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Linear Algebra Span

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data

    Determine whether the set spans ℜ3. If the set does not span ℜ3 give a geometric description of the subspace it does span.

    s = {(1, 0, 3), (2, 0, -1), (4, 0, 5), (2, 0, 6)}

    2. Relevant equations


    3. The attempt at a solution
    I am having trouble with the second part of this problem (geometric shape). First off here is what I did to show that S does not span ℜ3.

    First I define A, B, C, D as scalars and say;
    A + 2B + 4C + 2D = X
    0 + 0 + 0 + 0 = Y
    3A - B + 5C + 6D = Z

    I set up the augmented matrix

    | 1 2 4 2 : X |
    | 0 0 0 0 : Y |
    | 3 -1 5 6 : Z |

    rref (using ti89 titanium) ~
    | 1 0 2 2 : 0 |
    | 0 1 1 0 : 0 |
    | 0 0 0 0 : 1 |

    I can conclude from this result that S does not span ℜ3.

    Now for the geometric shape.

    I immediately see that the first and last vectors in the set are scalar multiples which means that they aer parallel.

    Next I note that for all vectors y=0. My first thought is that the geometric shape is the plane formed by the x z axis to check I create a new set in ℜ2 on the xz plane

    W = {(1, 3), (2, -1), (4, 5), (2, 6)}

    Then repeat the above process to see is W spans xz plane

    My augmented matrix becomes
    | 1 2 4 2 : X |
    | 3 -1 5 6 : Z |

    rref (using ti89 titanium) ~

    | 1 0 2 2 : (X + 2Z) / 7 |
    | 0 1 1 0 : (3X - Z) / 7 |

    From this I can say that W spans xz in ℜ2.

    Is this correct? If not, where am I lacking in my understanding? My big hold up is the only example we have seen is a line. I am not convinced that this is not a line in space.
     
    Last edited: Feb 20, 2016
  2. jcsd
  3. Feb 20, 2016 #2

    Mark44

    Staff: Mentor

    This is more work than you need to do. The fourth vector is a multiple of the first, so it suffices to check whether the first three vectors in your set are linearly independent.
    How did Z turn into 1?
    The x-z plane is a subset of ##\mathbb{R}^3##, so the set of four vectors spans a plane in ##\mathbb{R}^3##.
    Right. The vectors span a plane in space.
     
  4. Feb 20, 2016 #3
    Z turns to 1 because of the way a TI89 handles rref of an augmented matrix.

    Thank you for your help.
     
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