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Linear algebra question

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data

    True or False: The linear combinations [itex]a_{1}v_{1} + a_{2}v_{2} [/itex] and
    [itex]b_{1}v_{1} + b_{2}v_{2} [/itex] can only be equal if [itex] a_{1} = b_{1} [/itex] and [itex] a_{2} = b_{2} [/itex]

    2. Relevant equations



    3. The attempt at a solution
    I have determined that this statement is false if at least of the vectors is a zero vector. What about if none of them are zero vectors?

    BiP
     
  2. jcsd
  3. Oct 1, 2012 #2

    LCKurtz

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    What happens if you set them equal? Show us what you have tried.
     
  4. Oct 1, 2012 #3
    I obtain [tex] (a_{1}v_{1} - b_{1}v_{1} + a_{2}v_{2} - b_{2}v_{2}) = 0 [/tex] where () indicates a vector.
    What next?

    BiP
     
  5. Oct 1, 2012 #4

    LCKurtz

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    Write it as ##(a_1-b_1)v_1 + (a_2-b_2)v_2=0## and think about about whether that implies ##a_1=b_1## and ##a_2=b_2## or not.
     
  6. Oct 1, 2012 #5
    Since if we let the two vectors be equal, then we get [tex] a_{1}-b_{1} = b_{2}-a_{2} [/tex] which does not necessarily imply that both the LHS and RHS are 0. In fact, they could equal any arbitrary constant.

    Does this prove that the statement is false even when both vectors are nonzero?

    BiP
     
  7. Oct 1, 2012 #6

    LCKurtz

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    One might call that a trivial case, where in a sense you don't have two vectors because they are the same. But yes, in that case it is false. But there is more to the story. Are there cases where it is true? You have only talked about when one vector is 0 or the two vectors are equal, neither of which one might call the more interesting case. What about other situations? And, by the way, you never stated whether there were hypotheses on the two vectors like, for example, neither is the 0 vector or other conditions.
     
  8. Oct 2, 2012 #7
    To disprove the statement, we need only one counterexample, which I have given by assuming that the vectors are equal. That should be sufficient, should not it?

    BiP
     
  9. Oct 2, 2012 #8

    LCKurtz

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    That is why I asked you if there are any hypotheses on the vectors. I would guess that the hypotheses might be they are two distinct vectors, neither of which is 0. Then what is the answer? The thing is, you are missing the point of that problem. Even though you have a counterexample as stated, there is more to be understood.
     
  10. Oct 2, 2012 #9
    Interesting, if the vectors are assumed to be (2,1) and (1,1), then the problem statement is infact true. I am going to see whether this is true for any ordered pair. My intuition now tells me that the statement is true in the case of nonzero unequal ordered pairs, but for higher order tuples I think it is false. Interesting indeed! Let's see if I can prove it.

    Also, what are your thoughts LC?

    BiP
     
  11. Oct 2, 2012 #10

    LCKurtz

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    My thoughts are to see if you can discover the answer as to when it is true and when it is false. It has to do with a standard concept in linear algebra.
     
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