1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook