proof of theorem A1

 Theorem A1 A parallel projection is an element of A(2). Proof Suppose that t is the parallel projection from Π to Π' along v, and that Π and Π' are equipped with axes. We introduce axes for R3 by taking the x- and y-axes of Π', and then the z-axis is the normal to Π', suitably directed. Let f be the transformation which maps P in R3 to the point where the line through P, parallel to v, meets Π'. Thus, f maps R3 to R2. It is easy to see that t is the restriction of f to Π. (1) f is a linear transformation. Since v is not parallel to Π', the xy-plane in R3, its z-component is non-zero. We may as well scale v so that it is (m,n,1). The line through P(a,b,c) parallel to v then has equation r = p + kv = (a+km,b+kn,c+k). This meets Π' where c+k=0, i.e. k=-c, so that f(a,b,c)=(a-cm,b-cn) - the image is on Π'. i.e. R2. Thus f is linear. (2) the algebraic form of t. Let A,B,C be the points on Π whose Π-coordinates are (0,0),(1,0),(0,1), respectively, and let a, u and w be the vectors OA, AB and AC. Then the point P with Π-coordinates (α,β) has position vector p = a + αu + βw. Since f is linear, f(p) = f(a) + αf(u) + βf(w). Since P is on Π, t(p) = f(p) = f(a) + αf(u) + βf(w). Finally, f(a) + αf(u) + βf(w) = A(α,β)T+f(a), where A is the matrix with columns f(u), f(w), so that t is an element of A(2).