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Proof of linear dependence

  1. Oct 16, 2005 #1
    This may be a really simple proof but its giving me grief.

    If {[itex]v_1, v_2, v_3[/itex]} is a linearly dependent set of vectors in [itex]\mathbb{R}^n[/itex], show that {[itex]v_1, v_2, v_3, v_4[/itex]} is also linearly dependent, where [itex]v_4[/itex] is any other vector in [itex]\mathbb{R}^n[/itex].

    Any hints on where to start? I started out by writing out all the claims that could be made by taking the first set of vectors to be linearly independent, but that didn't get me terribly far. :confused:
     
  2. jcsd
  3. Oct 16, 2005 #2

    TD

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    The definition of lineair dependance (or at least an equivalent statement) is that you can form the zero-vector with a lineair combination of your vectors without having all 0 coefficients. Now, you can do that for your first set, would you be able to do it too for the second one?
     
  4. Oct 16, 2005 #3
    i don't like this question because it doesn't exlude a v_4 that is in the span of the first set of vectors.

    in that case, the second set WOULD be dependent!

    :grumpy:
     
  5. Oct 16, 2005 #4
    But thats exaclty what you want to prove, so your proof is finished!
     
  6. Oct 16, 2005 #5
    I think I got it. The rule basically says the weights c_1, c_2, ... c_n can't all be equal to zero at the same time. If the first set is already linearly dependent, i.e. [itex]c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 = 0[/itex], then we can take c_4 = 0, which would make [itex]c_1\vec{v}_1 + c_2\vec{v}_2 + c_3\vec{v}_3 + c_4\vec{v}_4 = 0[/itex] as well, therefore the second set is also linearly dependent.
     
  7. Oct 17, 2005 #6

    TD

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    That's what I meant :smile:
     
  8. Oct 17, 2005 #7
    oh!

    i thought it said INdependent!

    :rofl:
     
  9. Oct 17, 2005 #8
    yep, that's it. good job.
     
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