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 $A$.

  • $\begingroup$ What type of projection do you have in mind? $\endgroup$
    – amd
    Commented Sep 16, 2017 at 0:42
  • $\begingroup$ You already know how to project to the coordinate planes. But the result is true for any plane. Hints: WLOG, pick a plane that passes through the origin. Then, if n is the unit vector perpendiclar to that plane, note that the the projection is $\mathbf{x^p}=\mathbf{x}-\mathbf{n}(\mathbf{n} \cdot \mathbf{x})$ $\endgroup$ Commented Sep 16, 2017 at 3:27
  • $\begingroup$ Thank you for taking your time to answer. I need a projection in any 2D plane. For instance, if x=[x1,x2,x3,x4], I need the projection in the plane x1-x2. $\endgroup$ Commented Sep 18, 2017 at 3:03
  • $\begingroup$ I am not sured if the solution in the links I mentioned above are complete. I do not see how to get the equation of an ellipse from that. I really appreciate all the help you can provide. $\endgroup$ Commented Sep 18, 2017 at 3:12
  • $\begingroup$ @JairoGiraldo Got it. Wrote you an answer. $\endgroup$ Commented Sep 23, 2017 at 13:49

1 Answer 1


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.

  • $\begingroup$ Glad you found it helpful. Please do accept, upvote answers that worked for you. You can do that by clicking the check mark (and up arrow) to the left of the answer. $\endgroup$ Commented Nov 7, 2017 at 23:15

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