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 | |||

-> |

- closure

∀ c ∈ F, v ∈ V, cv ∈ V

- associativity

∀ c_{1},c_{2} ∈ F, v ∈ V, c_{1}(c_{2}v) = (c_{1}c_{2})v

- identity

1 ∈ F, ∀ v ∈ V, 1v = v1=v

- distributivity

i) ∀ c ∈ F, v_{1},v_{2} ∈ V, c(v_{1}+v_{2}) = cv_{1}+cv_{2}

ii) ∀ c_{1},c_{2} ∈ F, v ∈ V, (c_{1}+c_{2})v = c_{1}v+c_{2}v

let c_{2}=0, (c_{1}+0)v = c_{1}v+0v => c_{1}v=c_{1}v+0v, c_{1}v is vector in vector space, both sides apply additive inverse of vector space,

=>(-c_{1}v)+c_{1}v=(-c_{1}v)+c_{1}v+0v => 0 = 0v, note the 0 in the left side represents zero vector because two vectors (-c_{1}v)+c_{1}v 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

- closure

∀ c ∈ F, v ∈ V=F, cv ∈ V=F

- associativity

∀ c_{1},c_{2} ∈ F, v ∈ V=F, c_{1}(c_{2}v) = (c_{1}c_{2})v

- identity

1 ∈ F, ∀ v ∈ V=F, 1v = v1=v

- distributivity

i) ∀ c ∈ F, v_{1},v_{2} ∈ V, c(v_{1}+v_{2}) = cv_{1}+cv_{2} 显然满足Field的distributivity

ii) ∀ c_{1},c_{2} ∈ F, v ∈ V, (c_{1}+c_{2})v = c_{1}v+c_{2}v，这个可以先利用Field multiplicative communitive, 然后再用前面的distributivity

(c_{1}+c_{2})v = v(c_{1}+c_{2}) = vc_{1}+vc_{2} = c_{1}v+c_{2}v

## # V=F^{n}

F^{n}, 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 F^{n} 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) ∈ F^{n}

v¯=(x1¯,x2¯,........xn¯) ∈ F^{n}

v+v¯=(x1+x1¯,x2+x2¯,........xn+xn¯) ∈ F^{n}

closure, prove by definition of addition

∀ v1,v2,v3 ∈ F

^{n}

(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), .................)

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

additive inverse

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

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

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

- communitative

v=(x1,x2,........xn) ∈ F^{n}

v¯=(x1¯,x2¯,........xn¯) ∈ F^{n}

v+v¯=(x1+x1¯,x2+x2¯,........xn+xn¯) ∈ F^{n}

v¯+v=(x1¯+x1,x2¯+x2,........xn¯+xn) ∈ F^{n}

v+v¯=v¯+v

### # scalar Multiplication

define scalar multiplication

c ∈ F, v ∈ F^{n},

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

- closure

∀ c ∈ F, v ∈ V=F^{n}, cv ∈ V=F^{n}

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 F^{n}: { (x1,x2,........xn) | xi ∈ F },

cv= ( cx1, cx2,......... cxn) ∈ V=F^{n}

- associativity

∀ c_{1},c_{2} ∈ F, v ∈ V=F^{n}, c_{1}(c_{2}v) = (c_{1}c_{2})v

c_{1}(c_{2}v) = c_{1}(c_{2}x1, c_{2}x2,......... c_{2}xn) = (c_{1}(c_{2}x1), c_{1}(c_{2}x2),......... c_{1}(c_{2}xn))

(c_{1}c_{2})v = ( (c_{1}c_{2})x1, (c_{1}c_{2})x2........... (c_{1}c_{2})xn)

同样的by the definition of F^{n}: { (x1,x2,........xn) | xi ∈ F }, 并且c_{1},c_{2} ∈ F，所以

c_{1}(c_{2}x1)= (c_{1}c_{2})x1。。。。。。。。。。。c_{1}(c_{2}xn)= (c_{1}c_{2})xn

- identity

1 ∈ F, ∀ v ∈ V=F^{n}, 1v = v1=v

- distributivity

i) ∀ c ∈ F, v_{1},v_{2} ∈ V, c(v_{1}+v_{2}) = cv_{1}+cv_{2} 证明方法同上，利用Field的distributivity

ii) ∀ c_{1},c_{2} ∈ F, v ∈ V, (c_{1}+c_{2})v = c_{1}v+c_{2}v，这个可以先利用Field multiplicative communitive, 然后再用前面的distributivity

(c_{1}+c_{2})v = v(c_{1}+c_{2}) = vc_{1}+vc_{2} = c_{1}v+c_{2}v

## # V=R^{3}

R^{3}={(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