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Linear Algebra- Dependent or independent

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Let(v1, v2, v3) be three linearly independent vectors in a vector space V. Is the set {v1-v2, v2-v3, v3-v1} linearly dependent or independent?

    2. Relevant equations
    Linearly independent is when c1v1+c2v2+...+ckvk=0
    and c1=c2=...ck=0
    3. The attempt at a solution
    c1(v1-v2)+ c2(v2-v3)+ c3(v3-v1)=0

    c1-c3=0
    -c1+c2=0
    -c2+c3=0

    therefore c1=c2=c3 and since c1, c2 and c3 are zero because for the first set of independent vectors I got c1v1+c2v2+c3v3=0 all c1=c2=c3=0,
    which means this is the case for the second set and it must be linearly independent.

    This is what i got but my answer key says the second set is linearly dependent. I'm having trouble seeing why.
    Thanks for any help
     
  2. jcsd
  3. Apr 25, 2016 #2

    Samy_A

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    Can you explain in detail how you concluded from c1=c2=c3 that c1, c2 and c3 are zero?
     
  4. Apr 25, 2016 #3
    from my set of equations i concluded that c1=c2=c3. So then I had made the assumption that c1=c2=c3 was zero because from the given set v1, v2, v3 being linearly indepedent, which would mean that c1v1+c2v2+c3v3=0 must have c1=c2=c3 from the definition of linear dependence. I assumed they were the same constants, is this wrong to assume?
     
  5. Apr 25, 2016 #4

    Samy_A

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    ##v_1,v_2,v_3## being linearly independent means that if ##d_1v_1+d_2v_2+d_3v_3=0##, then ##d_1=0,d_2=0,d_3=0##.
    There is no reason whatsoever to assume that the numbers ##c_1,c_2,c_3## that you chose for the set {##v_1-v_2, v_2-v_3, v_3-v_1##} must also work for ##v_1,v_2,v_3##.
     
  6. Apr 25, 2016 #5
    but even still for my set of {v1-v2,v2-v1....} the c1=c2=c3 are all equall and I had set my equations to equal zero, then the only way this will be true is if they all equal zero, right?
     
  7. Apr 25, 2016 #6

    Samy_A

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    Why? All you found is that ##c_1=c_2=c_3##.
    That seems sufficient to have ##c_1(v_1-v_2)+c_2(v_2-v_3)+c_3(v_3-v_1)=0##.
     
  8. Apr 25, 2016 #7
    Ok, I'm seeing it clearer now. Thank you
     
  9. Apr 25, 2016 #8

    Mark44

    Staff: Mentor

    I would add that because the vectors are linearly independent, there can be no other solutions for the constants ##d_1, d_2,## and ##d_3##.

    If the three vectors were linearly dependent, the equation ##d_1v_1+d_2v_2+d_3v_3=0## would have an infinite number of solutions for the constants, including ##d_1=0,d_2=0,d_3=0##. This is a fine point that often eludes new students of linear algebra.
     
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