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Describing span(x) geometrically

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    There is a vector space with real entries, in ℝ3 with the subset X = [tex]
    \begin{pmatrix}
    2\\
    -1\\
    -3
    \end{pmatrix}\\
    ,
    \begin{pmatrix}
    4\\
    0\\
    1
    \end{pmatrix}\\
    ,
    \begin{pmatrix}
    0\\
    2\\
    7
    \end{pmatrix}

    [/tex]
    and you have to describe span(x) geometrically.

    2. Relevant equations
    In answering this question, I found the span of the subset through [tex]a
    \begin{pmatrix}
    2\\
    -1\\
    -3
    \end{pmatrix}\\
    + b
    \begin{pmatrix}
    4\\
    0\\
    1
    \end{pmatrix}\\
    + c\begin{pmatrix}
    0\\
    2\\
    7
    \end{pmatrix}=
    \begin{pmatrix}
    x\\
    y\\
    z
    \end{pmatrix}
    [/tex]

    3. The attempt at a solution
    This formed the matrix:
    [tex]
    \begin{pmatrix}
    2 & 4 & 0 & | & x\\
    -1 & 0 & 2 & | & y\\
    -3 & 1 & 7 & | & z
    \end{pmatrix}
    [/tex]
    Using row operations, I then made the matrix into reduced row echelon form and it was non-trivial with the final result being:
    [tex]
    \begin{pmatrix}
    1 & 2 & 0 & | & x/2\\
    0 & 1 & 1 & | & x/4+y/2\\
    0 & 0 & 0 & | & -x/4-7y/2+z
    \end{pmatrix}
    [/tex]
    So, we can then interpret the span geometrically as the plane in ℝ3 with the equation [itex]-1/4x - 7y/2 + z=0[/itex]
    Is this right?
     
    Last edited: Mar 24, 2013
  2. jcsd
  3. Mar 24, 2013 #2

    vela

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    You made a mistake somewhere. Those three vectors are linearly independent, so you shouldn't end up with a row of zeros. But if that were the correct matrix, then yes, the span would be the plane you said.
     
  4. Mar 24, 2013 #3
    Sorry, I didn't realise but I put the first entry as 3 instead of -3 and now I have corrected it. Is that correct now?
     
  5. Mar 24, 2013 #4

    vela

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    Yes, that matches what I get.
     
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