Chapter 4 General Vector Spaces Face Recognition ( á¹¢å ± 彐嬀)

Example 2 (1/2)x x2 x 3 y yy639 121113639 1000101.54.57.5 • Slide 59: b Span{column space(A)} No solution• Overdetermined () problem• Try to find a least-squares solution: (Another application)min ( x + y –6) 2 + (– x + y –3) 2 + (2 x + 3 y –9) 2 1 1 1 12 3 xy 639 1 1 1 12 3 xy 639 Geometric Interpretation ()Aa• y Span{column space(A)} No solution Least-squares solution a = pinv(A) y =y 100010 aa12 234 23 is the closest we can get to a true solution Projectiony y230 2 100 3100010010, 2 3 004 234100 0,230 0 04 004 , 010 2 34 0LA 04_69LA 04_71Example 2 (2/2)x x2 x 3 y yy63 A 9 1 1 1 12 3 LA 04_70• rank (A) = 2 full column rank least-squares solution is 6 11 5 17 4 pinv ( A ) Y 32 30 0 12 6 3 9 • (least-squares) Error:(0.536)2(0.533)2(199)27.5• Compare with () other solutions:22— (2, 3): Error (2 3 6) ( 2 3 3)— (0, 3): Error 9 (4 9 9)2 21Example 7• Find the projection matrix for the plane x –2y – z = 0 in R 3 .• Find the projection of the vector (1, 2, 3) onto this plane.LA 04_72Solution: Let W be the subspace of vectors that lie in thisplane W = {(x, y, z)} = {(2y + z, y, z)} = { y (2, 1, 0) +z (1, 0, 1)}, the space generated by the vectors (2, 1, 0)and (1, 0, 1).• Let A be the matrix having these vectors as columns. 2 1 A 1 0 0 1 • (1) Projection matrix (2) Projection of any vector.A pinv(A)16 521222125 16 521222125123202W't