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Vector space question

  1. Sep 21, 2009 #1
    Just wondering. Suppose we some plane, any plane like [tex]S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \} [/tex] where F is either [tex]\mathbb{R}[/tex] or [tex]\mathbb{C}[/tex] . We know that S is a vector space (passes the origin).

    We know that [tex](0,0,0)[/tex] is the additive identity and it should be unique by virtue of the field we're working with. But say we had any arbitrary vector [tex](a,b,c)[/tex]. Wouldn't [tex](-a, -b, -c)[/tex] or [tex](0, 3, -5)[/tex] also be counted as an additive identity as well?
  2. jcsd
  3. Sep 21, 2009 #2
    (-a, -b, -c) would be the inverse under addition.

    the other (0,3,-5), forgot what's it called, I think the kernel if I'm not mistaken
  4. Sep 21, 2009 #3
    But wouldn't they also be the additive identity /zero vector as well, violating the fact zero vectors of vector spaces are unique?
  5. Sep 21, 2009 #4
    Given an identity 0, and a vector v

    v + 0 = v


    v + (0,3,-5) does not equal v

    so it's not an identity, it's a vector that maps to zero, but it's not zero.
  6. Sep 21, 2009 #5
    Doesn't it though for v in S?

    Let [tex]v = (a,b,c) = a + 5b + 3c \in S[/tex]. Then [tex](a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v[/tex]

    So we found an element [tex]w = (0, 3, -5)[/tex] such that [tex]v + w = v[/tex] which by definition, we call w the additive inverse?
  7. Sep 21, 2009 #6
    I was wrong about the kernel thing, because you need to have a map for that.

    You just have a vector space that is constrained to S:

    \ x_1 + 5x_2 + 3x_3 = 0 \

    meaning that the components of every element of S have to satisfy this equation.
  8. Sep 22, 2009 #7
    Not quite sure if I entirely understand. Wouldn't any element in S then be called an additive identity since adding it to any vector v would just simply give us v.?
  9. Sep 22, 2009 #8


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    I think you may be a little confused. Since you are in a 3 dimensional space, each vector contains three components, including the zero vector. So let
    [tex] \mathbf{v} = \left(a,b,c\right) \in S[/tex], and let
    [tex] \mathbf{w} = \left(0,3,-5\right) [/tex], where of course
    [tex] \mathbf{w} \in S [/tex] as well. When we add these, we get
    [tex] \mathbf{v} + \mathbf{w} = \left(a+0, b+3, c-5 \right) [/tex].

    For example, if [tex] \mathbf{v} = \left(1,-2,3\right) [/tex]
    then [tex] \mathbf{v} + \mathbf{w} = \left(1, 1, -2 \right) [/tex].

    Also note that if [tex] \mathbf{u} = \left(0,-3,5\right) [/tex],
    [tex] \mathbf{v} + \mathbf{u} = \left(0+0, 3-3, -5+5 \right)
    = \left(0,0,0 \right) = \mathbf{0} [/tex]. Note that the
    vector [tex]\mathbf{0} \in S [/tex]
    is not the same as the scalar [tex]0 \in F[/tex].

    I hope that helps,


  10. Sep 22, 2009 #9


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    NO! (a, b, c) is NOT equal to a+ 5b+ 3c and a+ 5b+ 3c is NOT in S. "(a,b,c)= a+ 5b+ 3c doesn't even make sense. One is a member of R3, the other of R1- they can't be equal! And a+ 5b+ 3c is not in S because S only contains things in R3 and a+ 5b+ 3c is not.

    [/quote]Then [tex](a + 5b + 3c) + (0 + 5(3) + 3(-5)) = (a + 5b + 3c) + (0) = a + 5b + 3c = v[/tex]

    So we found an element [tex]w = (0, 3, -5)[/tex] such that [tex]v + w = v[/tex] which by definition, we call w the additive inverse?[/QUOTE]
    IF v= (a, b, c) is in S, then a+ 5b+ 3c= 0. If w= (0, 3, -5) then v+ w= (a, b+3, c- 5) which is NOT equal to (a, b, c). Frankly, it appears that you have no clue what
    [tex]S = \{ (x_1, x_2, x_3) \in F^{3} \ : \ x_1 + 5x_2 + 3x_3 = 0 \} [/tex]
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