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.
We are given the determinant of a linear transformation - this has nothing to do with a column vector. In fact, if you were to write down the matrix for , it would be .
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 ?
Here's a nice but somewhat technical way to prove this : let be the usual stereographic projection; then is an automorphism of , i.e. a Möbius transformation, which has one or two fixed points; but the fixed points of come in pairs, so the case where has only one fixed point is excluded.
(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.)