# Showing positive definiteness in the projection of ellipsoid

I would like to show that the projection onto the $$xoy$$ plane of the centered ellipsoid given by the definition $$\mathbf{x'}\mathbf{A}\mathbf{x}=1$$ where we have a positive definite $$\mathbf{A}= \left[\begin{array}{rrr} a & d & e \\ d & b & f \\ e & f & c \\ \end{array} \right]$$ is an ellipse. I know this has been covered in posts such as this, however, the answers there end by showing that the figure we need is the shadow of an ellipsoidal section (by a plane). Then two statements are made that I know from intuition are true, but are not proved, namely.

1. An ellipsoidal section is an ellipse
2. The shadow of an ellipse is an ellipse.

Is there a way instead to show from the definition in 2-d space that it is indeed an ellipse in 2-D, the set of all $$\mathbf{x}=(x,y)$$ satisfying $$\mathbf{(x-v)'}\mathbf{B}\mathbf{(x-v)}=1$$ where $$B$$ is a positive definite $$2 \times 2$$ matrix and $$\mathbf{v} \in \mathbb{R}^2$$ is the center of the ellipse?

Here is what I have so far. Similar to Andrew Hwang's answer and Christian Blatter's answer, I need that the z-component of $$\nabla f$$ should vanish on the curve I seek, where $$f(\mathbf{x})=\mathbf{x'}\mathbf{A}\mathbf{x}-1$$. Solving this along with the original equation of the ellipsoid, I get that the equation of the curve I seek is $$\mathbf{x'}\mathbf{B}\mathbf{x}=1$$ where $$\mathbf{B}= \left[\begin{array}{rr} (a-\frac{e^2}{c}) & (d-\frac{ef}{c}) \\ (d-\frac{ef}{c}) & (b-\frac{f^2}{c}) \\ \end{array} \right]$$ EDITED: The matrix B previously had a mistake in it - the off-diagonal element was stated as $$(\frac{d}{2}-\frac{ef}{c})$$ instead of $$(d-\frac{ef}{c})$$. Now fixed. My question is, how do I show that this $$B$$ is positive definite? The info I have is that from Sylvester's criterion applied to $$A$$, I know

1. $$a > 0$$
2. $$ab - d^2 >0$$
3. $$a(bc-f^2) + d(ef-cd) + e(df-eb) >0$$

I also know that $$b$$ and $$c$$ have to be positive.

Note: As an example of the kind of answer I was trying: I was able to prove for a different problem (that the section of the ellipsoid by a plane $$z=l$$ is an ellipse). Here the curve is given by $$\mathbf{(x-v)'}\mathbf{B}\mathbf{(x-v)}=1-cl^2+\frac{l^2(af^2+be^2)}{ab-d^2}+\frac{efl^2d^3}{(ab-d^2)^2}$$ with $$\mathbf{v}=(\frac{l(fd-eb)}{ab-d^2},\frac{l(ed-af)}{ab-d^2})$$.In this case, $$B$$ was $$\mathbf{B}= \left[\begin{array}{rr} a & d \\ d & b \\ \end{array} \right]$$ and I know that is positive definite by Sylvester's criterion applied to the original $$A$$ matrix.

• You have in fact much more inequalities like $ac-e^2>0$, $bc-f^2>0$, etc... Sep 21, 2017 at 7:18
• @JeanMarie Thanks, but why do you say that? Is it because of the use of Sylvester's criterion for positive semidefinite matrices? In that case, ALL principal minors, and not just the leading principal minors are non-negative. That would give me $ac-e^2 \geq 0$, $bc-f^2 \geq 0$ etc. Many more. Of course, they are also true for a p.d. matrix like A, but I would end up being able to only prove positive semidefiniteness for B. But we know that the shadow is an ellipse, i.e . B is actually p.d. Sep 21, 2017 at 8:50
• And indeed, only two quantities need to be shown as positive to show B is p.d.. One is $ac-e^2$. And we already know that is $\geq 0$. The other (call it G) is $G=(ac-e^2)(bc-f^2)-\frac{1}{4}(cd-2ef)^2$. Intuition says that is is so. Not only are the terms $(ac-e^2)$ and $bc-f^2$ positive, they are also "big" relative to $(cd-2ef)^2$ because A is diagonally dominant ($a$, $b$ and $c$ are big relative to the other entries in A). So G looks like it should be positive, but I am waving hands all over the place now. Sep 21, 2017 at 9:17

First of all, a graphical illustration (Matlab program given below). Answer to the question "Why is $B$ an spd (symmetric positive definite) matrix ?"

It is because $\mathbf{B}$ can be written, in a very natural way, as a "Schur complement" (http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.AppP.d/IFEM.AppP.pdf) with respect to matrix $\mathbf{A}$.

Let us partition matrix $\mathbf{A}$ into the following way:

$$\tag{1}\mathbf{A}= \left(\begin{array}{rr|r} a & d & e \\ d & b & f \\ \hline e & f & c \\ \end{array} \right)$$

yielding the following Schur complement:

$$\tag{2}\begin{pmatrix}a&d\\d&b\end{pmatrix}-\tfrac{1}{c}\begin{pmatrix}e\\f\end{pmatrix}\begin{pmatrix}e&f\end{pmatrix}= \left(\begin{array}{rr} a-\frac{e^2}{c} & d-\frac{ef}{c} \\ d-\frac{ef}{c} & b-\frac{f^2}{c} \\ \end{array} \right)$$

which coincides with the given matrix $\mathbf{B}$.

Now use the fact that a Schur complement of a spd matrix is itself spd (see "Properties" in (https://en.wikipedia.org/wiki/Schur_complement)).

Edit: a complementary view (see (https://stats.stackexchange.com/q/30588/147896)) is to interpret matrix $\mathbf{A}$ as a covariance matrix of a certain multivariate normal random variable $(X,Y,Z) \in \mathbb{R^3}$ and $\mathbf{B}$ (https://stats.stackexchange.com/q/30588/147896) as the covariance matrix associated with marginal distribution of $(X,Y) \in \mathbb{R^2}.$

Matlab program for the "cigar and smoke ring" figure:

clear all;close all;hold on;axis equal;
p=0.05;
[X, Y, Z] = meshgrid(-1:p:1, -1:p:1, -1.5:p:1.5);
r=1;a=3;b=1;c=2;d=2;e=1;f=2;
A=[a b d
b c e
d e f];% caution: the coefficients' order has been changed
isosurface(X, Y, Z, a*X.^2 + 2*b*X.*Y + c*Y.^2 +...
f*Z.^2 + 2*d*X.*Z+ 2*e*Y.*Z, r^2);