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Vector Space Problem

  1. Jun 26, 2005 #1
    Is the following a vector space:
    the set of all ordered triples of real numbers, {x1,x2,x3)}, usual addition, and r(x1,x2,x3)=(0,0,0), all numbers r

    I think this is a vector space since it is the vector (0,0,0), but I'm not sure how to show the work for it.

    Thanks in adv.
     
  2. jcsd
  3. Jun 26, 2005 #2
    what is the definition of a vector space?

    check each of the criteria against what you have, and if all of the criteria are satisfied then it is a vector space.
     
  4. Jun 26, 2005 #3

    matt grime

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    why is it the vector (0,0,0)? it states it is the set of ALL triples...
     
  5. Jun 26, 2005 #4

    mathwonk

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    you need to read your definition of vector space. In my book it is the following:

    A "vector space" over the field k, is an abelian group (V,+) together with a way to multiply elements of V by elements of k, such that if x,y are in k, and v,w are in V then (x+y)v = xv+yv, 1v = v, x(v+w) = xv+xw, and (xy)v = x(yv).


    does your example satisfy these?
     
  6. Jun 26, 2005 #5
    For r(x1,x2,x3)=(0,0,0) to be true, either x1, x2, and x2 have to all zeros or r has to be zero. That's the only two things that work.
     
  7. Jun 26, 2005 #6

    matt grime

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    no, that isn't what the question states (as you have written it, read what you wrote, you declared the underlying set to be ALL the elements in R^3). you are assuming that the "old" scalar mult is being used. it isn't, it is a "new" scalar mult. albeit one that fails to actually be a proper scalr mult (see mathwonk's post) and thus the set RxRxR with this "new" scalar mult is not a vector space.

    yes, with the old mult the only way for r((x,y,z) to be zero for all r is if (0,0,0) is the only element of R^3 we are considering, but that isn't what we're doing, nor are we asking, given these new rules, what subsets of R^3 are a vector space.

    EXAMPLE, given the subset of RxR consisting of all pairs (x,a) where a is a FIXED number with the "new" addition (x,a)+(y,a)=(x+y,a) and scalr mult. of k(x,a)=(kx,a) is a vector space. but if were to assume that the opertions were the old ones then this is only true if a=0.
     
    Last edited: Jun 26, 2005
  8. Jun 26, 2005 #7
    I kinda get it now.
    So since the vector doesn't satisfy the fellowing identity of 1*u=u, where u is a non zero triple (x1,x2,x3), it is not a vector space?
     
  9. Jun 26, 2005 #8

    matt grime

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    yes.

    it is confusing, i imagine, because you are automaticallly associating the usual operations since R^3 is such a well known vector space.

    i come across this all the time in group theory when my students assert that the integers with the operation

    x@y=x+y-1

    where + is the usual addition is not a grou becuase 0 is not the identity of @.

    remember the rules are applied to the operations as given, thus the identity with respect to @ is 1
     
  10. Jun 26, 2005 #9
    So by that axiom 1*u=u, you can also show that r(x1,x2,x3)=(2*r*x1,2*r*x2,2*r*x3) is not a vector space since 1*u in this case is (2*x1,2*x2,2*x3)?
     
    Last edited: Jun 27, 2005
  11. Jun 27, 2005 #10
    Yes, Anyone?
     
  12. Jun 27, 2005 #11

    matt grime

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    you've shown that an axiom is falsified, what more was there to say to you?
     
  13. Jun 27, 2005 #12
    I just wasn't so sure about if the above was right and was asking if I was doing it correctly since the concept of vector space is still new to me.
     
  14. Jun 27, 2005 #13

    matt grime

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    it's nothing to do with vector spaces; it is deductive reasoning.

    You know all X's must satisfy Y, you know Z doesn't satisfy Y, therefore Z isn't an X. no mention of vector spaces.
     
  15. Jun 27, 2005 #14

    mathwonk

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    you must get used to the fact that we want you to decide for yourself if you are right. i.e. to get to where you look at your argument to see if it is logically sound, rather than asking us to confirm it for you.
     
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