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Show that these vectors are in a vector space?

  1. Mar 17, 2012 #1
    How do I show that for any vectors u,v, and w in a vector space V, the set of the vectors {u-v, v-w, w-u} is a linearly dependent set?

    do it in general!
     
  2. jcsd
  3. Mar 17, 2012 #2

    HallsofIvy

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    Isn't the point for you to "do it in general"? Vectors are "dependent" if and only if some linear combination of them (other than the trivial one with all multipliers equal to 0) is equal to 0. Can you find such a linear combination?
     
  4. Mar 17, 2012 #3
    Just take the set of vectors, build a 3x3 matrix out of them and calculate its determinant. If the determinant is not equal to cero, then they are linearly independent. If it is cero, then they are linearly dependent.

    EDIT: I have explained it better in this thread (actually the question is almost exactly the same)
     
  5. Mar 17, 2012 #4

    The teacher wants the answer in a general form, and I don't know how to find the determinant of that.

    if I write something like this, is it correct?
    {u-v, v-w, w-u} = {V1, V2, V3}

    To prove this is linearly dependent, one of the vectors can be written as the sum of the other two vectors.

    V3 = -V2-V1

    Therefore, it is linearly dependent.
     
  6. Mar 17, 2012 #5

    lavinia

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    just add them up
     
  7. Mar 18, 2012 #6
    Would this be correct?

    a(u-v)+b(v-w)+c(w-u) = 0
    au-av+bv-bw+cw-cu = 0

    au-cu = 0
    -au+bv = 0
    -bw+cw = 0

    augmented matrix =

    1 0 -1 0
    -1 1 0 0
    0 -1 1 0

    RREF on the calculator =

    1 0 -1 0
    0 1 -1 0
    0 0 0 0

    Since the soultion is not a trivial solution, it means the set is dependent.
     
  8. Mar 19, 2012 #7

    HallsofIvy

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    The definition of 'dependent' vectors is that there exist a set if coefficients, not all 0, so that the linear combination is the 0 vector.

    Is it possible to choose A, B, C, not all 0, so that A(u- v)+ B(v- w)+ C(w- u)= 0.
    Hint- there is a very simple choice for A, B, and C.
     
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