Chapter 4 General Vector Spaces Face Recognition ( á¹¢å± å½å¬)

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4.7 Kernel (), Range (), and theRank/Nullity () TheoremLA 04_77 Definition: Let U and V be vector spaces. Let u and v bevectors in U and let c be a scalar. A transformation T: U V is said to be linear if T(u + v) = T(u) + T(v), T(cu) = c T(u) Example 2: Let P n be the vector space of real polynomialfunctions of degree n. Prove TP 2 P 1 is linearT(ax 2 + bx + c) = (a + b) x + c— Proof: Let ax 2 + bx + c, px 2 + qx + r P 2 .— (1) <strong>Vector</strong> addition: T((ax 2 + bx + c) + (px 2 + qx + r)) =(a + p + b + q) x + (c + r) = (a + b) x+c + (p + q) x + r= T (ax 2 + bx + c) + T (px 2 + qx + r)—(2) Scalar multiplication: Assume k is any scalar.T ( k ( a x 2 + b x + c ) ) = ( k a + k b ) x + k c= k ( (a + b) x + c ) = k T ( a x 2 + b x + c )LA 04_79Definition: Let T: U V be a linear transformation.•The kernel () of T = { u U | T u = 0 V }. Denotedker(T). ()•The range () of T : { T u : u U }.Denoted range(T).Example 3 D D = d D P ndxDxDn(4x3nxn3x122x1) f g P nc D ( f g ) Df D ( cf ) cD ( f• Others: Integration (), Mean value () of a randomvariable ()Ef[f)12Dg gdx fdx gdx , cfdx c g ] E [ f ] E [ g ], E [ cf ] cEx26x[fdxf2]Theorem 4.26 : Let T: U V be a linear transformation.(a) The kernel of T is a subspace of U.(b) The range of T is a subspace of V.Proof:(a) If u 1 , u 2 ker(T), that is T u 1 = T u 2 = 0, then(1) T ( u 1 + u 2 ) = T u 1 + T u 2 = 0 + 0 = 0,(2) T ( c u 1 ) = c T u 1 = c 0 = 0(b) If v 1 , v 2 Range (T), that is, there exist u 1 , u 2 such thatT u 1 = v 1 , T u 2 = v 2 . Then(1) v 1 + v 2 = T u 1 + T u 2 = T (u 1 + u 2 )(2) c v 1 = c T u 1 = T ( c u 1 )LA 04_78LA 04_80
Example 4: T ( x, y, z ) = ( x, y, 0 )LA 04_81 Kernel:— T (x, y, z) = (x, y, 0) (),T (x, y, z) = (0, 0, 0) () x = 0, y = 0— Thus, ker(T) = {(0, 0 , z)},all vectors on the z axis.— Check T ( 0, 0, z ) = ( 0, 0, 0 ) Range(T) = {(x, y, 0)},all vectors that lie in the xy plane.LA 04_83.Example 5: 1 2 3 A 0 1 1 A is a 3 3 matrix. Thus A defines a 1 1 4 linear operator T: R3 R 3 . T(x) = A x• Kernel: All vectors x = (x 1 , x 2 , x 3 ) in R 3 such that T(x) = 0. Thus 1 2 3 x 10 5 r 0 1 1 x2 0 X r 1 1 4 x 0 r 3ker(T) is a one-dimensional subspace of R 3 with basis (5, 1, 1) t .• Range: range spanned by the column vectors of A, or row vectorof A t .10 1 10 1 10 1 10 1 A t 2 1 1 0 1 1 01 1 01 1 3 1 4 0 1 1 0 1 1 0 0 0 • The basis vectors (1, 0, 1) t and (0, 1, 1) t span range( T ). Thus therange of T is Range(T) = {a(1, 0, 1) t + b(0, 1, 1) t = (a, b, a+ b) t }Theorem 4.27The range of T: R n R m , defined by T(u) = Au, is spanned by thecolumn vectors of A. (The dim of the column space is the rank.)LA 04_82Proof: Let v be a vector in the range u such that T(u) = v.Express u in terms of the standard basis u = a 1 e 1 + … + a n e n• Thus v = T(a 1 e 1 + … + a n e n ) = a 1 T(e 1 ) + … + a n T(e n )where 1 a11a12 a1 n1 a110 a21a22 a2 n0 a21T ( e1) A , T ( en) A 0 a a a 0 a• The column vectors of A, namely, T(e 1 ), …, T(e n ), span the rangeof A.m1m2mnm1001aaa12nnmnTheorem 4.28Let T: U V be a linear transformation. Thendim ker(T) + dim range(T) = dim domain(T)•dim ker(T) is called the nullity (). Another way to saythis rank/nullity theorem: nullity(T) + rank(T) = dim domain(T)Example:A100010001000LA 04_84Solution•Matrix A is in reduced echelon form.• The column vectors are linear independent. Thus rank(A) = 3 dim range(T) = 3.• The domain of T is R 4 ; dim domain(T) = 4.• Therefore, dim ker(T) = 4 3 = 1 (Basis {(0, 0, 0, 1) T })
Page 2 and 3: Vector Spaces of Matrices (1/2)•
Page 4 and 5: Section 4.2 to 4.4• Consider R 2
Page 6 and 7: Example 7Show that the function h (
Page 8 and 9: Theorem 4.8Let {v 1 , …, v n } be
Page 10 and 11: Rank• Theorem 4.13: The row space
Page 12 and 13: 4.6 Orthonormal Vectors and Project
Page 14 and 15: ExampleLA 04_53 v 1 = (1 0 0), v 2
Page 16 and 17: Example 6Find the distance of the p
Page 18 and 19: Example 2 (1/2)x x2 x 3 y yy639 12
Page 22 and 23: 4.9 Transformations and Systems of
Page 24: Demo Tacoma BridgeLA 04_93

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Chapter 4 General Vector Spaces Face Recognition ( á¹¢å± å½å¬)