Finding polar set

Question

I am trying to solve this question but am not able to understand how to approach it: What is the polar of an ellipsoid described by the equation:

{$(z_1, . . . , z_d) ∈ R^d: a_1z_1^2 + · · · + a_dz_d^{2} ≤ 1$}, where $a_1, . . . , a_d$ are positive.

I know what generally polar of a set means. Moreover, I am following Algebraic And geometric ideas in the theory of discrete optimization to understand more about the polar of a subset. This link gives a beautiful explanation of how to find polar of various set What are good examples of polar sets in $\mathbb R^2$? I'm trying to approach it the way you solve for a disc but I am still unable to do it. Any help is appreciated

Thank you.

Does this answer your question? Deriving the equation of the polar axis-aligned ellipsoid — Amit Rajaraman, Jun 23 '21 at 04:33
@AmitRajaraman Hey! I actually saw this but I got confused with all the equation and got lost. Actually, I am quite new. So I need a simple approach to understand this. I did start with understanding singleton and line as I have attached the link. But I am not still there — Logo, Jun 23 '21 at 05:24
@AmitRajaraman I personally did not understand why they are taking Cauchy Schwartz inequalities. I see a lot of question-related to polar use it and even taking the norm. — Logo, Jun 23 '21 at 05:27
What is your definition of the polar of a set? The definition given in the question I linked is the one I use. The ellipsoid you have described is the same as in the other question -- $\langle x,e_i\rangle$ is your $z_i$ and $\alpha_i$ is your $1/\sqrt{a_i}$. It might be easier to follow my answer if you first try to figure out why the polar of the unit ball is the unit ball itself. Cauchy-Schwarz inequalities are common because they upper bound $\langle x,y\rangle$. Choosing $x,y$ such that one is in the body, the other is in the polar, and their inner product is $1$ gives you what you want. — Amit Rajaraman, Jun 23 '21 at 05:38
@AmitRajaraman the definition of polar set: $A^\circ = {y \in R^d : \langle x,y \rangle \leq 1 }$. Yup, I'm trying to understand actually why the polar of a unit ball is a unit ball. Like I know it is self-dual. But I am trying to dig deep about a polar of a centered ball of any radius — Logo, Jun 23 '21 at 05:49
@AmitRajaraman One more doubt if you don't mind. why did you take $v^\circ=\frac{v}{|v|}$? — Logo, Jun 23 '21 at 05:53
I thought it might be possible to solve the problem for the ellipsoid using the solution for the unit ball. I want to show that for any point $v$ in the polar, $\lVert v\rVert\leq 1$. I did this by showing that $\langle v , v / \lVert v\rVert \rangle \leq 1$, which is just $\langle v,v^\circ\rangle \leq 1$ (as mentioned at the beginning, $v^\circ$ is the largest vector in the direction of the given point that is in the polar). Try going through my solution when $\alpha_i = 1$ (the unit ball), then seeing how it generalizes to a general ellipsoid on applying a linear transform. — Amit Rajaraman, Jun 23 '21 at 06:04

score 3 · Answer 1 · answered Jun 23 '21 at 15:23

The set $C$ you are considering is

$$ C = \{ (x_1,x_2,\ldots,x_d)\in\mathbb{R}^d \,|\,a_1x_1^2+\cdots + a_dx_d^2 \leq 1\}$$

Then its polar is the set

$$ C^\odot = \{ (x_1,x_2,\ldots,x_d)\in\mathbb{R}^d \,|\,\tfrac{1}{a_1}x_1^2+\cdots + \tfrac{1}{a_d}x_d^2 \leq 1\},$$ which is another ellipsoid. (Note that if $a_1=\cdots=a_d=1$, then you see that $C=C^\odot$ is the unit ball - how beautiful!)

More generally, $$C = \{ x\in\mathbb{R}^d \,|\, \langle x,Qx\rangle \leq 1\} \quad \Rightarrow \quad C^\odot = \{ x\in\mathbb{R}^d \,|\, \langle x,Q^{-1}x\rangle \leq 1\},$$ provided that $Q$ is positive definite.

All this can be found in R.T. Rockafellar's classical Convex Analysis book on page 136.

Finding polar set

1 Answers1