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Math Help - Basis for space which contains the image of a matrix

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    Basis for space which contains the image of a matrix

    Find a basis for \mathbb{R}^3 which contains a basis of im(A), where

    A=\begin{pmatrix}1&2&3&4\\ 2&-4&6&-2\\-1 & 2& -3 & 1 \end{pmatrix}

    After row reduction

    U\sim\begin{pmatrix}1&0&3&1.5\\ 0&1&0&1.25\\0 & 0& 0 & 0 \end{pmatrix}

    So I've found the basis for the image of A

    \textrm{basis(im(A))}=\left\{ \begin{pmatrix}1\\2\\-1\end{pmatrix}, \begin{pmatrix}2\\-4\\2\end{pmatrix}\right\}

    How would I find the basis for \mathbb{R}^3 which contains the image of A



    EDIT: Ok so I've just seen the way my textbook does it, and they row reduce the matrix A alongside the identity matrix and then choose the columns from A and the identity matrix (which are the pivot columns in the reduced form).

    Why does this method work?

    (A|I)=\begin{pmatrix}1&2&3&4&1&0&0\\ 2&-4&6&-2&0&1&0\\-1 & 2& -3 & 1&0&0&1 \end{pmatrix}
    Last edited by acevipa; October 8th 2010 at 04:23 AM.
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