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I got problem in understanding the Im finding method

  1. Feb 9, 2008 #1
    i am puzzled about the method of finding the Im

    in one axample i see the method of thaking the column withound the term of the zero
    row
    and exract the vectors from the column

    the other way is taking the "b" term that make the whole row zeros
    and extract from it

    the third way i making the coulmns zeros

    i am realy confused about the method of finding Im
    because there is 2 example the contaridct each other

    http://img504.imageshack.us/my.php?image=img8271ip1.jpg
     
  2. jcsd
  3. Feb 9, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You seem to be consistently having trouble distinguishing between "kernel" and "image". If A is a linear transformation from vector space U to vector space V, then the kernel of A is a subspace of U and the image of A is a subspace of V.

    In the problem before, where you were seeking the kernel, you found A0, a vector in V, when you should have been solving Au= 0 to get a vector in U.

    Now, you appear to be solving the equation
    [tex]\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array}\right][/tex]
    when you should be calculating
    [tex]\left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 1\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} x+ y \\ 0 \\ 2x+ z\end{array}\right][/tex]

    It is obvious that the the second component is always 0 and is easy to see that, since x, y, z are arbitrary, that the first and third components can be any numbers: the image is the set of vectors <x, 0, z> or, equivalently, the subspace with basis <1, 0, 0> and <0, 0, 1>.
     
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