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Math Help - null space

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

    Let A=\begin{bmatrix}<br />
  1 & 2 & 3 & 6\\<br />
  2 & 6 & 9 & 18\\<br />
  1 & 2 & 6 & 12<br />
\end{bmatrix}.

    Find a basis for the null space of A.

    I got the answer \begin{bmatrix}<br />
0\\<br />
0\\<br />
-2\\<br />
1<br />
\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 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 \begin{bmatrix}<br />
0\\<br />
0\\<br />
-2t\\<br />
t<br />
\end{bmatrix}<br />
where t is a real number?
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  5. #5
    MHF Contributor

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    Yes, or, to phrase it slightly differently, 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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