typed by Charles Qin June 2006 1 How Mathematicians Study

(ii) idRn ∈ G, recall idRnT = T = T idRn for any T ∈ G (iii) If T ∈ G, then T is bijective and T−1 ∈ G Proof. (i) and (ii) are fairly easy exercises. For (iii), we have that T ∈ G is surjective by definition, so only need to check one to one. For x,y ∈ R, Tx = Ty =⇒ 0 = ‖Tx−Ty‖ = ‖x−y‖ =⇒ x = y. So T is bijective, and T−1 is surjective. T−1 is an isometry since for x,y ∈ R, ‖T−1x− T−1y‖ = ‖TT−1x− TT−1y‖ = ‖x− y‖. Lastly T−1(F ) = T−1(T (F )) = F . So T−1 ∈ G. ∗The following notes were based on Dr Daniel Chan’s MATH3711 lectures in semester 1, 2006