前提：先要搞清楚 fields和prime fields
Vector space V over Fields F:
V={………..}, the elements of the V is called Vectors
| + | <- | 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
| * | <- | 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
The Field is a vector space over itself
看看是否满足前面的定义：
显然 obey the axioms of Abelian Group，因为V=F，Field的加法结构就是Abelian Group
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
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 } |
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
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
| 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