So I am very new to the idea of polar set and trying to solve a few questions to get a hang of it. Below are the two questions that I am trying to solve:
The polar of a ball of radius $r$ with centre at the origin.
My approach is: The polar of $A$ can be defined as: $A^\circ = \{y \in R^d : \langle x,y \rangle \leq 1 \}$ and its also the intersection of all the halfspaces satisfying the above equation. Then intuitively I believe it will be a ball. Even I read in this https://ocw.mit.edu/courses/mathematics/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/lecture-notes/MIT18_409F09_scribe11.pdf that it will be a ball of $1/r$ radius. But it does not show how exactly it is done. I saw a similar question on the stack where they use some standard norm in the equation but I exactly did not get why they use it.
My second question is :
what will be polar of that is a hyperplane passing through the origin.
I do think these are very simple question but since I am new to this that's I am not able to grab the concept properly and I am finding it difficult to solve. Any explanation or help is appreciated.
Note that $A^\circ = \cap_{x \in A}\{y \in R^d | \langle x,y \rangle \leq 1 \} = \cap_{x \in A}\{x\}^\circ$. It should be clear that $A^\circ$ is always convex and closed.
For intuition, note that since $\{x\}^\circ = \{y | \langle x,y \rangle \leq 1 \} $, we see that if $x =0$ then $\{x\}^\circ = \mathbb{R}^d$ otherwise $\{x\}^\circ$ is a half space with normal $x$ and shortest distance from the origin to the boundary is ${1 \over \|x\|}$. It may help to sketch $\{x\}^\circ$ for a few points in the plane.
If $r>0$, then it is straightforward to check that $(r A)^\circ = {1 \over r} A^\circ$. Hence it is sufficient to find the polar of $A = \{ x | \|x\| \le 1 \}$.
Another useful result is that if $L$ is an invertible linear transformation, then $(LA)^\circ = (L^T)^{-1} A^ \circ$. This is straightforward to show from the definition.
There are various approaches one can take here. One simple one is to notice from the previous result that if $A$ is invariant under rotations by orthogonal matrices, then so is $A^\circ$. In particular, $A^\circ$ must be a closed ball and it is easy to check that the radius is one, hence $A^\circ = A$.
For the second question, $A = \{ a \}^\bot = \{ y | \langle y , a \rangle = 0 \}$ for some $a$. In particular, $A$ is a subspace. It is not too hard to check from the definition that if $A$ is a subspace, then $A^\circ = A^\bot$, and in this case we have $A^\circ = (\{ a \}^\bot)^\bot = \operatorname{sp} \{a\} = \{ \lambda a | \lambda \in \mathbb{R} \}$.
Clarification:
Suppose $L$ is an invertible linear transformation. Then \begin{eqnarray} (LA)^\circ &=& \{ y | \langle y , Lx \rangle \le 1, \text{ for all } x \in A \} \\ &=& \{ y | \langle L^Ty , x \rangle \le 1, \text{ for all } x \in A \} \\ &=& \{ (L^T)^{-1} z | \langle z , x \rangle \le 1, \text{ for all } x \in A \} \\ &=& (L^T)^{-1} A^ \circ \end{eqnarray}
Suppose $A$ is a subspace. Suppose $y \in A^\circ$. Then $\langle y , x \rangle \le 1$ for all $x \in A$. Since $A$ is a subspace, $t x \in A$ for all $t$ and so $t \langle y , x \rangle \le 1$ for all $t$. Then only way this can be true is if $\langle y , x \rangle = 0$. Hence $y \in A^\bot$ and so $A^\circ \subset A^\bot$.
For the other way around, it is immediate that if $y \in A^\bot$ then $y \in A^\circ$ (since $0 < 1$).