前提：先要搞清楚 fields和prime fields

Vector space Definition

Vector space V over Fields F:

V={...........}, the elements of the V is called Vectors

1. Addition Composition Law： must obey the axioms of Abelian Group

+	<-	V	->
<-
V
->

Note: additive identity 0 here represents zero vector in the vector space, while 0 also can represents additive identity in Field in below, don't confuse each other

2. Scalar multiplication Law:

*	<-	V	->
<-
F
->

1. closure

∀ c ∈ F, v ∈ V, cv ∈ V

2. associativity

∀ c₁,c₂ ∈ F, v ∈ V, c₁(c₂v) = (c₁c₂)v

3. identity

1 ∈ F, ∀ v ∈ V, 1v = v1=v

4. distributivity

i) ∀ c ∈ F, v₁,v₂ ∈ V, c(v₁+v₂) = cv₁+cv₂

ii) ∀ c₁,c₂ ∈ F, v ∈ V, (c₁+c₂)v = c₁v+c₂v

let c₂=0, (c₁+0)v = c₁v+0v => c₁v=c₁v+0v, c₁v is vector in vector space, both sides apply additive inverse of vector space,

=>(-c₁v)+c₁v=(-c₁v)+c₁v+0v => 0 = 0v, note the 0 in the left side represents zero vector because two vectors (-c₁v)+c₁v add together results in another vector

V=F

The Field is a vector space over itself

看看是否满足前面的定义：

1. Addition Composition Law：

显然 obey the axioms of Abelian Group，因为V=F，Field的加法结构就是Abelian Group

2. Scalar multiplication Law:

都很显然

1. closure

∀ c ∈ F, v ∈ V=F, cv ∈ V=F

2. associativity

∀ c₁,c₂ ∈ F, v ∈ V=F, c₁(c₂v) = (c₁c₂)v

3. identity

1 ∈ F, ∀ v ∈ V=F, 1v = v1=v

4. distributivity

i) ∀ c ∈ F, v₁,v₂ ∈ V, c(v₁+v₂) = cv₁+cv₂ 显然满足Field的distributivity

ii) ∀ c₁,c₂ ∈ F, v ∈ V, (c₁+c₂)v = c₁v+c₂v，这个可以先利用Field multiplicative communitive, 然后再用前面的distributivity

(c₁+c₂)v = v(c₁+c₂) = vc₁+vc₂ = c₁v+c₂v

V=Fⁿ

Fⁿ, n ∈ N={1,2,3.........}

take the arragnements of the little n elements of the field, so the set is going to underlie this vector space Fⁿ is going to be all n tuples of elements from the filed capital F, denote:

{ (x1,x2,........xn) | xi ∈ F }

Addition

define addition:

v=(x1,x2,........xn) ∈ Fⁿ

v¯=(x1¯,x2¯,........xn¯) ∈ Fⁿ

v+v¯=(x1+x1¯,x2+x2¯,........xn+xn¯) ∈ Fⁿ

1. closure, prove by definition of addition

2. ∀ v1,v2,v3 ∈ Fⁿ

(v1+v2)+v3 = v1+(v2+v3)

v1=(x1,x2,........xn)

v2=(y1,y2,........yn)

v3=(z1,z2,........zn)

(v1+v2)+v3 = ( (x1+y1)+z1, (x2+y2)+z2...........,(xn+yn)+zn )

v1+(v2+v3) = ( (x1+(y1+z1), .................)

3. identity = (0,0............0)

4. additive inverse

v=(x1,x2,........xn)

-v = (-x1,-x2,........-xn)

v+(-v) =(x1+(-x1), x2+(-x2),...........) = (0,0............0)

5. communitative

v=(x1,x2,........xn) ∈ Fⁿ

v¯=(x1¯,x2¯,........xn¯) ∈ Fⁿ

v+v¯=(x1+x1¯,x2+x2¯,........xn+xn¯) ∈ Fⁿ

v¯+v=(x1¯+x1,x2¯+x2,........xn¯+xn) ∈ Fⁿ

v+v¯=v¯+v

scalar Multiplication

define scalar multiplication

c ∈ F, v ∈ Fⁿ,

cv=c(x1,x2.......xn) = ( cx1, cx2,......... cxn)

1. closure

∀ c ∈ F, v ∈ V=Fⁿ, cv ∈ V=Fⁿ

cv= ( cx1, cx2,......... cxn), c, x1,x2...xn∈ F， when we multiple two elements in the field we got another element in the field, so cx1, cx2..........cxn are all elements in the field F, by the definition of Fⁿ: { (x1,x2,........xn) | xi ∈ F },

cv= ( cx1, cx2,......... cxn) ∈ V=Fⁿ

2. associativity

∀ c₁,c₂ ∈ F, v ∈ V=Fⁿ, c₁(c₂v) = (c₁c₂)v

c₁(c₂v) = c₁(c₂x1, c₂x2,......... c₂xn) = (c₁(c₂x1), c₁(c₂x2),......... c₁(c₂xn))

(c₁c₂)v = ( (c₁c₂)x1, (c₁c₂)x2........... (c₁c₂)xn)

同样的by the definition of Fⁿ: { (x1,x2,........xn) | xi ∈ F }, 并且c₁,c₂ ∈ F，所以

c₁(c₂x1)= (c₁c₂)x1。。。。。。。。。。。c₁(c₂xn)= (c₁c₂)xn

3. identity

1 ∈ F, ∀ v ∈ V=Fⁿ, 1v = v1=v

4. distributivity

i) ∀ c ∈ F, v₁,v₂ ∈ V, c(v₁+v₂) = cv₁+cv₂ 证明方法同上，利用Field的distributivity

ii) ∀ c₁,c₂ ∈ F, v ∈ V, (c₁+c₂)v = c₁v+c₂v，这个可以先利用Field multiplicative communitive, 然后再用前面的distributivity

(c₁+c₂)v = v(c₁+c₂) = vc₁+vc₂ = c₁v+c₂v

V=R³

R³={(x1,x2,x3)|xi ∈ R}

the intuitive way of visualizing the vector space is that you can view each one of these as being an arrow basically in three-dimentionaly space, basically you can view each of these vectors as a position vecor where the x coordinate is x1, y coordinate is x2, z coordinate is x3,

what is so wonderful about this is not only does it give us a picture for each vector but also gives us a picture of what it means to add two vectors together and scalar multiply vector