1. 3.II. Homomorphisms 3.II.1. Definition 3.II.2. Range Space and Nullspace

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 π: R 3 → R 2 by is a homomorphism. Proof:

3. 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.

4. h: R 3 → R by is not linear & hence not a homomorphism. since is linear & a homomorphism. is not linear & hence not a homomorphism.

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.

6. Theorem 1.9:A homomorphism is determined by its action on a basis. Let  β 1, …, β 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(β k ) = w k  k Proof: Define h : V →W by Then → h is a homomorphism Let g be another homomorphism s.t. g(β k ) = w k. Then → h is unique

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.

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. It’s 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)

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.

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

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

12. Example 2.5:Projection π: R 3 → R 2 by = Vertical line Example 2.6:Homomorphism h: R 2 → R 1 by = Line with slope  1

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 π: R 3 → R 2  R 3 is like R 2 Vectors add like their shadows.

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.

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

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)

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

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 →

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  β 1, …, β n  is a basis for V then  h(β 1 ), …, h(β n )  is a basis for R (h) Proof: See Hefferon, p.191

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)].