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Homework Help: Basis and dimension of the solution space

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Find (a) a basis for and (b) the dimension of the solution space of the homogeneous system of equations.

    x - 2y + z = 0
    y - z + w = 0
    x - y + w = 0

    2. Relevant equations

    3. The attempt at a solution
    (a)
    [1 -2 1 0] => [1 0 -1 2]
    [0 1 -1 1] => [0 1 -1 1]
    [1 -1 0 1] => [0 0 0 0]
    basis = {<1,0,-1,2>, <0,1,-1,1>}

    (b) for b i make
    x = z - 2w
    y = z - w

    would i set w = s and z = t?
    if so.
    [x] = [ t - 2s] = [1] + [-2]
    [y] = [ t - s] = t[1] + s[-1]
    [z] = [ t ] = [1] + [0]
    [w] = [ s ] = [0] + [1]

    so the dimension would be the number of vectors
    so dimension = 2?

    im uncertain about that dimension part
     
  2. jcsd
  3. May 25, 2010 #2

    LCKurtz

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    I didn't check your arithmetic, but assuming it is OK try writing your solution like this:

    [tex]\left ( \begin{array}{c}x\\y\\z\\w\end{array}\right) =
    t\left ( \begin{array}{c}1\\1\\1\\0\end{array}\right)
    +s\left ( \begin{array}{c}-2\\-1\\0\\1\end{array}\right)[/tex]
    and the answer should become apparent.
     
  4. May 25, 2010 #3

    vela

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    You found that the solution of the system is given by

    [tex]\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = t\begin{bmatrix}1\\1\\1\\0\end{bmatrix}+s\begin{bmatrix}-2\\-1\\0\\1\end{bmatrix}[/tex]

    The vectors multiplying t and s are a basis for the solution space. Your answer to (a) is the basis for the row space of the matrix, not the solution space.

    The dimension is just the number of vectors in the basis, so in this case, it's 2.
     
  5. May 25, 2010 #4
    oh so for (a) i got it mixed up with the basis for the row space
    the correct answer is { <1,1,1,0>, < -2,-1,0,1>}
     
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