If $P$ and $p$ are pole and polar with respect to a polarity with matrix $C$; then $P’$ and $p’$, their images under a collineation, will be pole and polar with respect to the polarity with matrix $C’$.
This statement is from a book I just read from Judith N Cederberg. I was wondering how I can prove it?
Well let’s suppose the collineation has matrix $T$, so $P’=kTP$ for some scalar k and $p’=spT^{-1}$. Since $P$ and $p$ are pole-polar, then it has that $s_1p^t=CP$ for the matrix $C$. I want to show that $s_2(p’)^t=C’P’$. I am not sure how to show it. If $P$ was a self conjugate point, then I should be able to show it. But what if $P$ is an interior point?
Ignoring the scalar factors, and writing $\sim$ for homogeneous equivalence instead, you wrote correctly:
$$P'\sim TP\qquad p'=pT^{-1}\qquad p^t\sim CP\qquad p'^t\sim C'P'$$
Let's try to solve this for $C'$.
$$C'P'\sim p'^t\sim (pT^{-1})^t=(T^{-1})^tp^t\sim (T^{-1})^tCP\sim (T^{-1})^tC(T^{-1})P'$$
So with $C'\sim(T^{-1})^tC(T^{-1})$ you will get the desired behavior. Now you not only know that polarity with respect to some conic is preserved, you get the resulting conic after the transformation. If you want you can check for incidence: if $P$ lies on $C$, i.e. if $P^tCP=0$, then $P'^tC'P'=0$ as well. Easy to see as $T$ and $T^{-1}$ cancel each other. So $C'$ is indeed the conic made up of the images of points on the original conic $C$.
Note that you write collineation but assume the existence of a matrix. This is valid for some projective spaces like the real projective plane, where indeed every collineation is a projective transformation. But if the underlying field has a non-trivial automorphism, like for example the complex numbers have, then there are collineations which are not projective transformations and hence not expressible as a matrix multiplication. Thus I would have preferred the term projective transformation here.