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Thread: null space

  1. #1
    Senior Member Sambit's Avatar
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    Question null space

    Let $\displaystyle A=\begin{bmatrix}
    1 & 2 & 3 & 6\\
    2 & 6 & 9 & 18\\
    1 & 2 & 6 & 12
    \end{bmatrix}$.

    Find a basis for the null space of A.

    I got the answer $\displaystyle \begin{bmatrix}
    0\\
    0\\
    -2\\
    1
    \end{bmatrix}$

    Is it correct?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yes it is correct. By the way the matrix has Rank 3 so the dim of the nullspace must be 1 and you found 1 vector in the null space.
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  3. #3
    MHF Contributor

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    So that, strictly speaking, you have NOT yet solved the problem. The "null space" is a one-dimensional subspace of $\displaystyle R^4$ and you have found one vector in that space. How would you represent any vector in the null space?
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  4. #4
    Senior Member Sambit's Avatar
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    any vector in the null space.....do you mean $\displaystyle \begin{bmatrix}
    0\\
    0\\
    -2t\\
    t
    \end{bmatrix}
    $ where $\displaystyle t$ is a real number?
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  5. #5
    MHF Contributor

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    Yes, or, to phrase it slightly differently, $\displaystyle t\begin{bmatrix}0 \\ 0 \\ -2 \\ 1\end{bmatrix}$, the set of all real multiples of your original vector= the subspace spanned by that vector.
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