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3.II. Homomorphisms 3.II.1. Definition 3.II.2. Range Space and Nullspace
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### Transcript of 3.II. Homomorphisms 3.II.1. Definition 3.II.2. Range Space and Nullspace.

• Slide 1
• 3.II. Homomorphisms 3.II.1. Definition 3.II.2. Range Space and Nullspace
• Slide 2
• 3.II.1. Definition Definition 1.1: Homomorphism A function between vector spaces h: V W that preserves the algebraic structure is a homomorphism or linear map. I.e., Example 1.2:Projection map : R 3 R 2 by is a homomorphism. Proof:
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• Example 1.3: by Example 1.4:Zero Homomorphism h: V W byv 0 Example 1.5:Linear Map g: R 3 R by is linear & a homomorphism.
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• h: R 3 R by is not linear & hence not a homomorphism. since is linear & a homomorphism. is not linear & hence not a homomorphism.
• Slide 5
• Lemma 1.6: A homomorphism sends a zero vector to a zero vector. Lemma 1.7: Each is a necessary and sufficient condition for f : V W to be a homomorphism: 1.and 2. Example 1.8: g: R 2 R 4 byis a homomorphism.
• Slide 6
• Theorem 1.9:A homomorphism is determined by its action on a basis. Let 1, , n be a basis of a vector space V, and w 1, , w n are (perhaps not distinct) elements of a vector space W. Then there exists a unique homomorphism h : V W s.t. h( k ) = w k k Proof: Define h : V W by Then h is a homomorphism Let g be another homomorphism s.t. g( k ) = w k. Then h is unique
• Slide 7
• Example 1.10 specifies a homomorphism h: R 2 R 2 Definition 1.11: Linear Transformation A linear map from a space into itself t : V V is a linear transformation. Remark 1.12: Some authors treat linear transformation as a synonym for homomorphism. Example 1.13:Projection P: R 2 R 2 is a linear transformation.
• Slide 8
• Example 1.14:Derivative Map d /dx: P n P n is a linear transformation. Example 1.15:Transpose Map is a linear transformation of M 2 2. Its actually an automorphism. Lemma 1.16: L (V,W) For vector spaces V and W, the set of linear functions from V to W is itself a vector space, a subspace of the space of all functions from V to W. It is denoted L (V,W). Proof: Straightforward (see Hefferon, p.190)
• Slide 9
• Exercise 3.II.1 1. Stating that a function is linear is different than stating that its graph is a line. (a) The function f 1 : R R given by f 1 (x) = 2x 1 has a graph that is a line. Show that it is not a linear function. (b) The function f 2 : R 2 R given by does not have a graph that is a line. Show that it is a linear function. 2. Consider this transformation of R 2. What is the image under this map of this ellipse.
• Slide 10
• 3.II.2. Rangespace and Nullspace Lemma 2.1: Let h: V W be a homomorphism between vector spaces. Let S be subspace of V. Then h(S) is a subspace of W. So is h(V). Proof: s 1, s 2 V and a, b R, QED Definition 2.2: Rangespace and Rank The rangespace of a homomorphism h: V W is R (h) = h(V ) = { h(v) | v V } dim[ R (h) ] = rank of h
• Slide 11
• Example 2.3:d/dx: P 3 P 3 Rank d/dx = 3 Example 2.4:Homomorphism h: M 2 2 P 3 by Rank h = 2 Homomorphism: Many-to one map h: V W Inverse image
• Slide 12
• Example 2.5:Projection : R 3 R 2 by = Vertical line Example 2.6:Homomorphism h: R 2 R 1 by = Line with slope 1
• Slide 13
• Isomorphism i: V n W n V is the same as W Homomorphism h: V n W m V is like W 1-1ontobijection f: V Wf (V) W f 1 : f (V) V f (V) W f 1 : W V Example 2.7:Projection : R 3 R 2 R 3 is like R 2 Vectors add like their shadows.
• Slide 14
• Example 2.8:Homomorphism h: R 2 R 1 by Example 2.9:Homomorphism h: R 3 R 2 by Range is diagonal line in x-y plane. Inverse image sets are planes perpendicular to the x-axis.
• Slide 15
• A homomorphism separates the domain space into classes. Lemma 2.10: Let h: V W be a homomorphism. If S is a subspace of h(V), then h 1 (S) is a subspace of V. In particular, h 1 ({0 W }) is a subspace of V. Proof:Straightforward (see Hefferon p.188 ) Definition 2.11:Nullspace or Kernel The nullspace or kernel of a linear map h: V W is the inverse image of 0 W N (h) = h 1 (0 W ) = { v V | h(v) = 0 W } dim N (h) = nullity
• Slide 16
• Example 2.12:d/dx: P 3 P 3 by Example 2.13: h: M 2 2 P 3 by Theorem 2.14: h: V W rank(h) + N (h) = dim V Proof: Show B V \ B N is a basis for B R (see Hefferon p.189)
• Slide 17
• Example 2.15:Homomorphism h: R 3 R 4 by Rank h = 2Nullity h = 1 Example 2.16:t: R R byx 4x R (t) = R Rank t = 1 N (t) = 0 Nullity t = 0
• Slide 18
• Corollary 2.17: Let h: V W be a homomorphism. rank h dim V rank h = dim V nullity h = 0(isomorphism if onto) Lemma 2.18:Homomorphism preserves Linear Dependency Under a linear map, the image of a L.D. set is L.D. Proof:Let h: V W be a linear map. with some c k 0
• Slide 19
• Definition 2.19: A linear map that is 1-1 is nonsingular.(1-1 map preserves L.I.) Example 2.20:Nonsingular h: R 2 R 3 by gives a correspondence between R 2 and the xy-plane inside of R 3. Theorem 2.21: In an n-D vector space V, the following are equivalent statements about a linear map h: V W. (1) h is nonsingular, that is, 1-1 (2) h has a linear inverse (3) N (h) = { 0 }, that is, nullity(h) = 0 (4) rank(h) = n (5) if 1, , n is a basis for V then h( 1 ), , h( n ) is a basis for R (h) Proof: See Hefferon, p.191
• Slide 20
• Exercises 3.II.2 1. For the homomorphism h: P 3 P 3 given by Find the followings: (a) N (h)(b) h 1 ( 2 x 3 )(c) h 1 ( 1+ x 2 ) 2. For the map f : R 2 R given by sketch these inverse image sets: f 1 ( 3), f 1 (0), and f 1 (1). 3. Prove that the image of a span equals the span of the images. That is, where h: V W is linear, prove that if S is a subset of V then h([S]) = [h(S)].