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Math Help - Dimension of a subspace of R^4

  1. #1
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    Dimension of a subspace of R^4

    My question asks me to consider a matrix A
    \left( \begin{array}{cccc} 1 & 2 & 2 & 3 \\ 2 & 5 & 4 & 8 \\ -1 & -3 & -2 & -5 \\ 0 & 2 & 0 & 4 \end{array}\right)
    Then find a basis for the nullspace of A, and hence then dimension of the nullspace;
    which I find to be \{ \left( \begin{array}{c} -2 \\ 0 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right) \}.
    Since there are two basis vectors the dimension is 2, right?

    My problem then is "Using the rank-nullity theorem or otherwise, determine the dimension of the subspace of R^{4} spanned by the four columns of A".

    R-N states that dimension null = #cols - rank...
    So dimension null =2 , #cols = 4, so I'm guessing rank = 2...

    I'm confused by the mention of the "dimension of the subspace of R^{4}" though.
    Is the answer they're looking for simply 2? And how ought I to phrase this to explain my answer, rather than merely subtracting one number from another?


    Thanks in advance
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  2. #2
    Senior Member
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    Don't overcomplicate things

    Yes, you've gotten the answer already. R^4 merely refers to the 4 dimensional vector space with each vector component being in the reals. Each one of your rows or columns is a point in that space.
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