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A0 = 0 (0 of a vector space)

  1. Jul 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that a0 = 0

    2. Relevant equations



    3. The attempt at a solution

    Let V be a vector space on a field F. Let x be a member of V and a be a member of F.

    Consider that the 0 vector is the unique vector such that

    x + 0 = x

    Now, apply a scalar multiplication by a to both sides of the equation. Because scalar multiplication is distributive in all vector spaces,

    ax + a0 = ax

    Thus, we see that a0 has the same property of the 0 vector in V. Since the 0 vector in V is also unique, it must be the case that

    a0 = 0

    QED


    Can I do that?
     
  2. jcsd
  3. Jul 17, 2013 #2

    LCKurtz

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    I don't think you quite have it. Have you shown that for ALL y you have y + a0 = y?
     
  4. Jul 17, 2013 #3

    But if

    x + 0 = x

    is true for all x (which it is by definition of the zero vector) then

    ax + a0 = ax

    is also true for all x, and it is true for all a (because scalar multiplication is distributive in all vector spaces), so then it is always true, so it is true for any ax (or y).

    Right?
     
    Last edited: Jul 17, 2013
  5. Jul 17, 2013 #4

    LCKurtz

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    It may be trivial, but you haven't shown that any y can be written in the form ax.
     
  6. Jul 17, 2013 #5
    Oh, ok.

    So, I need to justify that ax is also an element of V? Am I understanding that correctly?

    If so, then can I just say that in all vector spaces, for each element a in F and each element x in V, there is a unique element ax in V, thus, let ax = y, which is a known element of V (and then the result follows)?
     
    Last edited: Jul 17, 2013
  7. Jul 17, 2013 #6

    LCKurtz

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    You have shown that a0 works as an additive identity on all elements of the form ax. If every y in V can be written that way you have it as an additive identity on all of V, which is what you want. Think about ##y = a(\frac 1 a y)##.

    As an alternative approach to the problem think about a(0+0) in the first place.
     
  8. Jul 17, 2013 #7
    Ok, instead of saying "the result follows" I'll just write it out.

    "for each element a in F and each element x in V, there is a unique element ax in V, thus, let ax = y, which is a known element of V (and then the result follows)?"

    ax = y
    =>
    x = (1/a)y

    (1/a) is the multiplicative inverse of a and is therefore a member of F. y is already known as a member of V. Therefore any element x of V can be expressed as a scalar multiplication of a member of the field and another member of V.

    That's what I was getting at, would you say that is adequate?

    Thanks again.
     
  9. Jul 17, 2013 #8

    Zondrina

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    Using LC's hint :

    a0 = a(0+0) = a0 +a0

    So you have :

    a0 = a0 + a0

    Hmm what property of vector spaces would help you here now? There aren't too many to consider so I'm sure you will see it any moment.
     
  10. Jul 17, 2013 #9
    I'd rather just complete the attempt I've done myself, this is the proof my book gives.
     
  11. Jul 17, 2013 #10

    LCKurtz

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    No, I don't think so. At the very least, it is confusing. Summarizing your argument so far you have shown that for any ##a\in F## and ##{\bf x} \in V## that ##a{\bf x} + a{\bf 0} = a{\bf x}##. What you need to show is that for any ##{\bf y}\in V## you have ##{\bf y} + a{\bf 0} = {\bf y}##. You have to start with ##\bf y##. You can't start with ##a{\bf x}## and call it ##\bf y##. Look at my suggestion in post #6 again.
     
  12. Jul 17, 2013 #11

    Zondrina

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    Well you know that ##a \in F## and I would recommend writing ##0## as ##0_V## as to be more explicit.

    So really you want to show ##a0_V = 0_V##.

    ##F## is a field of scalars. Since ##V## is a vector space, it is closed under addition and scalar multiplication. You know that ##\forall v \in V## and ##\forall a \in F## that ##av \in V## because ##V## is closed.

    Now, we've shown there are basically infinitely many vectors of the form ##av## inside of ##V##. So if you want to proceed your way from before, add ##av## to both sides :

    ##a0_V + av = 0_V + av##

    Now because ##0_V## is the unique vector such that ##0_V + v = v, \forall v \in V## ( Prove this if you havent! ) we can proceed :

    ##a0_V + av = av##

    I'll let you finish that one if you want to. Your argument in your last post wouldn't work though. All you did was set y=ax and say y was in V. LC stated if you can show every ##y \in V## can be written as some ##ax##, then you know that ##a0## will work perfectly as an identity.
     
  13. Jul 17, 2013 #12
    Why can't I call ax, y? It's just an element of V.

    Looking at y=a((1/a)y) makes me fall back to the same reasoning. If y is an element of V then so is (1/a)y, so a((1/a)y) is an expression for y that is a scalar multiplication of an element of F and an element of V. But that's exactly what I did before. What is wrong with this reasoning?
     
  14. Jul 18, 2013 #13

    verty

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    1MileCrash: I think you can't say for each ##a## because you are given ##a##. Given ##a##, you must show that a##0_v = 0_v##. I can see a much quicker proof using components.

    -edit- Hmm, I suppose this component idea is limited (without choice) to finite-dimensional vector spaces, better to prove this in general.
     
    Last edited: Jul 18, 2013
  15. Jul 18, 2013 #14

    LCKurtz

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    Because you have to start with an arbitrary ##{\bf y}\in V##. You have all the pieces but your argument should be written like this:

    Suppose ##{\bf y}\in V## and ##a \in F##.

    Let ##{\bf x} =\frac 1 a{\bf y}##

    ##{\bf x} + {\bf 0} = {\bf x}##

    ##a{\bf x} + a{\bf 0} = a{\bf x}##

    ##a\frac 1 a{\bf y} +a{\bf 0} = a\frac 1 a{\bf y}##

    ##{\bf y} +a{\bf 0} ={\bf y}##

    Therefore ##a{\bf 0}## is the unique additive identity.
     
  16. Jul 18, 2013 #15
    Oh, I see, that comes out very nicely. I just thought that since x was an arbitrary vector, then so is ax or y. I don't see a difference between letting ax=y and letting x=(1/a)y.

    Thanks again.
     
  17. Jul 18, 2013 #16
    EDIT: nevermind, got it backwards.
     
  18. Jul 19, 2013 #17

    verty

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    There is still a mistake here. Given a, we aren't told that it isn't 0.

    Gotta keep your apples in order. If you start with a given ##a \in F##, you must be faithful to that.
     
  19. Jul 19, 2013 #18
    I have no idea what you are saying. Could you explain how I have not done that? Or what that means?

    All I did was define a to be an element of F so that I could use it in an expression without the reader not knowing what it is. What do you mean by "starting" with it?
     
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