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Linear Algebra (Span)

  1. Jul 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that B={(1,-1,-1),(1,0,1),(0,-1,1)} spans R^3.

    (Actually, the problem asks to show that B is a basis for R^3. This would require that I prove linear independence AND that it spans R^3. The first is easy, but I'm not sure about the second.)

    2. Relevant equations


    3. The attempt at a solution

    I know how to prove that these vectors are linearly independent, but does that prove that B spans real space?
  2. jcsd
  3. Jul 19, 2009 #2
    It has to do with the notion of "dimension" ... If R^3 has dimension 3, and you have a linearly independent set of 3 vectors, then that set spans R^3.
  4. Jul 19, 2009 #3
    Oh, I see. So the fact that they are 3 linearly independent vectors means that they do span R^3. Got it. Thanks!
  5. Jul 19, 2009 #4


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    If you have that theorem:
    "A basis for vector space has three properties:
    1) It spans the space
    2) It is a set of independent vectors
    3) The number of vectors in the basis is equal to the dimension of the vector space"
    and you know that R3 has dimension 3, then you can say "this is a set of 3 independent vectors so it spans R3."

    If you do not have all of those facts, you could do it directly:
    Show how to find, for any real numbers, x, y, z, real numbers, a, b, c, such that
    a(1,-1,-1)+ b(1,0,1)+ c(0, -1, 1)= (x,y,z). That is the same as showing that the three equations, a+ b= x, -a- c= y, -a+ b+ c= z, has a solution (not necessarily unique) no matter what x, y, and z are.
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