Matrix Theory study notes[3]

linear space

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 , 2 x + 1 ) ∣ x ∈ N } V=\{(2x,2x+1)|x \in \mathbb{N}\} V={(2x,2x+1)∣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 , 2 a + 1 ) + ( 2 b , 2 b + 1 ) = ( 2 ( a + b ) , 2 ( a + b ) + 1 ) (2a,2a+1)+(2b,2b+1)=(2(a+b),2(a+b)+1) (2a,2a+1)+(2b,2b+1)=(2(a+b),2(a+b)+1)
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,-2a-1) for the element (2a,2a+1).
the scalar multiplication can be defined as follows.
α ∈ R , α ( 2 a , 2 a + 1 ) = ( 2 α a , 2 α a + 1 ) ∈ V \alpha \in \mathbb{R},\alpha (2a,2a+1) =(2\alpha a,2\alpha a+1)\in V α∈R,α(2a,2a+1)=(2αa,2αa+1)∈V
α ( ( 2 a , 2 a + 1 ) + ( 2 b , 2 b + 1 ) ) = α ( 2 a , 2 a + 1 ) + α ( 2 b , 2 b + 1 ) \alpha((2a,2a+1) +(2b,2b+1) )=\alpha(2a,2a+1)+\alpha(2b,2b+1) α((2a,2a+1)+(2b,2b+1))=α(2a,2a+1)+α(2b,2b+1)
( α + β ) ( 2 a , 2 a + 1 ) = α ( 2 a , 2 a + 1 ) + β ( 2 a , 2 a + 1 ) (\alpha+\beta)(2a,2a+1) =\alpha (2a,2a+1) +\beta (2a,2a+1) (α+β)(2a,2a+1)=α(2a,2a+1)+β(2a,2a+1)
α ( β a ) = ( α + β ) a \alpha(\beta a)=(\alpha+\beta)a α(βa)=(α+β)a
1 a = a 1a=a 1a=a