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

Section 4.2 to 4.4• Consider R 2• Section 4.4: ( 1 , 0 ), ( 0 , 1 ), 2• Mean 2 things:— (1) Section 4.2: For any vector ( a , b ) in R 2 , it couldbe represented () as a linear combination () in terms of () this basis.( a , b ) = a ( 1, 0 ) + b ( 0 , 1 )— (2) Section 4.3: Basis is linearly independent ()c ( 1, 0 ) + d ( 0 , 1 ) = 0 c = d = 0— That is, we could represent every vector withminimal () spanning vectors.• Applications: After Section 4.5LA 04_13Example 2Express the vector (4, 5, 5) as () a linear combination of thevectors (1, 2, 3), (1, 1, 4), and (3, 3, 2).SolutionExamine the following identity () for values of c 1 , c 2 , c 3 .c1(1, 2, 3) c2( 1 , 1, 4) c3(3, 3, 2) (4, 5, 5)LA 04_15 c1 c2 3 c3 4 2 c1 c2 3 c3 5 c1 2 r 3, c2 r 1, c3 3 c 4 c 2 c 5123 r ( 4, 5, 5) ( 2 r 3)(1, 2, 3) ( r 1)( 1,1, 4) r (3,3,2)• r = 3 gives (4, 5, 5) = 3 (1, 2, 3) + 2 (1,1,4) + 3 (3,3,2)• r = 1 gives (4, 5, 5) = 5 (1, 2, 3) 2 (1,1,4) (3,3,2)4.2 Linear Combinations ()• Definition: Let v 1 , v 2 , …, v m be vectors in a vector space V.The vector v in V is a linear combination of v 1 , v 2 , …, v mif there exist scalars c 1 , c 2 , …, c m such thatv = c 1 v 1 + c 2 v 2 + … + c m v mLA 04_14• Example: (5, 4, 2) = (1, 2, 0) + 2 (3, 1, 4) – 2 (1, 0, 3)• Example: (3, 4, 2) = a ( 1, 0, 0) + b ( 0, 1, 0) is impossible(), so the vector (3, 4, 2) is not a linear combinationof ( 1, 0, 0) and ( 0, 1, 0).LA 04_16DefinitionThe vectors v 1 , v 2 , …, v m are said to span () a vector space ifevery vector in the space can be expressed as a linearcombination of these vectors.Example 3Show that the vectors (1, 2, 0), (0, 1, 1), and (1, 1, 2) span R 3 .Proof: Let (x, y, z) be an arbitrary element of R 3 . Determinewhether we can writeThus( x , y , z ) c1(1, 2, 0) c2(0,1, 1) c32cc11cc222ccc333xyz(1,1,2)