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Linear independence in normed vector space

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data
    If we have a normed vector space, and a sequence of vectors
    [itex]\{\mathbf{v}_k\}_{k=1}^{N}[/itex] in the normed vector space.
    If there exists a constant B>0 such that the following holds for all scalar coefficients [itex]c_1,c_2\cdots c_N[/itex]
    [itex]B\sum\limits_{k=1}^N |c_k|^2 \leq ||\sum\limits_{k=1}^Nc_k\mathbf{v}_k||^2[/itex]
    Show that the vectors are linearly independent.

    2. Relevant equations
    Triangle equality [itex] ||a+b|| \leq ||a||+||b|| [/itex]
    [itex]||\alpha v|| =|\alpha|\cdot ||v||[/itex]
    [itex] ||v|| = 0 -> \mathbf{v} = \mathbf{0}[/itex]

    3. The attempt at a solution
    I remember the definition of linear independence: [itex] k_1+v_1+k_2v_2\cdots k_Nv_N = 0[/itex] for non trivial vectors and all coefficients.
    i use the triangle inequality and scalar multiplication:
    [itex]||\sum\limits_{k=1}^Nc_k\mathbf{v}_k||^2 = \sum\limits_{k=1}^N|c_k|^2||\mathbf{v}_k||^2[/itex]
    Which combined by the inequality stated in the problem implies that at least one of the vectors are different from the zero-vector
     
    Last edited: Feb 1, 2012
  2. jcsd
  3. Feb 1, 2012 #2

    Dick

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    Try a proof by contradiction. Suppose the vectors are linearly dependent. Then what?
     
  4. Feb 1, 2012 #3
    Shouldn't i try supposing the opposite e.g. that they are dependent?
    But if they are independent then the coefficients = 0 is the only solution to this equation:
    v1c1+v2c2..vncn = 0
     
  5. Feb 1, 2012 #4

    Dick

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    Yes, suppose they are dependent. So that v1c1+v2c2+..+vncn=0 with at least one of the c's nonzero. Then what does your inequality tell you?
     
  6. Feb 1, 2012 #5
    Hmm. if all but c1 is zero:
    B|c1|^2 < = ||c1v1||^2 = |c1|^2||v1||^2 <=>
    Divide by the nonzero nummerical value of |c1|^2
    and B<= ||v1||^2
    Hmm? I still really can't see it
     
  7. Feb 1, 2012 #6
    Aah, wait it tells me that the rhs. allways is greater than zero cause B>0?
    So the only solution would be if all the coefficients is zero, which i believe proves the linear independence
     
    Last edited: Feb 1, 2012
  8. Feb 1, 2012 #7

    Dick

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    Go back a step. You assumed the v's are linearly dependent. That means v1c1+v2c2+..+vncn=0 with not all of the c's equal to zero. So what is ||v1c1+v2c2+..+vncn||? What can you say about (c1^2+c2^2+...+cn^2)? What does that imply about B?
     
  9. Feb 1, 2012 #8
    ||v1c1+...|| is not zero i guess. and(c1^2+c2^2...) is also not zero, then B <= ||v1+v2...||^2. Im sorry just started at a pretty abstract course.
     
  10. Feb 1, 2012 #9
    So i would conclude that the condition for linear dependence isn't true here, because the all the coefficients would have to be zero for the inequality and the dependence condition to be true.
     
  11. Feb 1, 2012 #10

    Dick

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    Just slow down. You are jumping all over the place. You aren't following the 'proof by contradiction' track. Reread post 7 and answer my first question.
     
  12. Feb 1, 2012 #11
    Okay, ||v1c1+v2c2...vncn|| should then also be zero, because of the third "norm condition" i stated in the relevant equations.
    And (c1^2+c2^2+c3^2) is not zero, because of the assumption of dependence.
    Edit:
    That gives me, that B should be zero if the vectors are dependent, and that proves the opposite, thus they are linearly independent?
     
    Last edited: Feb 1, 2012
  13. Feb 1, 2012 #12

    Dick

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    So what does that tell you about B? Remember, you are looking for a contradiction.
     
  14. Feb 1, 2012 #13
    Yay That B should be zero and there lies the contradiction. You made it possible for a tired upcoming physics engineer from Denmark to sleep tonight. Thanks a lot :)
     
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