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Finding a basis for a subspace

  1. Nov 4, 2006 #1
    Let U be a proper subspace of R^4 and let it be given by the equations:

    1) x1+x2+x3+x4=0
    2) x1-x2+2x3+x4=0

    how do i find a basis for this subspace???

    I got that (0,1,2,0) is one of the basis vectors since x2=2x3, therefore whatever we pick for x2, x3 will be twice that value.

    i also got that x4=-x1-1.5x3, but does this require two more basis vectors or one?

    ie, i'm asking, it seems that every x value can be determined once x1 and x2 are determined, therefore it should have 2 basis vectors, but i cant quite put it into that form
     
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 4, 2006 #2
    I think that you have 2 leading variables and 2 dependent variables (parameters), i.e. your solution subspace is in dimension 2, with 2 vectors generating it.
    reduced row echelon form gives you
    x1 + 1.5x3 + x4 = 0
    x2 - 0.5x3 = 0

    from the first equation you have
    x1 = - 1.5x3 - x4
    from the second one you have
    x2 = 0.5x3

    and x3, x4 are parameters, so you can let x3 = t and x4 = s

    Now

    (x1,x2,x3,x4) = (-1.5t+s, 0.5t, t, s) =
    t(-1.5, 0.5, 1, 0) + s(1, 0, 0, 1)

    so the vectors you need to find are:

    (-1.5, 0.5, 1, 0) and (1, 0, 0, 1).

    I hope I didn't mess up with the numbers...
     
  4. Nov 4, 2006 #3

    Office_Shredder

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    It's 2*x2 = x3 actually
     
  5. Nov 4, 2006 #4
    ok i found that basis, now how do i find an orthonormal basis?????
     
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