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Linear Algebra - independent vectors spanning R^3

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that S={u,v,w} is a basis of R3.
    a) Is {u-v,u+v} linearly independent? Why or why not?
    b) Does {u+v,u-v} span R3? Why or why not?

    2. Relevant equations

    3. The attempt at a solution
    a) Because we know that u,v,w are linearly independent and span R3, c1u+c2v+c3w = 0.
    To test u+v, u-v, consider a(u+v)+b(u-v)=0. So, au+av+bu-bv=0, so u(a+b)+v(a-b)=0. Since u and v are linearly independent, a+b=0 and a-b=0. So 2b=0, so b=0, so a=0. So u+v, u-v are linearly independent.
    b) Because there are only 2 vectors, u and v, the rank can only be 2, so it cannot span R3.

    I'm pretty sure I got a) correct, but I'm not sure about b). Thanks!
  2. jcsd
  3. Oct 24, 2011 #2


    Staff: Mentor

    Both a and b are right.
  4. Oct 24, 2011 #3


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    You got the first one right. The second one is essentially correct, but your terminology is off. When you say "rank", you're referring to a matrix or a linear transformation; here you have just two vectors. The span of {u+v, u-v} is a subspace of dimension 2.
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