let [tex]\mathbb{R}^2[/tex] be a set containing all possible columns:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\left( \begin{array}{cc} a \\ b \right) [/tex]

where a, b are arbitrary real numbers.

show under scalar multiplication and vector addition [tex]\mathbb{R}^2[/tex] is indeed a vector space over the real number field.

I will check the eight axioms:

[tex]X=\left( \begin{array}{cc} a \\ b \right) [/tex]

[tex]Y=\left( \begin{array}{cc} c \\ d \right) [/tex]

[tex]Z=\left( \begin{array}{cc} e \\ f \right) [/tex]

[tex]X,Y,Z \Epsilon \mathbb{R}^2[/tex]

Vector addition is associative:

X+(Y+Z)=(X+Y)+Z

[tex]\left[ \begin{array}{cc} a+c+e \\ b+d+f \right] [/tex]

[tex]=\left[ \begin{array}{cc} a+c+e \\ b+d+f \right] [/tex]

Vector addition is commutative:

X+Y=Y+X

[tex]\left( \begin{array}{cc} a+c \\ b+d \right) [/tex]

[tex]=\left( \begin{array}{cc} c+a \\ b+d \right) [/tex]

Vector addition has an identity element:

[tex]\Theta=\left( \begin{array}{cc} 0 \\ 0 \right)[/tex]

[tex]\Theta+X=X[/tex]

[tex]\left( \begin{array}{cc} 0 \\ 0 \right) +\left( \begin{array}{cc} a \\ b \right) = \left( \begin{array}{cc} a \\ b \right)[/tex]

Inverse Element:

[tex]X=\left[ \begin{array}{cc} a \\ b \right] [/tex]

[tex]W=\left[ \begin{array}{cc} -a \\ -b \right] [/tex]

X+W=0

[tex]\left[ \begin{array}{cc} a \\ b \right] +\left[ \begin{array}{cc} -a \\ -b \right] = \left[ \begin{array}{cc} 0 \\ 0 \right][/tex]

Distributivity holds for scalar multiplication over vector addition:

[tex]\alpha(X+Y)=\alpha X+\alpha Y [/tex]

[tex]\alpha\letf(\left[ \begin{array}{cc} a \\ b \right] +\left[ \begin{array}{cc} c \\ d \right]\right)=\alpha\letf(\left[ \begin{array}{cc} a+c \\ b+d \right]=\alpha X+\alphaY[/tex]

Distributivity holds for scalar multiplication over field addition:

[tex](\alpha+\beta)X=\alphaX+\betaX[/tex]

[tex](\alpha+\beta)\left[ \begin{array}{cc} a \\ b \right] =\left[ \begin{array}{cc} a(\alpha+\beta) \\ b(\alpha+\beta) \right]=\left[ \begin{array}{cc} a\alpha+a\beta \\ b\alpha+b\beta) \right] =\alphaX+\betaX[/tex]

Scalar multiplication is compatible with multiplication in the field of scalars:

a(bX)=(ab)X=abX

[tex]\alpha\left(\beta \left[ \begin{array}{cc} a \\ b \right] \right) = \alpha\left(\left[ \begin{array}{cc} a\beta \\ b\beta \right] \right) = \left[ \begin{array}{cc} a\alpha\beta \\ b\alpha\beta \right] [/tex]

[tex]\alpha(\beta X)=(\alpha \beta) X= \alpha \beta X[/tex]

Scalar multiplication has an identity element:

[tex]F=\left[ \begin{array}{cc} 1 \\ 1 \right][/tex]

such that FX=X

I dont know if this is what I have to do to show R^2 is a vector space. did I do this correct?

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# Vector space

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