Linear Algebra

348 Chapter Four. DeterminantsOT 1U 1V 1T 2U 2V 2Consider the pairs of corresponding sides: the sides T 1 U 1 and T 2 U 2 , the sidesT 1 V 1 and T 2 V 2 , and the sides U 1 V 1 and U 2 V 2 . Desargue’s Theorem is thatwhen we extend the three pairs of corresponding sides to full lines, they intersect(shown here as the point TU, the point TV, and the point UV), and further,those three intersection points are collinear.TVT UUVWe will prove this theorem, using projective geometry. (We’ve drawn theseas Euclidean figures because it is the more familiar image. To consider themas projective figures, we can imagine that, although the line segments shownare parts of great circles and so are curved, the model has such a large radiuscompared to the size of the figures that the sides appear in this sketch to bestraight.)For this proof we need a preliminary lemma [Coxeter]: if W, X, Y, Z are fourpoints in the projective plane (no three of which are collinear) then there arehomogeneous coordinate vectors ⃗w, ⃗x, ⃗y, and ⃗z for the projective points, and abasis B for R 3 , satisfying this.⎛ ⎞⎛ ⎞⎛10⎜ ⎟⎜ ⎟⎜0⎞⎛⎟⎜1⎞⎟Rep B (⃗w) = ⎝0⎠ Rep B (⃗x) = ⎝1⎠ Rep B (⃗y) = ⎝0⎠ Rep B (⃗z) = ⎝1⎠0011For the proof, because W, X, and Y are not on the same projective line, anyhomogeneous coordinate vectors ⃗w 0 , ⃗x 0 , and ⃗y 0 do not line on the same planethrough the origin in R 3 and so form a spanning set for R 3 . Thus any homogeneouscoordinate vector for Z is a combination ⃗z 0 = a · ⃗w 0 + b · ⃗x 0 + c · ⃗y 0 .Then, we can take ⃗w = a · ⃗w 0 , ⃗x = b ·⃗x 0 , ⃗y = c · ⃗y 0 , and ⃗z = ⃗z 0 , where the basisis B = 〈⃗w,⃗x, ⃗y〉.Now, to prove Desargue’s Theorem use the lemma to fix homogeneouscoordinate vectors and a basis.⎛⎜1⎞⎛ ⎞⎛ ⎞⎛00⎟⎜ ⎟⎜ ⎟⎜1⎞⎟Rep B (⃗t 1 ) = ⎝0⎠ Rep B (⃗u 1 ) = ⎝1⎠ Rep B (⃗v 1 ) = ⎝0⎠ Rep B (⃗o) = ⎝1⎠0011