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Thread: 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:

    $\displaystyle \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}$
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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:

    $\displaystyle \begin{pmatrix}1 & -1 & 2 & 1 & 0\\2 & 0 & 1 & -1 & 3\\5 & -1 & 3 & 0 & 3\\4 & -2 & 5 & 1 & 3\\1 & 3 & -4 & -5 & 6\end{pmatrix}$
    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
    $\displaystyle
    \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}
    $
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  4. #4
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    Quote Originally Posted by antman View Post
    Reduced matrix
    $\displaystyle
    \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}
    $
    Given a $\displaystyle m\times n $ matrix $\displaystyle A$ the nullspace is the set of all $\displaystyle \bold{x}\in \mathbb{R}^n$ so that $\displaystyle A\bold{x} = \bold{0}$ where $\displaystyle \bold{0}$ is zero vector in $\displaystyle \mathbb{R}^m$. To find the nullspace you basically need to solve the homogenous equation $\displaystyle A\bold{x} = \bold{0}$. If you convert this problem in Gaussian-Jordan elimination, it means you need to solve the equation $\displaystyle (A|\bold{0})$ where $\displaystyle \bold{0}$ 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):
    $\displaystyle
    \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}
    $
    This matrix is telling us that (the last two coloums do not tell us anything):
    $\displaystyle -x_1 = 0 $
    $\displaystyle x_2 - 3x_4 + 6x_5 = 0$
    $\displaystyle x_3 -x_4 + 3x_5 = 0 $

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

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

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

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