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Linear Algebra (Linearly Independent vs. Linearly Dependent)

  1. Jul 11, 2009 #1
    1. The problem statement, all variables and given/known data

    If S = {v1, v2, v3} where v1, v2 and v3 are in [itex]\mathbb{R}^2[/itex], then the vectors v1, v2 and v3 are linearly independent.

    2. Relevant equations
    None


    3. The attempt at a solution
    I thought the answer was true, but I know the correct answer is false and I'm not sure why. Here was my reasoning for true:

    If this implies that these three vectors span [itex]\mathbb{R}^3[/itex], then they must be linearly independent. If they weren't, then they would only span [itex]\mathbb{R}^2[/itex] or [itex]\mathbb{R}^3[/itex] space.

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    Hootenanny
     
    Last edited by a moderator: Jul 11, 2009
  2. jcsd
  3. Jul 11, 2009 #2
    So is the problem asking you whether or not this statement is true? Is there any other information given regarding what S is?
     
  4. Jul 11, 2009 #3
    Sorry. Yes, this is a true/false question. S is a set of vectors [tex]v_1, v_2 \text{ and } v_3 \text{ in }\mathbb{R}^3 \text{ space}[/tex]. That's all that is given.
     
  5. Jul 11, 2009 #4
    Okay. Cool. So if you think it is false, then you need to try and come up with a counterexample. That is, you need to find three vectors in R3 such that the three vectors are not linearly independent, i.e. they are linearly dependent.

    I think what you have as a solution to it being true is wrong because it isn't given that S spans R3. If it was given, then what you have would be correct, but that isn't given information.
     
  6. Jul 11, 2009 #5
    Oh, I see. So the keyword is span, I suppose. I guess you can still imagine a 3-D space with these vectors even if the vectors don't span that space. OK, that makes sense.

    Thanks! I posted two other T/F questions - do you think you could take a look?
     
  7. Jul 11, 2009 #6
    Can you find three distinct vectors in R^3 that are linearly dependent?
     
  8. Jul 11, 2009 #7
    The first picture that comes to my mind is a vector centered at the origin and two identical vectors, the first multiplied by -C and the second multiplied by C. However, this would create three vectors parallel to each other, so they would only span R^2.

    What is a better way to think about it?
     
  9. Jul 11, 2009 #8

    HallsofIvy

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    One of the things you should have learned already is that, in a vector space of dimension n, there cannot be a set of fewer than n vectors that spans the space nor can there be a set of more than n vectors that are independent.
     
  10. Jul 11, 2009 #9
    Thanks for the comment, but as I mentioned in my second post, I made a mistake in my original post. The vectors v1, v2 and v3 are in R^3, not R^2.
     
  11. Jul 11, 2009 #10
    Well if you have three vectors that are just multiples of each other (i.e. they are linearly dependent), then no two out of the three are linearly independent, which means that they couldn't span [itex]\mathbb{R}^2[/itex]. They would actually only span [itex]\mathbb{R}[/itex]. The example you gave is definitely a counterexample to your original problem though. Good job.
     
  12. Jul 11, 2009 #11
    I see. Thanks!
     
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