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Math Help - Span of a subset

  1. #1
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    Span of a subset

    I just need to know how to start this... I'm looking through my notes but can't figure out how to begin. The wording is also a little bit confusing.

    Find a finite subset of each of the following subspaces for which the subspace is the span of the subset.

    {x bar is an element of R3 such that A(x bar) = theta bar} where A=
    1 1 2 1 -1 and theta bar is the zero vector in R3.
    2 2 1 2 -1
    3 3 2 3 -1
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  2. #2
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    Re: Span of a subset

    In this problem, the subspace in question is the nullspace of Ax=0. It is an elementary linear algebra problem to find the parametric solution for the system Ax=0. The vectors you get there will span the nullspace.
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  3. #3
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    Re: Span of a subset

    okay, what I did was the following:

    Let the matrix
    a
    b
    c be an element of U. Then the matrix:
    d
    e


    1 1 2 1 -1 a 0
    2 2 1 2 -1 X b = 0
    3 3 2 3 -1 c 0
    d
    e


    Then put that in an augmented matrix and row reduced to get

    1 1 0 1 0
    0 0 1 0 0
    0 0 0 0 1

    I got that e=0 and c=0 but now I don't know where to go or even if I'm doing this the right way
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  4. #4
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    Re: Span of a subset

    Then put that in an augmented matrix and row reduced to get

    1 1 0 1 0
    0 0 1 0 0
    0 0 0 0 1
    This translates to the equations a+b+d=0 and c=e=0 as you have noted. Then

    \bar{x}=\begin{bmatrix} a  \\ b \\c \\ d \\ e \end{bmatrix}=\begin{bmatrix} -b-d  \\ b \\0 \\ d \\0 \end{bmatrix}=b\begin{bmatrix} -1  \\ 1 \\0 \\ 0 \\0 \end{bmatrix}+d\begin{bmatrix} -1  \\ 0 \\0 \\ 1 \\0 \end{bmatrix}

    with b,d free. Since every \bar{x} is of this form, it follows that the two vectors I wrote down above span the nullspace of A.
    Thanks from widenerl194
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  5. #5
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    Re: Span of a subset

    Okay, this make so much sense now. But where did you get those two vectors from?
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  6. #6
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    Re: Span of a subset

    Nevermind I figured it out! Thanks so much!!
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  7. #7
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    Re: Span of a subset

    \begin{bmatrix} -b-d  \\ b \\0 \\ d \\0 \end{bmatrix}=\begin{bmatrix} -b  \\ b \\0 \\ 0 \\0 \end{bmatrix}+\begin{bmatrix} -d  \\ 0 \\0 \\ d \\0 \end{bmatrix}=b\begin{bmatrix} -1  \\ 1 \\0 \\ 0 \\0 \end{bmatrix}+d\begin{bmatrix} -1  \\ 0 \\0 \\ 1 \\0 \end{bmatrix}.

    Does this make more sense?
    Thanks from widenerl194
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