4.2 Isomorphisms of Vector Spaces

4.2 Isomorphisms of Vector Spaces

Motivation

P² := {a₀ + a₁x + a₂x²|a_i ∈ R}

l

R³ = {(a₀, a₁, a₂)|a_i ∈ R}

형태는 다르게 보이지만, 구조는 같은 vector space 들을 골라내자.

함수를 나타내는 다양한 표현 방식을 보자.

S T f

a b c

ㄱ ㄴ ㄷ

(I) f (a) = 1, f (b) = 7, f (c) = 4 (II) f : a → ㄱ

b → ㄷ c → ㄴ (III) (a,ㄱ) ∈ f

(b,ㄷ) ∈ f

(c,ㄴ) ∈ f

f = {(a,ㄱ), (b,ㄷ), (c,ㄴ)} ⊂ A× ㄱ

Definition 4.2.1.

A mapping(function, transformation) from S to T is a set f = {(s, t)|s ∈ S, t ∈ T } such that

for s ∈ S, there exists exactly one t ∈ T s.t. (s, t) ∈ f or f (s) = t;

or s₁ = s₂ in S ⇒ f (s₁) = f (s₂) in T let f ⊂ S × T,

say f : well-defined (as a function) if

S₁ = S₂ ⇒ f (S₁) = f (S₂); 함수가 정의되었는지 확인할 때 사용

S T

“exactly one t ∈ T ” means:

(i) f (s) 값은 존재해야 한다.

(ii) f (s) 값은 두개이상이 되면 안된다.

Definition 4.2.2.

f is one-to-one (injective) if for a, b ∈ S, f (a) = f (b) ⇒ a = b or

a 6= b ⇒ f (a) 6= (b)

S T

Definition 4.2.3.

f is onto (surjective) if for every t ∈ T , there is s ∈ S s.t. f (s) = t

S T

Definition 4.2.4.

f is bijective if f is one-to-one and onto

Definition 4.2.5.

f : U → W : function on vector spaces say f has a homomorphy property if

f (au + bv) = af (u) + bf (v) for any a, b : scalar & u, v ∈ U

“f has a homomorphy property ” means

the vector space structure of U is identical to the vector space structure of V (actually the image f (u))

Definition 4.2.6.

let f : U → V : mapping

define f is an isomorphism if 0. f is well-defined 1. f is one-to-one.

2. f is onto.

3. f has homomorphy.

call such U, V are isomorphic

Example 4.2.7. define f : R³ −→ P² as

(a₁, a₂, a₃) 7→ a₃ + (a₂ − a₁)x + (a₁ + a₃)x² show f : isomorphism

(0) show well-defined i.e.

u = v ⇒ f (u) = f (v)

let u, v ∈ R³, u = (a₁, a₂, a₃), v = (b₁, b₂, b₃) assume u = v i.e.

a₁ = b₁, a₂ = b₂, a₃ = b₃, −(?)

then f (u) = a₃ + (a₂ − a₁)x + (a₁ + a₃)x² and f (v) = b₃ + (b₂ − b₁)x + (b₁ + b₃)x²

form (?), a₃ = b₃, a₂ − a₁ = b₂ − b₁, a₁ + a₃ = b₁ + b₃

∴ f (u) = f (v)

(1) show one-to-one i.e.

f (u) = f (v) ⇒ u = v assume f (u) = f (v) i.e.

(??) a₃ = b₃, a₂ − a₁ = b₂ − b₁, a₁ + a₃ = b₁ + b₃ from (??), a₁ = b₁, a₂ = b₂, a₃ = b₃

∴ u = v

hence f : one-to-one

(2) show f : onto, i.e.

∀w ∈ P₂, ∃ u ∈ R³ s.t. f (u) = w let w ∈ P₂, write w = α₁ + α₂x + α₃x² let u = (a₁, a₂, a₃) ∈ R³

want f (u) = a₃ + (a₂ − a₁)x + (a₁ + a₃)x²

= α₁ + α₂x + α₃x²

a₃ = α₁, a₂ − a₁ = α₂, a₁ + a₃ = α₃

⇒ a₃ = α₁, a₁ = α₃ − α₁, a₂ = α₁ + α₂ − α₃ let u = (−α₁ + α₃, α₁ + α₂ − α₃, α₁)

then f (u) = w

∴ f : onto.

(3) show f : homomorphy, i.e.

f (au + bv) = af (u) + bf (v) let u = (a₁, a₂, a₃), v = (b₁, b₂, b₃)

f (au + bv) = f (aa₁ + bb₁, aa₂ + bb₂, aa₃ + bb₃)

= (aa₃ + bb₃) + (aa₂ + bb₂ − aa₁ − bb₁)x + (aa₁ + bb₁ + aa₃ + bb₃)x²

= aa₃ + (aa₂ − aa₁)x + (aa₁ + aa₃)x² + bb₃ + (bb₂ − bb₁)x + (bb₁ + bb₃)x²

= a(a + (a − a )x + (a + a )x²) + b(b + (b − b )x + (b + b )x²)

= af (a₁, a₂, a₃) + bf (b₁, b₂, b₃)

= af (u) + bf (v)

∴ f : homomorphy therefore f : isomorphism

Example 4.2.8. define h : R −→ P2 as (a₁, a₂, a₃) 7→ a₃ + (a₃ − a₁)x + a₃x² not isomorphism.

because, not one-to-one, h(0, 0, 0) = h(0, 1, 0) = 0

(성립하지 않는경우 구체적인 반례제시; counterexample)

Theorem 4.2.9.

V : n − dim vector space over F Fⁿ = {(a₁, · · · , a_n)|a_i ∈ F}

then V ≈ Fⁿ; isomorphic (structure 상 일치한다.)

idea of proof. let B = {v¹, · · · , v_n} basis for V for u ∈ V , can write uniquely

u = a₁v₁ + a₂v₂ + · · · + a_nv_n

where a_i ∈ F, aⁱ : unique define f : V → Fⁿ

u 7→ (a₁, a₂, · · · , a_n)

(0) well-defined?

let u = P a_iv_i, v = P b_iv_i ∈ V assume u = v

then u − v = 0 = P(a_i − b_i)v_i

since {v_i} : linearly independent, a_i − b_i = 0, ∀ i

∴ aⁱ = b_i, ∀ i

∴ (a¹, a₂, · · · , a_n) = (b₁, b₂, · · · , b_n)

⇒ f (u) = f (v)

(1) one-to-one?

f (u) = f (v) ⇒ a_i = b_i, ∀

⇒ P a_iv_i = P b_iv_i

⇒ u = v

(2) onto?

let a = (a₁, · · · , a_n) ∈ Fⁿ

choose u = a₁v₁ + a₂v₂ + · · · + a_n then f (u) = (a₁, · · · , a_n) = a

(3) homomorphic?

f (au + bv) = f (aP a_iv_i + b P b_iv_i)

= f (P(aa_i + bb_i)v_i)

= (a · a₁ + b · b₁, · · · , a · a_n + b · b_n)

= a(a₁, · · · , a_n) + b(b₁, · · · , b_n)

= af (u) + bf (v)

∴ homorphism

therefore f : isomorphism and V ≈ Fⁿ