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Prove that if av=0, a or v =0

  1. Jul 7, 2009 #1
    Man. I have always taken a statement like this for granted, so I have no idea how to prove it.

    1. The problem statement, all variables and given/known data
    [tex]\text{Prove that if }a\in\mathbf{F}\text{ and }v\in V\text{ and }av=0\text{ then a=0 or v=0}[/tex]

    Seriously? What about saying

    [tex]\text{Let }v=(x_1,...,x_n) \text{ be a vector space over \textbf{F} and }a\in\mathbf{F}[/tex]

    Suppose that [tex]av=0[/tex]

    then

    [tex]a(x_1,...,x_n)=\mathbf{0}\Rightarrow (ax_1,...,ax_n)=(0,...,0)[/tex]

    therefore a=0 or v=0.


    Seems like I just said a whole lot...but I am not sure if I actually proved anything. Is using the definition of scalar mult over V enough? Or am I missing something else?
     
    Last edited: Jul 7, 2009
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  3. Jul 7, 2009 #2

    Office_Shredder

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    v is an arbitrary vector in an arbitrary vector space, it doesn't need to be in Fn

    Start like this: If a=0, then av=0 (prove this). Then, assume a is NOT zero, and av=0. Can you prove v is the zero vector?
     
  4. Jul 7, 2009 #3
    First of all, you have a slight typo in your proof. In the second to last line, you have av = (x_1, ..., x_n), which is probably not what you intended.

    While your steps can be easily turned into a proof, your last line (beginning with "therefore") does not justify the claim. One way to complete the proof is by simple case examination. If a = 0, then you are done, since you end up with (0*x_1, ... , 0*x_n) = (0, ..., 0). Now what if a =/= 0? Remember, you need to use the properties of a field to justify your conclusions, or else reaching the last equality before your last line is kind of pointless.
     
  5. Jul 7, 2009 #4
    Oh..I was thinking that Fn is how we represent an arbitrary vector space. How do we represent an arbitrary vector space?

    Righto! Typo fixed. Once I figure out the proper way to represent an arbitrary vector space, I will continue.
     
  6. Jul 7, 2009 #5
    Okay. I know that it may seem like I am getting hung up on little things, but how do you represent the elements of an arbitrary vector space if not by (x1,...,xn)?

    I would just say that V is an arbitrary vector space, but than when I want to actually apply some operation on V, I need to use the elements of V. Know what i mean?

    So, if I say that V is an arbitrary vector space then v=(......)<--- what goes in here? :smile:
     
    Last edited: Jul 7, 2009
  7. Jul 7, 2009 #6
    Anyone help me out on this notational issue? :smile:
     
  8. Jul 7, 2009 #7

    HallsofIvy

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    You could choose some basis, say [itex]{v_1, v_2, \cdot\cdot\cdot, v_n}[/itex] and then write [itex]v= x_1v_1+ x_2v_2+ \cdot\cdot\cdot+ x_nv_n[/itex]. But you don't "need to use the elements of V". That is almost never a good way to prove general statements about vector spaces. You need to use the definitions for a general vector space. The definition of "scalar product" requires (a+ b)v= av+ bv and a(u+ v)= au+ av as well as a(bv)= (ab)v. Suppose a is NOT 0. Then, since a is a non-zero number, it has a reciprocal (1/a). (1/a)(av)= ((1/a)a)v= v. And therefore?
     
  9. Jul 7, 2009 #8
    so Halls, it's better to just deal with the symbolic representation of the a vector in the vector space rather than with its elements?

    Okay then. Now with regard to your approach to the proof: Suppose that a is non-zero. Therefore (1/a) exists.
    (1/a)(av)=((1/a)a)v=v
    I really don't see where we are going with this :redface: v=v ... how does that help?
     
  10. Jul 7, 2009 #9

    Office_Shredder

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    If av=0V (the zero vector in V)

    then

    1/a(av) = (1/a)0v

    Hence

    v=?
     
  11. Jul 7, 2009 #10
    Okay. Halls. Is this what you mean?

    [tex]\text{Assume that a and v }\ne 0\text{ therefore }1/a \text{ exists. Also assume that av=0}[/tex]

    Multiplying both sides by 1/a we have:

    (1/a)av=0
    ((1/a)a)v=0
    v=0 which is a contradiction.

    Does this do it? I really have no idea how to use a contradiction.
     
  12. Jul 7, 2009 #11

    jgens

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    I don't think that's what he's looking for. Clearly, if av = 0 and a = 0 the proof is trivial. Now suppose that a != 0. If av = 0 then (a-1)(a)(v) = 0(a-1) . . .
     
  13. Jul 7, 2009 #12
    I don't see how this is any different from what I wrote?

     
  14. Jul 7, 2009 #13

    jgens

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    It's not really that much different, it's just clearer and more concise if you don't do this with a contradiction. You don't need to assume that v != 0, it should follow from the proof.
     
  15. Jul 7, 2009 #14
    AH!! Okay! I am with you now. And then i assume I would just do the same thing we v, yes?
     
  16. Jul 7, 2009 #15

    jgens

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    Yup! That should work out.
     
  17. Jul 7, 2009 #16
    Thanks everyone. Get ready for a lot more of these!!! :smile:
     
  18. Jul 8, 2009 #17

    HallsofIvy

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    What's different is that he did NOT say "Assume that a and V [itex]\ne 0[/itex]" as you did before. Assume that [itex]a\ne 0[/itex] so that 1/a exists. (1/a)(av)= ((1/a)a)v= v= 0. And, since av= 0, that is the same as (1/a)(0)= 0 so v= 0. That's all. Either a= 0 or [itex]a\ne 0[/itex] and then v= 0.
     
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