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

Example 7Show that the function h (x) = 4 x 2 + 3 x – 7 lies in () thespace Span {f, g} generated by f (x) = 2 x 2 – 5 and g(x) = x + 1.SolutionCheck if there exist scalars c 1 and c 2 such thatc 1 (2 x 2 –5) + c 2 ( x + 1) = 4 x 2 + 3 x –7Equating () corresponding coefficients ():x 2 : 2 c 1 = 4x : c 2 = 3x 0 : –5 c 1 + c 2 = –7Unique solution c 1 = 2, c 2 = 3. Thus2 (2 x 2 –5) + 3 ( x + 1) = 4 x 2 + 3 x –7Example 2(a)• Consider f(x) = x 2 + 1, g(x) = 3x –1, h(x) = – 4x + 1 of thevector space P 2 of polynomials of degree 2.• Show that { f, g, h } is linearly independent.Solutionc 1 f + c 2 g + c 3 h = c 1 ( x 2 + 1 ) + c 2 (3 x 1 ) + c 3 (4 x + 1 ) = 0.xxx20: : : c113c cc22c4 c33 000• Unique solutionc 1 = 0, c 2 = 0, c 3 = 0LA 04_21LA 04_234.3 Linear Dependence () andIndependence ()Definition(a) The set of vectors { v 1 , …, v m } in a vector space V issaid to be linearly dependent if there exist scalars c 1 , …,c m , not all zero, such that c 1 v 1 + … + c m v m = 0LA 04_22• Not all zero: (1, 0, …, 0), (1, 1, …, 1), (1, 1, …, 1)(b) The set of vectors { v 1 , …, v m } is linearly independent ifc 1 v 1 + … + c m v m = 0 can only be satisfied () whenc 1 = 0, …, c m = 0.Theorem 4.4 A set consisting of two or more vectors in a vectorspace is linearly dependent if and only if it is possible toexpress one of the vectors as a linear combination of the othervectors.Proof () Let the set { v 1 , v 2 , …, v m } be linearly dependent.Therefore, there exist scalars c 1 , c 2 , …, c m , not all zero, such thatc 1 v 1 + c 2 v 2 + … + c m v m = 0Assume that c 1 0. Then c2 cmv1 v2 c1 c1 vmLA 04_24() Assume that v 1 is a linear combination of v 2 , …, v m exist scalars d 2 , …, d m , such thatv 1 = d 2 v 2 + … + d m v m 1 v 1 + (d 2 ) v 2 + … + ( d m ) v m = 0Linearly dependent because the coefficients NOT all zero