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Showing linear independence, correct logic?

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let u and v be two nonzero vectors in R^2. If there is no c E R such that u = cv, show that {u, Bv} is a basis of R^2 and that R^2 is a direct sum of the subspaces generated by U = <u> and V = <v> respectively.

    2. Relevant equations

    Clearly, u and v are linearly independent. So...

    3. The attempt at a solution

    I want to show r1u + r2Bv = 0 implies r1 = r2 = 0. If B=0 (since v cannot equal 0), then the second term drops out and r1u =0. Because u is non-zero, r1=0. Back to the original statement, if r1=0, then the first term drops out and r2Bv = 0. Since v is non-zero, and assume B is non zero as well, r2 = 0. But what if B IS zero? I'm getting a little confused by this logic...
  2. jcsd
  3. Feb 17, 2009 #2


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    If B=0 then {u,Bv}={u,0} is NOT a basis. So, yes, you can assume B is not zero. So assume r1*u+r2*B*v=0. If r1=0 then r2*B*v=0. What can you conclude about r2? If r1 is NOT equal to zero then u=-r2*B*v/r1. What's wrong with that?
  4. Feb 17, 2009 #3
    Then u is no longer non-zero! It makes sense, now. Thanks very much!
  5. Feb 17, 2009 #4


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    No, you aren't quite getting it. The problem with writing u=-r2*B*v/r1 is that you were given that there is no number c such that u=c*v.
  6. Feb 17, 2009 #5
    Oh ok, I see it. I should have read the question more carefully. Thanks for all the help!
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