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Vector space property

  1. Oct 3, 2016 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    Prove that in any vector space V, we have:

    ##\alpha \overrightarrow a = \overrightarrow 0 \Rightarrow \alpha = 0 \lor \overrightarrow a = \overrightarrow 0##

    2. Relevant equations

    I already proved:

    ##\alpha \overrightarrow 0 = \overrightarrow 0##
    ##0 \overrightarrow a = \overrightarrow 0##

    3. The attempt at a solution

    Suppose ##\alpha \neq 0##

    Then: ##\alpha \overrightarrow a = \overrightarrow 0##
    ##\Rightarrow \alpha^{-1} (\alpha \overrightarrow a) = \alpha^{-1} \overrightarrow 0##
    ##\Rightarrow (\alpha^{-1} \alpha) \overrightarrow a = \overrightarrow 0##
    ##\Rightarrow 1 \overrightarrow a = \overrightarrow 0##
    ##\Rightarrow \overrightarrow a = \overrightarrow 0##

    The problem is. I don't know how to show that ##\alpha \overrightarrow a = \overrightarrow 0## can imply ##\alpha = 0## I can't suppose ##\alpha = 0##, because I need to prove that?

    Maybe something like this?

    ##\alpha \overrightarrow a = \overrightarrow 0##
    But ##0 \overrightarrow a = \overrightarrow 0##

    Thus: ##\alpha \overrightarrow a = 0 \overrightarrow a ##

    Comparing the two, we obtain ##\alpha = 0##

    Thanks in advance.
     
  2. jcsd
  3. Oct 3, 2016 #2

    Orodruin

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    I suggest you focus on the implications of assuming that ##\vec a \neq 0##.
     
  4. Oct 3, 2016 #3

    fresh_42

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    You can distinguish cases. Either ##\alpha = 0## or ##\alpha \neq 0##. One of the two has to happen.
     
  5. Oct 3, 2016 #4

    Math_QED

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    But can I assume ##\alpha = 0##? Then it follows trivially that ##\alpha = 0 \land \alpha \overrightarrow a = \overrightarrow 0 \Rightarrow \alpha = 0##
     
  6. Oct 3, 2016 #5

    fresh_42

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    Why not?
    $$A = A \wedge \text{ true } = A \wedge (B \vee \lnot B) = (A \wedge B) \vee (A \wedge \lnot B)$$
    and ##B=(\alpha = 0)## does the job.
     
  7. Oct 3, 2016 #6

    Math_QED

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    Nice. Thanks a lot.
     
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