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Linear (in)dependence of vector pairs and triples

  1. May 23, 2016 #1

    ibkev

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    1. The problem statement, all variables and given/known data
    Suppose that pairs of vectors ##v_1, v_2## and ##v_1, v_3## and ##v_2, v_3## are linearly independent. Must ##v_1, v_2, v_3## be linearly independent?

    2. Relevant equations
    So this means ##v_1 \neq av_2##, ##v_1 \neq bv_3## and ##v_2 \neq cv_3##

    3. The attempt at a solution

    For the sake of intuition, I started playing around with vectors in ##\mathbb {R}^3##. And I came up with values for ##v_1, v_2##, and ##v_3## that are independent as pairs but linearly dependent when together:

    (1,2,1)
    (0,1,1)
    (1,1,0)

    So I've answered the question by existence but this doesn't feel very satisfying, especially since I only shown this for ##\mathbb {R}^3## and not shown it in general. Do you have suggestions for a better proof?
     
  2. jcsd
  3. May 23, 2016 #2

    Mark44

    Staff: Mentor

    This is a good way to do the problem, although I would have started with three vectors in ##\mathbb{R}^2##. The question "Must ##v_1, v_2, v_3## be linearly independent?" is asking whether three such vectors are always independent. By showing three vectors that are pairwise linearly independent, but not independent as a whole, you have answered the question.

    If you want to extend this to a higher dimension space, just do essentially the same thing that you did to come up with your three vectors. Pick two vectors that aren't parallel (neither is a multiple of the other), and for the third vector use the sum of the other two. As long as the third vector isn't a scalar multiple of either of the other two, that's your example. The equation ##c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} = \vec{0}## will have a nonzero solution for the constants; namely ##c_1 = 1, c_2 = 1, c_3 = -1##.
     
  4. May 23, 2016 #3

    ibkev

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    Ok that makes a lot of sense - much appreciated Mark!
     
  5. May 23, 2016 #4

    Ray Vickson

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    Having an example is the best possible proof. You can extend it immediately to ##n## dimensions by appending ##n-3## zero components to the end of each of your vectors.
     
  6. May 23, 2016 #5

    ibkev

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    >What is the condition for the independence of three vectors?

    The condition for linear independence is that only the trivial linear combination (where all vector coefficients are zero) is zero.

    Is that what what you meant by your question?
     
  7. May 24, 2016 #6

    ehild

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    Yes. As @Mark44 said,
    ##c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} = \vec{0}## has only the trivial (all zero) solution if the vectors are independent. Assume it is not the case, and at least one of the coefficients (say c1) is not zero. Then you can choose it 1, and express ##\vec v_1 ## as linear combination of the other two vectors.
    ##\vec{v_1} = - c_2\vec{v_2} - c_3\vec{v_3} ##
    If both coefficients differ from zero, ##\vec v_1 ## is not the multiple of any of the other two vectors.
     
  8. May 26, 2016 #7

    ibkev

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    Ah - thanks for this! I'm still working on the switch to proof oriented math from computation...
     
  9. May 26, 2016 #8

    Ray Vickson

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    I don't think you really understand the issue here. When you seek a counterexample to a statement, just giving an example that violates that statement IS a proof---a 100% mathematically valid proof. There is no need to try anything more "abstract" or general.
     
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