Consider the 3 by 3 matrix A = [1 2 4; 3 6 12; 1 5 7] and the linear transformation TA : R3 -> R3 defined by y = TA(x) = Ax where x, y are in the standard coordinates. Let F : C(AT) -> C(A) xr -> Axr be the restriction of TA. We will examine this linear transformation from the row space to the column space. (a) Confirm that rank A = 2, hence dim C(AT) = dim C(A) = 2. (b) Find a basis <v1, v2> for C(AT) and a basis <w1, w2> for C(A). (c) Find the matrix representation B of F : C(AT) -> C(A) with respect to the bases <v1, v2> and <w1, w2> in Part (b). Namely, find the 2 by 2 matrix B that satisfies [F(v1) F(v2)] = [w1 w2] B. (d) Show that B is invertible. (For instance, find rank B.) Think about it: If we write F(xi1v1 + xi2v2) = eta1w1 + eta2w2, then, we have the following relation between coefficient vectors (coordinates for C(A) and C(AT)) [eta1; eta2] = B [xi1; xi2]. This formula tells that the linear transformation F : C(AT) -> C(A) is completely characterized by the matrix representation B. The fact that B is invertible reflects that F : C(AT) -> C(A) is bijective.