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.,
h a b ah bhu v u v , & ,a b Vu vR
Example 1.2: Projection map
x
x
y
y
z
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 by
z z az bz
1 2
1 2
0
ax bx
a y by
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
3. Example 1.3:
by1 2 3:f P P 2 2 3
0 1 2 0 1 2
1 1
2 3
a a x a x a x a x a x
by2 2 2:f M R
a b
a 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 → by
is linear & a homomorphism.
4. 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 h3 1 4
5 2
x
x y
y
x y
z
is linear & a homomorphism.
5 2
x
x y
y
x y
z
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. f f fv u v u and f a a fv v , &V au v R
2. k k k k
k k
cf c fv v &j jV cv R
Example 1.8:
g: 2 → 4 by
/ 2
0
3
x
x
y x y
y
is a homomorphism.
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:
k k
k
h chv βDefine h : V →W by
Then
k k k k
k k
a b a c b dh hv u β β k k k
k
h ac bd β
k k k k
k
ah c bh dβ βk k k
k
ac bd h β
ah bhv u → h is a homomorphism
Let g be another homomorphism s.t. g(βk ) = wk . Then
k k
k
c g β
k k
k
c h β k k
k
c w
k k
k
c wk k
k
g cgv β h v → h is unique
7. Example 1.10
1 1
0 1
h
0 4
1 4
h
specifies a homomorphism h: 2 → 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: 2 → 2
0
x x
y
is a linear transformation.
8. Example 1.14: Derivative Map d /dx: n → n
1
0 1
n n
k k
k k
k 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 2 2.
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)
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 : → 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.
2
x
x y
y
2. Consider this transformation of 2.
/2
/3
x x
y y
What is the image under this map of this ellipse.
2 2
1
4 9
x x y
y
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 ,
1 2 1 2a h b h h a b h Ss 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
11. Example 2.3: d/dx: 3 → 3
2
, ,
d
a bx cx a b c
dx
R R Rank d/dx = 3
Example 2.4: Homomorphism
2 3
2
a b
a b d cx cx
c d
h: 2 2 → 3 by
2 3
,h r sx sx r sR R Rank h = 2
Homomorphism: Many-to one map
h: V → W
Inverse image
1
h W V hw v v w
12. Example 2.5: Projection π: 3 → 2 by
x
x
y
y
z
1
x
x
y z
y
z
R = Vertical line
Example 2.6: Homomorphism h: 2 → 1 by
x
x y
y
1
x
h w x y w
y
= 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
1-1 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 2v v v v
Vectors add like their shadows.
14. Example 2.8: Homomorphism h: 2 → 1 by
x
x y
y
Example 2.9: Homomorphism h: 3 → 2 by
x
x
y
x
z
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 kernel of a linear map h: V → W is the inverse image of 0W
(h) = h 1(0W) = { v V | h(v) = 0W }
dim N (h) = nullity
16. Example 2.12: d/dx: 3 → 3 by
d
a a
dx
N R
Example 2.13: h: 2 2 → 3 by
2 3
2
a b
a b d cx cx
c d
,1
0
2
a b
h a b
a b
N R
2 3
2 0a b d cx cx →
2
2 3 0b cx d x → 0b c d
2 3 2
2 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)
17. Example 2.15: Homomorphism h: 3 → 4 by
0
0
x
x
y
y
z
0
,
0
a
h a b
b
R R
0
0h z
z
N R
0
0
0
x
x y
y
0 →
Rank h = 2 Nullity h = 1
Example 2.16: t: → by x 4x
(t) = Rank t = 1
(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.
k k V
k
c v 0 with some ck 0
→ k k V
k
h c hv 0
k k W
k
c w 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: 2 → 3 by
0
x
x
y
y
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
20. Exercises 3.II.2
1. For the homomorphism h: 3 → 3 given by
2 3 3
0 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
2
x
f x y
y
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)].
