1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector space properties: distributivity

  1. Oct 1, 2016 #1

    Math_QED

    User Avatar
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Then prove it, so you can use it.
     
  4. Oct 1, 2016 #3

    Math_QED

    User Avatar
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yes, you should be able to derive this easily

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

    Math_QED

    User Avatar
    Homework Helper

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Add the additive inverse of ##a##.
     
  8. Oct 1, 2016 #7

    Math_QED

    User Avatar
    Homework Helper

    Yeah, thought about that too. Why am I allowed to do that on both sides?
     
  9. Oct 1, 2016 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    What's wrong with it?
     
  10. Oct 1, 2016 #9

    Math_QED

    User Avatar
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Addition is a function.
     
  12. Oct 1, 2016 #11

    Math_QED

    User Avatar
    Homework Helper

    Yes, for vectors:

    ##+ : V \times V \rightarrow V## but how can I conclude that then?
     
  13. Oct 1, 2016 #12

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    What does it mean by definition that ##+## is a function?
     
  14. Oct 1, 2016 #13

    Math_QED

    User Avatar
    Homework Helper

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    Math_QED

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    Math_QED

    User Avatar
    Homework Helper

    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##
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Vector space properties: distributivity
Loading...