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Linear Independency

  1. Feb 4, 2010 #1
    The problem statement, all variables and given/known data
    Indicate whether the following is true or false. Explain your answer.

    If [tex]\overline{u}, \overline{v}, \overline{w}[/tex] are vectors in [tex]R^{n}[/tex] such that {[tex]\overline{u}, \overline{v}[/tex]} and {[tex]\overline{v}, \overline{w}[/tex]} are each linearly independent sets, then {[tex]\overline{u}, \overline{v}, \overline{w}[/tex]} is a linearly independent set.

    The attempt at a solution
    I think that the above is false because for {[tex]\overline{u}, \overline{v}[/tex]} and {[tex]\overline{v}, \overline{w}[/tex]} to each be linearly independent sets, they must have two entries for each vector, as this would give them trivial solutions only. Therefore, they would each be a 2 x 2 matrix. However, {[tex]\overline{u}, \overline{v}, \overline{w}[/tex]} is not a linearly independent set because it would form a 3 x 2 matrix. This would automatically have a free variable, and so infinite solutions would result.
     
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  3. Feb 4, 2010 #2

    Dick

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    Don't make this so complicated. Just give an example of three vectors where {u,v} and {v,w} are linearly independent, but {u,v,w} is not linearly independent. You can do it in R^2.
     
  4. Feb 4, 2010 #3
    Don't I have to do it in R^2?
     
  5. Feb 4, 2010 #4

    Dick

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    ?? Do it in R^n where n is whatever. You just can't do it in R^1. Because {u,v} is always linearly dependent in R^1.
     
  6. Feb 4, 2010 #5
    But isn't {u,v} a 2 x n matrix, where n must be two for the matrix to be linearly independent? If n was greater than 2, wouldn't you have more rows than columns, which means you would end up with free variables?
     
  7. Feb 4, 2010 #6

    Dick

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    That's where you are going wrong. {u,v} is not a matrix, it's just a list of two vectors. If u=(1,0,0), and v=(0,1,0) in R^3, is {u,v} linearly independent?
     
  8. Feb 4, 2010 #7
    {u,v} wouldn't be linearly independent since you end up with a row of all zeroes.
     
  9. Feb 4, 2010 #8

    Dick

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    To show they are linearly dependent you want to find a solution to c1*u+c2*v=(0,0,0), where c1 and c2 are not both zero. Can you find one?
     
  10. Feb 4, 2010 #9
    No. c1 = c2 = 0.
     
  11. Feb 4, 2010 #10

    Dick

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    Well, ok. So a 'row of zeros' has nothing to do with linear independence. Now back to the point. Can you find an example of three vectors where {u,v} and {v,w} are linearly independent and {u,v,w} is not.
     
  12. Feb 4, 2010 #11
    u = (1,0,1), v = (0,1,0), w = (0,0,1)

    So this is a guess and check type of question? Is there a quicker way about this?
     
  13. Feb 5, 2010 #12

    Dick

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    Mmm. You've got {u,v} and {v,w} independent AND {u,v,w} independent. Not really what you want. One more try, ok?
     
  14. Feb 5, 2010 #13
    Wait a second. This is a true or false question. Shouldn't the answer be "true" since {u,v,w} is independent?
     
  15. Feb 5, 2010 #14

    Dick

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    That's only one example. One example doesn't prove it's true. One counterexample will prove it's false. THINK ABOUT IT. You want {u,v} and {v,w} independent and {u,v,w} dependent. This isn't hard.
     
  16. Feb 5, 2010 #15

    Mark44

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    No linearly independent set can include the zero vector. So your example doesn't satisfy the hypothesis that {u, v} and {v, w} are lin. independent sets.

    Try Dick's suggestion of working with vectors in R2. Then it should be easy to find three vectors where {u, v} and {v, w} are linearly independent sets, while {u, v, w} is a linearly dependent set. In fact, it will be difficult NOT to find three vectors for which this is true.

    The statement you're working with is a sweeping generality for Rn, so if you can find a counterexample for a particular value of n -- say n = 2 -- then the entire statement is untrue.
     
  17. Feb 5, 2010 #16
    u = (0,2), v = (3,0), w = (0,1)
     
  18. Feb 5, 2010 #17

    Dick

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    Bingo!
     
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