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Math Help - Image/Column space of a Matrix

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    Image/Column space of a Matrix

    Hey there!
    I'm really struggling with this question . The solutions on it don't seem to match up with everything i've read/googled!

    Now, I understand how to get the kernel. But I have no idea what they are doing to get the image basis!

    I've looked around on wikipedia etc and read that you take the corresponding columns to the pivots - which to me would mean the basis would be: (1 0 3), (-2, 2, -4)...Not (1; 2; 1); (2; 0; 6)

    Please see the pdf attachment for the solutions.

    Any help would be appreciated!

    Thankyou!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Image/Column space of a Matrix

    Quote Originally Posted by AshleyT View Post
    Now, I understand how to get the kernel. But I have no idea what they are doing to get the image basis!
    In general, consider A:\mathbb{R}^m\to\mathbb{R}^n and suppose \{u_1,u_2,\ldots,u_r\} is a basis of \ker A and B=\{u_1,u_2,\ldots,u_r,u_{r+1},\ldots,u_m\} is a basis of \mathbb{R}^m .Then, \{Au_{r+1},\ldots,Au_m\} is a basis of \textrm{Im}A . Try to prove it (it is not difficult) and ask if you have any doubt.
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