Given an ellipsoid equation of the form \begin{equation}\label{eq_1}x'Ax=1\end{equation} where $A\in\mathbb{R}^{n\times n}$ is positive definite and non-diagonal and $x\in\mathbb{R}^n$. So, how can I obtain the projection or shadow of the ellipsoid into a 2D plane?
In these links:
part of this question was answered, but I need a general way to find an expression of the ellipse in the 2D plane for a given $A$.
Let $f(\mathbf{x})=\mathbf{x'}A\mathbf{x}-1$. The projection you seek into the $(x1,x2)$ plane is the set of points (see Christian Blatter's answer here) where the gradient $\nabla f$ has no $x_3,...,x_n$ components. Since $\nabla f = 2 A x$, this means you have n-2 linear equations relating $x_1, ..., x_n$. For example, in the 4 dimensional case, you will have 2 equations $$a_{13}x_1+ a_{23}x_2 + a_{33}x_3 + a_{43}x_4 = 0$$ and $$a_{14}x_1+ a_{24}x_2 + a_{34}x_3 + a_{44}x_4 = 0$$ This lets you solve for $x_3,...,x_n$ in terms of $x_1$ and $x_2$.
For the more general n-dimensional case, partition the matrix A into submatrices in this way: $$ \mathbf{A}= \left[\begin{array}{r|r} J & L' \\ \hline L & K \\ \end{array} \right]$$
where $J$ is a $2\times 2$ submatrix, $L$ is $(n-2) \times 2$, and $K$ is $(n-2) \times (n-2)$. Let $\mathbf{y}$ be the vector $(x_1,x_2)$ and let $\mathbf{z}$ be the vector $(x_3,...,x_n)$ The conditions I mentioned from the gradient would be $$ L\mathbf{y}+K\mathbf{z} =0$$. Solve this along with the equation of the original ellipsoid $$ \left( \begin{array}{r|r} \mathbf{y'} | \mathbf{z'} \\ \end{array}\right) \left[\begin{array}{r|r} J & L' \\ \hline L & K \\ \end{array} \right] \left( \begin{array}{r} \mathbf{y} \\ \hline \mathbf{z} \\ \end{array}\right) = 1$$
This gives the equation of the curve you seek as $$ \mathbf{y'} \left( J - L'K^{-1}L\right)\mathbf{y}= 1$$
Please see Jean Marie's answer to my question (specific 3-d case of the multi-dimensional case in your question) to see that $J - L'K^{-1}L$, a Schur complement, is also positive definite. This shows that the projection is an ellipse.
J is n-1 x n-1 sized and K is always 1 x 1 sized for the general case? Projection to nD plane excludes the only dimension, I sure. Projection of nD n-axis rotated ellipsoid on nD plane should be (n-1)D ellipsoid (which is ellipse in 2D). The case which is named "general n-dimensional case" is projection to 2D subspace, I think, not to the plane in the original nD space.
– Tomilov Anatoliy
Oct 30 '24 at 11:37
This is the same result as Mathemagical, but with a geometrical argument. The last step is a bit shaky so apologies.
As a warm up, how do we find the projection of a circle on the x-axis? The equation of a circle is given by $$x^2+y^2=1.$$ To find the projection, we scale the y-coordinate by a large number $\lambda$, i.e. $y\mapsto \lambda y$, and we take the limit. So we get \begin{align} x^2+\lambda^2y^2&=1 \implies x^2=1-\frac{1}{\lambda^2y^2} \end{align} The coordinates are in the ranges $x\in[-1,1]$ and $y\in[-\tfrac1\lambda,\tfrac1\lambda]$. By taking the limit, $y$ will now be $y\in\{0\}$, and the projection will be $x\in[-1,1]$.
Now let's look at the ellipse. I will do a projection on the xy-plane, but this should be generalizable to any plane. It is natural to look at it from the perspective of the inverse $\Sigma=A^{-1}$. This gives $$x'\Sigma^{-1}x=1.$$ Let us define a scaling matrix $$S_\lambda=\pmatrix{1&0&0\\0&1&0\\0&0&\lambda}.$$ The scaled version of the ellipsoid is written as $$x'S_\lambda \Sigma^{-1} S_\lambda x=1.$$ By taking the scaling matrix inside the inverse, we get $$x'(S_{1/\lambda}\Sigma S_{1/\lambda})^{-1}x=1,$$ where I used $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$. This is where my argument becomes a bit shaky, but in the limit that $\lambda\rightarrow\infty$, the scaling matrix becomes a projection matrix that makes the z-component zero and that makes inverse is undefined. But we know that in the end $z$ will be zero, so I will justify that to use a pseudo-inverse and just ignore the z-coordinate. So the equation will be \begin{align} \lim_{\lambda\rightarrow\infty}x'(S_{1/\lambda}\Sigma S_{1/\lambda})^+x&=1\\ x'_{xy}(P_z\Sigma P_z)^+x_{xy}&=1 \end{align} where $P_z$ projects out z, $x_{xy}=(x,y)$ and $(\cdot)^+$ is the pseudo-inverse. In words, calculate the inverse of $A$. Then select the xy-block, i.e. project out the z component. Finally, calculate the pseudo-inverse of this matrix. This is equivalent to simply calculating the inverse of the non-zero block. The process I described is equivalent to calculating the Schur complement of $A$ in the xy block, and for more details I will refer to Mathemagical's answer.
