3.II. Homomorphisms

3.II.1. Definition3.II.2. Range Space and Nullspace

3.II.1. Definition

Definition 1.1: HomomorphismA function between vector spaces h: V → W that preserves the algebraic structure is a homomorphism or linear map. I.e.,

h a b ah bh u v u v , & ,a b V u vR

Example 1.2: Projection mapx

xy

yz

π: 3 → 2 by is a homomorphism.

Proof: 1 2 1 2

1 2 1 2

1 2 1 2

x x ax bx

a y b y a y b y

z z az bz

1 2

1 2

0

ax bx

a y b y

1 2

1 2

0 0

x x

a y b y

1 2

1 2

1 2

x x

a y b y

z z

Example 1.3:

by1 2 3:f P P 2 2 30 1 2 0 1 2

1 1

2 3a a x a x a x a x a x

by2 2 2:f M R a ba d

c d

Example 1.4: Zero Homomorphism

h: V → W by v 0

Example 1.5: Linear Map

3 2 4.5

x

y x y z

z

g: 3 → byis linear & a homomorphism.

3 2 4.5 1

x

y x y z

z

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

0 1 1

0 0 0

0 0 0

h h

since

0 1

0 0 1 3 1 5

0 0

h h

3 1 4

5 2x

x yy

x yz

is linear & a homomorphism.

5 2x

x yy

x yz

is not linear & hence not a homomorphism.

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.

f f f v u v u and f a a fv v , &V a u v R

2.

k k k kk k

cf c f v v &j jV c v R

Example 1.8:

g: 2 → 4 by

/ 2

0

3

x

x

y x y

y

is a homomorphism.

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:

k kk

h ch v βDefine h : V →W by

Then k k k kk k

a b a c b dh h

v u β β k k kk

h ac bd β

k k k kk

ah c bh d β β k k kk

ac bd h β

ah bh v u → h is a homomorphism

Let g be another homomorphism s.t. g(βk ) = wk . Then

k kk

c g β

k kk

c h β k kk

c w

k kk

c w k kk

g cg v β h v → h is unique

Example 1.10

1 1

0 1h

0 4

1 4h

specifies a homomorphism h: 2 → 2

Definition 1.11: Linear TransformationA 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: 2 → 2

0

x x

y

is a linear transformation.

Example 1.14: Derivative Map d /dx: n → n

1

0 1

n nk k

k kk k

a x k a x

is a linear transformation.

Example 1.15: Transpose Map

a b a c

c d b d

is a linear transformation of 22.

It’s actually an automorphism.

Lemma 1.16: (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 (V,W).

Proof: Straightforward (see Hefferon, p.190)

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 : → 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 : 2 → given by

does not have a graph that is a line. Show that it is a linear function.

2x

x yy

2. Consider this transformation of 2.

/ 2

/ 3

x x

y y

What is the image under this map of this ellipse.

2 2

14 9

x x y

y

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 ,

1 2 1 2a h b h h a b h S s s s s QED

Definition 2.2: Rangespace and Rank

The rangespace of a homomorphism h: V → W is

(h) = h(V ) = { h(v) | v V }

dim[ (h) ] = rank of h

Example 2.3: d/dx: 3 → 3

2 , ,d

a bx cx a b cdx

R R Rank d/dx = 3

Example 2.4: Homomorphism

2 32a b

a b d cx cxc d

h: 22 → 3 by

2 3 ,h r sx sx r s R R Rank h = 2

Homomorphism: Many-to one map

h: V → W

Inverse image

1h W V h w v v w

Example 2.5: Projection π: 3 → 2 by

xx

yy

z

1

xx

y zy

z

R = Vertical line

Example 2.6: Homomorphism h: 2 → 1 byx

x yy

1 xh w x y w

y

= Line with slope 1

Isomorphism i: V n → W n V is the same as W

Homomorphism h: V n → W m V is like W

1-1 onto bijection

f: V → W f (V) W

f 1: f (V) → V

f (V) W f (V) W

f 1: W → V

Example 2.7: Projection π: 3 → 2 3 is like 2

1 2 1 2 v v v v

Vectors add like their shadows.

Example 2.8: Homomorphism h: 2 → 1 byx

x yy

Example 2.9: Homomorphism h: 3 → 2 by

xx

yx

z

Range is diagonal line in x-y plane.Inverse image sets are planes perpendicular to the x-axis.

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 h1(S) is a subspace of V.In particular, h1({0W }) 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 0W

(h) = h1(0W) = { v V | h(v) = 0W }

dim N (h) = nullity

Example 2.12: d/dx: 3 → 3 by

da a

dx

N R

Example 2.13: h: 22 → 3 by 2 32a b

a b d cx cxc d

,10

2

a bh a b

a b

N R

2 32 0a b d cx cx →

22 3 0b cx d x → 0b c d

2 3 22 3a bx cx dx b cx dx

→

2 0a b d c

→

Theorem 2.14:

h: V → W rank(h) + (h) = dim V

Proof: Show V \ is a basis for (see Hefferon p.189)

Example 2.15: Homomorphism h: 3 → 4 by0

0

xx

yy

z

0

,

0

a

h a bb

R R 0

0h z

z

N R0

0

0

x

x yy

0 →

Rank h = 2 Nullity h = 1

Example 2.16: t: → by x 4x

(t) = Rank t = 1

(t) = 0 Nullity t = 0

Corollary 2.17:Let h: V → W be a homomorphism.

rank h dim Vrank h = dim V nullity h = 0 (isomorphism if onto)

Lemma 2.18: Homomorphism preserves Linear DependencyUnder a linear map, the image of a L.D. set is L.D.

Proof: Let h: V → W be a linear map.

k k Vk

c v 0 with some ck 0

→ k k Vk

h c h v 0

k k Wk

c w 0 with some ck 0

Definition 2.19:A linear map that is 1-1 is nonsingular. (1-1 map preserves L.I.)

Example 2.20: Nonsingular h: 2 → 3 by0

xx

yy

gives a correspondence between 2 and the xy-plane inside of 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) (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 (h)

Proof: See Hefferon, p.191

Exercises 3.II.2

1. For the homomorphism h: 3 → 3 given by

2 3 30 1 2 3 0 0 1 2 3h a a x a x a x a a a x a a x

Find the followings:

(a) (h) (b) h 1( 2 x3 ) (c) h 1( 1+ x2 )

2. For the map f : 2 → given by

2x

f x yy

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)].