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Linearly independence proof

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that S = {v1, v2, v3} is linearly
    independent and
    w1 = v2
    w2 = v1 + v3
    and
    w3 = v1 + v2 + v3
    Determine whether the set T = {w1,w2,w3} is
    linearly independent or linearly dependent.

    2. Relevant equations

    Let c1, c2, c3=scalars

    c1w1+c2w2+c3w3=0
    c1v2+c2v1+c2v3+c3v1+c3v2+c3v3=0
    (c2+c3)v1+(c1+c3)v2+(c2+c3)v3=0

    c2+c3=0
    c1+c3=0
    c2+c3=0


    solving 1st equation gives: c2=-c3
    Plug into 3rd gives: -c3+c3=0 → 0=0 what does this mean?
     
  2. jcsd
  3. Sep 25, 2012 #2

    LCKurtz

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    It means c3 can be anything as long as c1 and c2 are = -c3. in particular not all the c's have to be 0. What does that tell you?
     
  4. Sep 25, 2012 #3
    It means T is linearly dependent because in order for it to be independent al c's have to be 0.

    So if it had been -c3-c3=0.
    c3=-c3
    Thus c3=0
    That would make it linearly independent right?
     
  5. Sep 25, 2012 #4

    LCKurtz

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    Yes. But in this case you can easily find particular values not all zero that work. For example ...?
     
  6. Sep 25, 2012 #5
    I'm not sure what you are asking. if -c3-c3=0 then c3=0
    c2=0 and c1=0
     
  7. Sep 25, 2012 #6

    LCKurtz

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    Woops, misunderstanding of what I meant. I didn't mean you weren't correct. I was referring to your actual problem, where you can find c's not all zero. To finish that problem you should really display three c's that work by plugging them in ##c_1w_1+c_2w_2 + c_3w_3## and getting ##0##.
     
  8. Sep 25, 2012 #7
    Oh I see what you're saying. c3 can be something like 2, then c1=c2=-2. Thus the system is linearly dependent for this case.
     
  9. Sep 25, 2012 #8

    LCKurtz

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    Right, because [I presume] you calculated ##-2w_1-2w_2+2w_3## and got zero. And you wouldn't say "for this case". They are linearly dependent period and these choices of the constants are one way of proving it.
     
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