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Math Help - basis for null space of matrix

  1. #1
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    basis for null space of matrix

    How do you find a basis for the null space of the matrix:

    \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}
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  2. #2
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    Quote Originally Posted by antman View Post
    How do you find a basis for the null space of the matrix:

    \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}
    The first thing you should do is reduce this by Gaussian-Jordan elimination.
    After you do this I tell you what to do next.
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  3. #3
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    Reduced matrix
     <br />
\begin{pmatrix} -1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & -3 & 6\\0 & 0 & 1 & -1 & 3\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{pmatrix}<br />
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  4. #4
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    Quote Originally Posted by antman View Post
    Reduced matrix
     <br />
\begin{pmatrix} -1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & -3 & 6\\0 & 0 & 1 & -1 & 3\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{pmatrix}<br />
    Given a m\times n matrix A the nullspace is the set of all \bold{x}\in \mathbb{R}^n so that A\bold{x} = \bold{0} where \bold{0} is zero vector in \mathbb{R}^m. To find the nullspace you basically need to solve the homogenous equation A\bold{x} = \bold{0}. If you convert this problem in Gaussian-Jordan elimination, it means you need to solve the equation (A|\bold{0}) where \bold{0} was an attached coloumn of zeros (corresponding to the fact that we are solving a homogenous system). Therefore, you end up with (notice the new coloum of zeros):
    <br />
\begin{pmatrix} -1 & 0 & 0 & 0 & 0&0\\0 & 1 & 0 & -3 & 6&0\\0 & 0 & 1 & -1 & 3&0\\0 & 0 & 0 & 0 & 0&0\\0 & 0 & 0 & 0 & 0&0\end{pmatrix}<br />
    This matrix is telling us that (the last two coloums do not tell us anything):
    -x_1 = 0
    x_2  - 3x_4 + 6x_5 = 0
    x_3 -x_4 + 3x_5 = 0

    Thus,
    x_1=0
    x_2 = 3x_4 - 6x_5
    x_3 = x_4 - 3x_5.

    If we let r,s\in \mathbb{R} then:
    x_1=0
    x_2 = 3r - 6s
    x_3 = r - 3s
    Are always in the nullspace.

    Thus, the nullspace is, \left\{ \begin{bmatrix}0\\3r-6s\\r-3s\\r\\s \end{bmatrix} \right\}

    Can you find a basis now?
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  5. #5
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    Would the basis just be {  <br />
\begin{pmatrix}0\\3\\1\\1\\0\end{pmatrix},\begin{p  matrix}0\\-6\\-3\\0\\1\end{pmatrix}<br />
}
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