Let be a linear transformation such that . If T is not the identity linear transformation and where .Then prove that T fixes exactly 2 points of S.
I'd like to point out a key fact. A linear transformation has a nonzero fixed point if and only if it has eigenvalue 1. Is it possible to show that this is the case, given that ?
(If we had , then all we could say is that is antiholomorphic, i.e. is conformal; for instance, reflection about a suitable plane passing through the origin in corresponds to , which is not conformal, and which has infinitely many fixed points, all of which lie on a circle.)