hw2sols_143f12 - Math 143 Fall 2012 HW2 Solutions(UAG 1.8...

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Math 143, Fall 2012. HW2 Solutions (UAG) 1.8 Let P 1 , ... , P 4 P 2 k be four points, no three of which are collinear. This means that any three of these points correspond to three lines in k 3 which are linearly independent. Let L 1 , L 2 , L 3 k 3 be the three lines corresponding to P 1 , P 2 , P 3 respectively, and let v i L i be nonzero vectors lying in these lines. Then, ( v 1 , v 2 , v 3 ) defines an (ordered) basis of k 3 so that by a (projective) linear isomorphism we may assume that P 1 = (1 : 0 : 0), P 2 = (0 : 1 : 0), P 3 = (0 : 0 : 1). Let 0 6 = v 4 L 4 , where L 4 k 3 is the line corresponding to the point P 4 P 2 k . Then, (with respect to the above linear automorphism) we may assume that v 4 = av 1 + bv 2 + cv 3 = ( a , b , c ) k 3 , where abc 6 = 0 since no three of our original points are collinear. By considering the linear transformation k 3 defined by the diagonal matrix diag( a - 1 , b - 1 , c - 1 ) we may assume that v 4 = v 1 + v 2 + v 3 = (1, 1, 1). Then, since this diagonal transformation has not changed the lines L 1 , L 2 , L 3
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