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Vector space

  1. Sep 11, 2006 #1
    let [tex]\mathbb{R}^2[/tex] be a set containing all possible columns:

    [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?
     
    Last edited: Sep 11, 2006
  2. jcsd
  3. Sep 11, 2006 #2
    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)
     
    Last edited: Sep 11, 2006
  4. Sep 12, 2006 #3
    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.
     
  5. Sep 13, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

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