_2P_3^{-1}P_2^T \right)$ that generates the ellipsoid $\mathcal{E}'$ is the Schur complement of the block $P_3$ of the matrix $P$.
What do you guys think about this approach? Does it seem correct?
Thanks for reading me.
