Suppose we have a $N\times N$ symmetric-positive-definite matrix $A$, representing an ellipsoïd in $N$ dimensional space. How to find the matrix $A_{xy}$ corresponding to orthogonal projection of ellispoid on $xy$-plan ?
I have already consulted this pages, they were not that helpful:
How to obtain the equation of the projection/shadow of an ellipsoid into 2D plane? https://math.stackexchange.com/a/1866994/1054066
edit after some suggestions : Schur complement method seems to work in 3 dimensions:
here's an exemple:
suppose $A = \begin{pmatrix} 1 & 0.5 & -1 \\ 0.5 & 2 & 0.5 \\ -1 & 0.5 & 3 \end{pmatrix}$
than Schur complement method says that $A_{xy} = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 2\end{pmatrix} - \frac{1}{3} \begin{pmatrix} -1 \\ 0.5\end{pmatrix} \begin{pmatrix} -1 & 0.5\end{pmatrix} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$
Here's the result :
but I dont see how to make it work in 5 dimansional space. The goal is to project 5D-ellipsoid on xy-plane.
Suppose M is a symmetric $N \times N$ matrix representing an ellipsoid : $$X^T M X = 1$$ with $X \in \mathbb{R}^N$.
1st method of projection : using a projection matrix P $$P_{xy} = \begin{pmatrix} 1 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 0\end{pmatrix} = \begin{pmatrix} \overrightarrow{i} \\ \overrightarrow{j} \end{pmatrix}$$ where $\overrightarrow{i}$ and $\overrightarrow{j}$ cartesian base vectors for $xy$-plane. For exemple if we want to project an object of 7-dimensional to 3-dimensional space (on any 3 arbitrary axis) than the projection matrix will take the form : $$ P_{7D \to 3D} = \begin{pmatrix} 1&0&0&0&0&0&0 \\ 0&0&0&0&1&0&0 \\ 0&0&1&0&0&0&0 \end{pmatrix} = \begin{pmatrix} \overrightarrow{e_1} \\ \overrightarrow{e_5}\\ \overrightarrow{e_3} \end{pmatrix}$$
The projection of M on $xy$-plane is an ellipse given by $M_{xy}$ : $$ M_{xy} = (P_{xy} M^{-1} P_{xy}^T)^{-1} $$
2nd method : using the Schur complement.
First we divide the matrix M into 4 blocs as follows : $$ M = \begin{pmatrix} A & B \\ D & C \end{pmatrix}$$ than Schur's complement gives $M_{xy}$ :
$$ M_{xy} = A - BC^{-1}D $$
here's an exemple :
$$ M = \begin{pmatrix} 1 & 0.5 & -1 \\ 0.5 & 2 & 0.5 \\ -1 & 0.5 & 3 \end{pmatrix}$$
1st method: projection matrix $$P_{xy} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$
$$M_{xy} = (P_{xy} M^{-1} P_{xy}^T)^{-1} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$$
2nd method : Schur complement $$M_{xy} = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 2 \end{pmatrix} - \begin{pmatrix} -1 \\ 0.5 \end{pmatrix} \frac{1}{3} \begin{pmatrix} -1 & 0.5 \end{pmatrix} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$$