Verifying Vector Space Properties of $\mathbb{R}^2$

[indigojoker]

let $$\mathbb{R}^2$$ be a set containing all possible columns:

$$\left( \begin{array}{cc} a \\ b \right)$$

where a, b are arbitrary real numbers.

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

I will check the eight axioms:

$$X=\left( \begin{array}{cc} a \\ b \right)$$
$$Y=\left( \begin{array}{cc} c \\ d \right)$$
$$Z=\left( \begin{array}{cc} e \\ f \right)$$

$$X,Y,Z \Epsilon \mathbb{R}^2$$

Vector addition is associative:

X+(Y+Z)=(X+Y)+Z
$$\left[ \begin{array}{cc} a+c+e \\ b+d+f \right]$$
$$=\left[ \begin{array}{cc} a+c+e \\ b+d+f \right]$$

Vector addition is commutative:

X+Y=Y+X
$$\left( \begin{array}{cc} a+c \\ b+d \right)$$
$$=\left( \begin{array}{cc} c+a \\ b+d \right)$$

Vector addition has an identity element:

$$\Theta=\left( \begin{array}{cc} 0 \\ 0 \right)$$

$$\Theta+X=X$$

$$\left( \begin{array}{cc} 0 \\ 0 \right) +\left( \begin{array}{cc} a \\ b \right) = \left( \begin{array}{cc} a \\ b \right)$$

Inverse Element:

$$X=\left[ \begin{array}{cc} a \\ b \right]$$
$$W=\left[ \begin{array}{cc} -a \\ -b \right]$$

X+W=0

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

Distributivity holds for scalar multiplication over vector addition:

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

Distributivity holds for scalar multiplication over field addition:

$$(\alpha+\beta)X=\alphaX+\betaX$$

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

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

a(bX)=(ab)X=abX

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

$$\alpha(\beta X)=(\alpha \beta) X= \alpha \beta X$$

Scalar multiplication has an identity element:

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

such that FX=X


I don't know if this is what I have to do to show R^2 is a vector space. did I do this correct?

 

[fourier jr]

to answer your last question, you said at the beginning you would check the 8 axioms. can you tell whether you did or not? looks ok to me but i think you should learn to check those things yourself. verifying axioms is usually pretty trivial & anyone should be able to do it in their sleep. (after a bit of practice of course)

 

[ircdan]

I think you should say why your statements are true. For example when you check associativity, say why it's true - it's true because of associativity of real numbers.

 

[HallsofIvy]

Scalar multiplication has an identity element:

$$F=\left[ \begin{array}{cc} 1 \\ 1 \end{array}\right]$$


such that FX=X

No, Scalar multiplication involves scalars! The identity is the NUMBER 1.

(By the way, you need a "\end{array}" before "\right]".)