let V is a non-empty set which has elements such as a,b,c and so on,All the elements in V are vector. let K is a number field consisted of number such as α , β , σ \alpha,\beta , \sigma α,β,σ and so on.
to define an operation of addition that there is a only sum a + b ∈ V a+b \in V a+b∈V when a , b ∈ V a,b \in V a,b∈V
a + b + c = a + ( b + c ) a+b+c=a+(b+c) a+b+c=a+(b+c)
a + b = b + a a+b=b+a a+b=b+a
the zero element exists ,which make a + 0 = a a+0=a a+0=a
there is a negative element . for every a ∈ V a \in V a∈V the existing of vector b b b result in a + b = 0 a+b=0 a+b=0, b b b is the negative element of a a a, b = − a b=-a b=−a.
the scalar multiplication is defined in set V.for a ∈ V , α ∈ K a \in V,\alpha \in K a∈V,α∈K,the only α a ∈ V \alpha a \in V αa∈V and has the following properties:
α ( a + b ) = α a + α b \alpha(a+b)=\alpha a+ \alpha b α(a+b)=αa+αb
( α + β ) a = α a + β a (\alpha+\beta)a=\alpha a+\beta a (α+β)a=αa+βa
α ( β a ) = ( α β ) a \alpha(\beta a)=(\alpha \beta)a α(βa)=(αβ)a
1 a = a 1 a=a 1a=a the above sistuations indentifies that the V is called as linear space on number field K.
we try to construct a linear space that is V = { ( 2 x , 3 x ) ∣ x ∈ N } V=\{(2x,3x)|x \in \mathbb{N}\} V={(2x,3x)∣x∈N} .
let a , b ∈ V a,b \in V a,b∈V,the sum of a and b , a + b ∈ V a+b \in V a+b∈V ( 2 a , 3 a ) + ( 2 b , 3 b ) = ( 2 ( a + b ) , 3 ( a + b ) ) (2a,3a)+(2b,3b)=(2(a+b),3(a+b)) (2a,3a)+(2b,3b)=(2(a+b),3(a+b))
it is glaringly obvious that a + b = b + a a+b=b+a a+b=b+a .
the zero element exists ,which is ( 0 , 0 ) (0,0) (0,0).
the negative element exists,which is (-2a,-3a) for the element (2a,3a).
the scalar multiplication can be defined as follows. α ∈ R , α ( 2 a , 3 a ) = ( 2 α a , 3 α a ) ∈ V \alpha \in \mathbb{R},\alpha (2a,3a) =(2\alpha a,3\alpha a)\in V α∈R,α(2a,3a)=(2αa,3αa)∈V
α ( ( 2 a , 3 a ) + ( 2 b , 3 b ) ) = α ( 2 a , 3 a ) + α ( 2 b , 3 b ) \alpha((2a,3a) +(2b,3b) )=\alpha(2a,3a)+\alpha(2b,3b) α((2a,3a)+(2b,3b))=α(2a,3a)+α(2b,3b)
( α + β ) ( 2 a , 3 a ) = α ( 2 a , 3 a ) + β ( 2 a , 3 a ) (\alpha+\beta)(2a,3a) =\alpha (2a,3a) +\beta (2a,3a) (α+β)(2a,3a)=α(2a,3a)+β(2a,3a)
α ( β a ) = ( α + β ) a \alpha(\beta a)=(\alpha+\beta)a α(βa)=(α+β)a
