conic sectionseigenvalues-eigenvectorslinear algebra
If I have two eigenvalue $\lambda_1$ and $\lambda_2$ and two associated normalized eigenvector $\mathbf e_1$ and $\mathbf e_2$ respectively, and I want to draw ellipse, How can I know which eigenvalue and eigenvector will construct the major axis and which one will be associated with minor axis ?
The ellipse looks like the following :
Let $e_1,e_2$ is the eccentricity of $E_1,E_2$ respectively.
First, we have $$e_1=\frac{\sqrt{a^2-b^2}}{a}.\tag1$$
Let $$E_2\ :\ \frac{x^2}{A^2}+\frac{y^2}{B^2}=1.$$ Here, note that $B\gt A.$
Since we have $$A=a,\ \ \ \sqrt{B^2-A^2}=b\Rightarrow B=\sqrt{a^2+b^2},$$ we have $$e_2=\frac{\sqrt{B^2-A^2}}{B}=\frac{b}{\sqrt{a^2+b^2}}.\tag2$$ Hence, from $(1),(2)$, we have $$\frac{\sqrt{a^2-b^2}}{a}=\frac{b}{\sqrt{a^2+b^2}}\Rightarrow a^2b^2=a^4-b^4$$$$\Rightarrow \left(\frac ba\right)^4+\left(\frac ba\right)^2-1=0\Rightarrow \left(\frac ba\right)^2=\frac{\sqrt 5-1}{2}.$$ Hence, the answer is $$\frac{\sqrt{a^2-b^2}}{a}=\sqrt{1-\left(\frac ba\right)^2}=\sqrt{1-\frac{\sqrt 5-1}{2}}=\sqrt{\frac{3-\sqrt 5}{2}}=\frac{\sqrt 5-1}{2}.$$
Hint:
Without loss of generality, one of the ellipses is the unit circle centered at the origin. (If not, you can translate its center to the origin, counter-rotate to bring its major axis horizontal and rescale non-isotropically by the axis lengths).
Then the problem reduces to checking if an ellipse is wholly contained in the unit circle.
Let the parametric equation of that ellipse be
$$\vec p=\vec p_c+\vec a\cos t+\vec b\sin t.$$ We need to guarantee the inequality
$$\vec p^2\le 1,$$ i.e.
$$\vec p_c^2+\vec a^2\cos^2t+\vec b^2\sin^2t+2\vec p_c\vec a\cos t+2\vec p_c\vec b\sin t\le 1$$ (we have $\vec a\vec b=0$).
The inequality is ensured by finding the maxima of the trigonometric polynomial and showing the they don't exceed $1$.
Unfortunately, this leads to a quartic equation, so that an analytical solution promises to be painful. The problem is very close to that of finding the shortest distance of a point to an ellipse, or to ellipse offsetting.
There is an analytical solution, but it is a little scary.
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$
"deflate" the circle while you "inflate" the ellipse; that means that you reduce the radius of the circle until it reduces to a point, while you compute the corresponding offset curve of the ellipse (https://en.wikipedia.org/wiki/Parallel_curve).
the offset curve has a known implicit equation, given in "Brief Atlas of Offset Curves, Juana Sandra", available online.
Best Answer
The major axis corresponding to the small eigenvalue, while the minor axis corresponding to the big eigenvalue. Here is a simple explanation,
An eclipse can be thought of a section of quadratic form $x^T Ax$, i.e. $x^T Ax=1$. ($A$ must be a postive definite matrix) In 2-dimentional case, $A$ is a 2 by 2 matrix. Now factorize A to eigenvalue and eigonvector.
$$A=\begin{pmatrix}e_1&e_2\end{pmatrix}\begin{pmatrix}\lambda_1&\\&\lambda_2\end{pmatrix}\begin{pmatrix}e_1^T\\e_2^T\end{pmatrix}$$
Now the equation for the ellipse become,
$$x^T\begin{pmatrix}e_1&e_2\end{pmatrix}\begin{pmatrix}\lambda_1&\\&\lambda_2\end{pmatrix}\begin{pmatrix}e_1^T\\e_2^T\end{pmatrix}x=1$$
$$\lambda_1 x^T e_1 e_1^T x+\lambda_2 x^T e_2 e_2^T x=1$$
There comes the familar ellipse equation,
$$\frac{(e_1^T x)^2}{(\frac{1}{\sqrt{\lambda_1}})^2}+\frac{(e_2^T x)^2}{(\frac{1}{\sqrt{\lambda_2}})^2}=1$$
Assuming $\lambda_1$ is smaller, from the equation, we can see that eigonvector $e_1$ and $e_2$ are corresponding to the major and minor axis direction, eigenvalue $\frac{1}{\sqrt{\lambda_1}}$ and $\frac{1}{\sqrt{\lambda_2}}$ are corresponding to the length of major and minor axis.