Let $S\subseteq \mathbb R^n$. Its polar is defined by $$S^\circ = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in S\} $$
Now, let's say we are interested in some affine image of $S$ (and its polar). Let $T = AS + b = \{Ax+b:x\in S\}$ be the affine image of interest. Then $$T^\circ = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in T\}$$
Is it possible to nicely characterize $T^\circ$ in terms of $S^\circ$ somehow? I know that $$x\in S \Longleftrightarrow Ax+b\in T$$ which gives me $T^\circ = \{z\in\mathbb R^n:z^Tx\le 1,\forall x\in T\} =\{z\in\mathbb R^n:z^T(Ax+b)\le 1,\forall x\in S\}$. Is this correct so far?
I'm not able to work with this expression further and obtain a nice expression for $T^\circ$ in terms of $S^\circ$. Intuitively I feel that $T^\circ = AS^\circ + b$ could be a good guess, but I'm not sure. Any ideas?
Thanks a lot!
Your derivations are correct. You can take a few more steps if $b = 0$: \begin{align} \{z\in\mathbb R^n:z^T(Ax+b)\le 1,\forall x\in S\} & =\{z\in\mathbb R^n: (A^Tz)^Tx\le 1,\forall x\in S\} \\ & =\{z\in\mathbb R^n: y^Tx\le 1,\forall x\in S, y=A^Tz\} \end{align} which is the inverse image of $S^\circ$ under $A^T$.
For $b\neq 0$ there is no simple transformation. For example, the polar of $[-1,1]$ is $[-1,1]$ while the polar of $[0,2]$ (corresponding to $A=b=1$) is $(-\infty,0.5]$.