3.II. Homomorphisms

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 homomorphismor linear map. I.e., Example 1.2: Projection map is a homomorphism. by π: R3 → R2 Proof:

3. Example 1.3: by by Example 1.4: Zero Homomorphism h: V → W by v 0 Example 1.5:Linear Map by g: R3 → R is linear & a homomorphism.

4. by is not linear & hence not a homomorphism. h: R3 → R 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: by is a homomorphism. g: R2 → R4

6. Theorem 1.9: A homomorphism is determined by its action on a basis. Let  β1 , … , βn be a basis of a vector space V , and w1 , …, wn are (perhaps not distinct) elements of a vector space W . Then there exists a unique homomorphism h : V →W s.t. h(βk ) = wk k Proof: Define h : V →W by Then → h is a homomorphism Let g be another homomorphism s.t. g(βk ) = wk . Then → h is unique

7. Example 1.10 specifies a homomorphism h: R2 → R2 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:R2 → R2 is a linear transformation.

8. Example 1.14: Derivative Map d /dx: Pn→ Pn is a linear transformation. Example 1.15: Transpose Map is a linear transformation of M2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 f1 : R →R given by f1(x) = 2x 1 has a graph that is a line. Show that it is not a linear function. (b) The function f2 : R2 →R given by does not have a graph that is a line. Show that it is a linear function. 2. Consider this transformation of R2. 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: s1 , s2 V and a, b  R, QED Definition 2.2: Rangespace and Rank The rangespaceof 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: P3 → P3 Rank d/dx = 3 Example 2.4: Homomorphism h: M22 → P3 by Rank h = 2 Homomorphism: Many-to one map h: V → W Inverse image

12. Example 2.5: Projection π: R3 → R2 by = Vertical line Example 2.6: Homomorphism h: R2 → R1by = 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 Example 2.7: Projection π: R3 → R2 R3 is like R2 Vectors add like their shadows.

14. Example 2.8: Homomorphism h: R2 → R1by Example 2.9: Homomorphism h: R3 → R2by 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({0W}) is a subspace of V. Proof: Straightforward (see Hefferon p.188 ) Definition 2.11: Nullspace or Kernel The nullspace or kernelof a linear map h: V → W is the inverse image of 0W N(h) = h1(0W) = { v V | h(v) = 0W } dim N (h) = nullity

16. Example 2.12: d/dx: P3 → P3 by → → Example 2.13: h: M22 → P3 by → → Theorem 2.14: h: V → W  rank(h) + N(h) = dim V Proof: Show BV \ BN is a basis for BR (see Hefferon p.189)

17. Example 2.15: Homomorphism h: R3 → R4by → Rank h = 2 Nullity h = 1 Example 2.16: t: R → Rby x 4x Rank t = 1 R(t) = R 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 ck 0 → with some ck 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: R2 → R3by gives a correspondence between R2 and the xy-plane inside of R3. 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  β1 , … , βn  is a basis for V then  h(β1 ), … , h(βn ) is a basis for R(h) Proof: See Hefferon, p.191

20. Exercises 3.II.2 1. For the homomorphism h: P3 →P3 given by Find the followings: (a) N(h) (b) h 1( 2  x3 ) (c) h 1( 1+ x2 ) 2. For the map f : R2 →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)].