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Homework Help: Vector space properties: distributivity

  1. Oct 1, 2016 #1

    Math_QED

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    1. The problem statement, all variables and given/known data
    I want to proof, using the axioms of a vector space, that:

    ##(\alpha - \beta)\overrightarrow a = \alpha \overrightarrow a - \beta \overrightarrow a##

    2. Relevant equations

    Definition vector space:

    3. The attempt at a solution

    ##(\alpha - \beta)\overrightarrow a = (\alpha + (-\beta))\overrightarrow a = \alpha \overrightarrow a + (-\beta) \overrightarrow a##

    I'm stuck here, I want to show that:

    ##(-\beta) \overrightarrow a = -(\beta \overrightarrow a)##

    I showed that both have the same symmetric element. But I did not show that symmetric elements are unique. Is there an easier way as this is an exercise and I think I'm overseeing something.

    EDIT: I cannot use ##(-1)\overrightarrow a = - \overrightarrow a##
     
    Last edited: Oct 1, 2016
  2. jcsd
  3. Oct 1, 2016 #2
    Then prove it, so you can use it.
     
  4. Oct 1, 2016 #3

    Math_QED

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    That's funny, because I already proved it but it uses this result.

    Can I derive, from the definition of vector space that:

    ##\overrightarrow a + \overrightarrow b = \overrightarrow a+ \overrightarrow c \Rightarrow \overrightarrow b = \overrightarrow c##. If yes, I can show uniqueness of inverse elements and then it's proven. Is there an easier way?
     
  5. Oct 1, 2016 #4
    Yes, you should be able to derive this easily

    This is the easiest way imo.
     
  6. Oct 1, 2016 #5

    Math_QED

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    I don't see how I can derive it that easily. Can you give me a hint?
     
  7. Oct 1, 2016 #6
    Add the additive inverse of ##a##.
     
  8. Oct 1, 2016 #7

    Math_QED

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    Yeah, thought about that too. Why am I allowed to do that on both sides?
     
  9. Oct 1, 2016 #8
    What's wrong with it?
     
  10. Oct 1, 2016 #9

    Math_QED

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    I would use the property then, which I did not prove:

    a + b = a + c => a + c + d = a + c + d
     
  11. Oct 1, 2016 #10
    Addition is a function.
     
  12. Oct 1, 2016 #11

    Math_QED

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    Yes, for vectors:

    ##+ : V \times V \rightarrow V## but how can I conclude that then?
     
  13. Oct 1, 2016 #12
    What does it mean by definition that ##+## is a function?
     
  14. Oct 1, 2016 #13

    Math_QED

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    It means that:

    ##+ \subset ((V \times V) \times V)## such that for every 2 vectors ##(\overrightarrow x, \overrightarrow y) \in (V \times V)## there is a unique vector ##\overrightarrow z \in V## such that ##+(\overrightarrow x, \overrightarrow y) = \overrightarrow z##
     
  15. Oct 1, 2016 #14
    And do you see that it means that if ##x=x'## and ##y = y'## then ##+(x,y) = +(x',y')##? To prove this, let ##z = +(x,y) = +(x',y')## and use uniqueness of ##z##.
     
  16. Oct 1, 2016 #15

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    Hm let me try:

    Proof: Let ##(x,y) \in V \times V##

    Then, there is a unique ##z \in V## such that ##z = +(x,y)##. But, since x = x' and y = y', we have that ##+(x',y') = z## too. Since z is unique, we conclude that ##+(x',y') = +(x,y)##
     
  17. Oct 1, 2016 #16

    Ray Vickson

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    How you can or should do it depends on exactly what vector-space axioms you start with. I have seen different books using different systems of axioms, so being very precise at this point is important.

    BTW: you cannot "proof" something, but you might be able to prove it, or to give a proof.
     
  18. Oct 1, 2016 #17

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    V is a commutative group for addition and for the scalar multiplication we have:

    ##1\overrightarrow a = \overrightarrow a##
    ##(\alpha \beta)\overrightarrow a = \alpha (\beta \overrightarrow a)##
    ##(\alpha + \beta)\overrightarrow a = \alpha \overrightarrow a + \beta \overrightarrow a##
    ##\alpha (\overrightarrow a + \overrightarrow b) = \alpha \overrightarrow a + \alpha \overrightarrow b##

    where ##\alpha, \beta \in \mathbb{K}, \overrightarrow a, \overrightarrow b \in V##
     
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