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Basis vectors and ortho solution spaces

  1. Oct 16, 2008 #1


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    I've got two homogenous equations: 3x + 2y + z - u = 0 and 2x + y + z +5u = 0. I'm trying to find a basis for these solutions. The solution vector x [x, y, z, u] is a solution if and only if it is orthogonal to the row vectors, in this case a and b ([3, 2, 1, -1], [2, 1, 1, 5)]. My understanding is that these row vectors span the space orthogonal to x, and they thus can be used as a basis. The answer given by the text I'm using gives for a basis these two vectors [1, -1, -1, 0] and [0, 6, -11, 1]. How were these basis vectors derived? I tried doing a linear combination of a and b to get the textbook answer but I couldn't get that to work. Help!

  2. jcsd
  3. Oct 16, 2008 #2


    Staff: Mentor

    I get a different basis, namely [-1, 1, 1, 0] and [-11, 17, 0, 1], but your vectors also are a basis for the solution space defined by the two equations.

    How did I get mine? I row reduced the 2 X 4 matrix whose rows were the coefficients of x, y, z, and u.

    I ended up with
    [1 0 1 11]
    [0 1 -1 -17]

    Call the reduced matrix A. Then A [x y z u]^T = 0
    From this you get [x y z u]^T = z[-1 1 1 0]^T + u[-11 17 0 1]^T

    The solution space is the 2-D subspace of R^4 spanned by/made up of linear combinations of the vectors you show or the vectors I show. The solution space is a hyperplane in R^4. Any two vectors that aren't parallel and that lie in this hyperplane would make a basis.

  4. Oct 16, 2008 #3


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    Hello EWW! :smile:

    You're getting confused …

    they can't be used as a basis for a space they are orthogonal to. :wink:
  5. Oct 17, 2008 #4


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    As tiny-tim said, you are confusing the "solution space" and the space orthogonal to the solutions space. Any linear combination of a and b will be orthogonal to the solution space, not in the solution space.

    EWW here's how I would do the problem: The solution space consists of vectors <x, y, z, u> satisfying 3x + 2y + z - u = 0 and 2x + y + z +5u = 0. Subtract the second equation from the first to eliminate z: x+ y- 6u= 0 so x= -y+ 6u. Put that back into the first equation: 3(-y+ 6u)+ 2y+ z- u= -y+ 17u+ z= 0 so z= y- 17u. Take y= 1, u= 0. Then x= -1 and z= 1. <-1, 1, 0, 1> is in the solution set. Take y= 0, u= 1. Then x= 6 and z= -17. <6, 0, -17, 1> is also in the solution set and the two vectors form a basis.

    If, instead, I had solved the equation x+ y- 6u= 0 for y, y= -x+ 6u, and put that back into the first equation I would have got z= -x-11u so that I have y and z in terms of x and u. Taking x= 1, u= 0 gives <1, -1, -1, 0> and taking x= 0, u= 1 gives <0, 6, -11, 1> the solution in your textbook.

    If you were to solve for x and y in terms of z and u, then take z=1, u=0 and z=0, u= 1, you would get Mark44's solution.
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