Linear Algebra

Topic: Projective Geometry 349Because the projective point T 2 is incident on the projective line OT 1 , anyhomogeneous coordinate vector for T 2 lies in the plane through the origin in R 3that is spanned by homogeneous coordinate vectors of O and T 1 :⎛⎜1⎞ ⎛ ⎞1⎟ ⎜ ⎟Rep B (⃗t 2 ) = a ⎝1⎠ + b ⎝0⎠1 0for some scalars a and b. That is, the homogeneous coordinate vectors ofmembers T 2 of the line OT 1 are of the form on the left below, and the forms forU 2 and V 2 are similar.⎛ ⎞⎛ ⎞⎛ ⎞2 11⎜ ⎟⎜ ⎟⎜ ⎟Rep B (⃗t 2 ) = ⎝t1 ⎠ Rep B (⃗u 2 ) = ⎝u 2 ⎠ Rep B (⃗v 2 ) = ⎝ 1 ⎠11v 2The projective line T 1 U 1 is the image of a plane through the origin in R 3 . Oneway to get its equation is to note that any vector in it is linearly dependent onthe vectors for T 1 and U 1 and so this determinant is zero.1 0 x0 1 y= 0 =⇒ z = 0∣0 0 z∣The equation of the plane in R 3 whose image is the projective line T 2 U 2 is this.t 2 1 x1 u 2 y= 0 =⇒ (1 − u 2 ) · x + (1 − t 2 ) · y + (t 2 u 2 − 1) · z = 0∣ 1 1 z∣Finding the intersection of the two is routine.⎛ ⎞t 2 − 1⎜ ⎟T 1 U 1 ∩ T 2 U 2 = ⎝1 − u 2 ⎠0(This is, of course, the homogeneous coordinate vector of a projective point.)The other two intersections are similar.⎛ ⎞⎛ ⎞1 − t 20⎜ ⎟⎜ ⎟T 1 V 1 ∩ T 2 V 2 = ⎝ 0 ⎠ U 1 V 1 ∩ U 2 V 2 = ⎝u 2 − 1⎠v 2 − 11 − v 2Finish the proof by noting that these projective points are on one projectiveline because the sum of the three homogeneous coordinate vectors is zero.Every projective theorem has a translation to a Euclidean version, althoughthe Euclidean result may be messier to state and prove. Desargue’s theoremillustrates this. In the translation to Euclidean space, we must treat separately