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Math Help - linear transformation

  1. #1
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    linear transformation

    Let  L: R^{2}->R^{2} be the linear transformation defined by L(\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix})=\begin  {pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}a_{1}  \\a_{2}\end{pmatrix}
    a) Is \begin{pmatrix}1\\2\end{pmatrix} in ker L?
    b) Is \begin{pmatrix}2\\-1\end{pmatrix} in ker L?
    c) Is \begin{pmatrix}3\\6\end{pmatrix} in range L?
    c) Is \begin{pmatrix}2\\3\end{pmatrix} in range L?

    Thank you for your help!
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  2. #2
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    Quote Originally Posted by antman View Post
    Let  L: R^{2}->R^{2} be the linear transformation defined by L(\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix})=\begin  {pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix}a_{1}  \\a_{2}\end{pmatrix}
    a) Is \begin{pmatrix}1\\2\end{pmatrix} in ker L?
    b) Is \begin{pmatrix}2\\-1\end{pmatrix} in ker L?
    c) Is \begin{pmatrix}3\\6\end{pmatrix} in range L?
    c) Is \begin{pmatrix}2\\3\end{pmatrix} in range L?

    Thank you for your help!
    Just compute L  \left[ \begin{pmatrix}1\\2\end{pmatrix} \right] and see if it is \begin{pmatrix}0\\0\end{pmatrix} for part a. Then do the same for the other ones.
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  3. #3
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    For range, I would just have to find a vector v such that L(v)=w, so w is in the range of L. So a is not in kernel L but b is and c is in range L but d is not? For the range, I set up a 2x3 matrix with the vector in question as column 3 then found rref. Is this correct? d would not be in range because in the reduced matrix -1 does not equal 0.
    Last edited by antman; May 1st 2009 at 12:31 PM.
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  4. #4
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    Quote Originally Posted by antman View Post
    For range, I would just have to find a vector v such that L(v)=w, so w is in the range of L. So a is not in kernel L but b is and c is in range L but d is not? For the range, I set up a 2x3 matrix with the vector in question as column 3 then found rref. Is this correct? d would not be in range because in the reduced matrix -1 does not equal 0.
    Notice that \begin{pmatrix}1&2\\2&4\end{pmatrix}\begin{pmatrix  }a_{1}\\a_{2}\end{pmatrix} = a_1 \begin{pmatrix}1\\2\end{pmatrix} + a_2 \begin{pmatrix}2\\4\end{pmatrix}.

    Thus, to be in the range it needs to be a linear combination of \left\{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}2\\4\end{pmatrix} \right\}.

    There are of course many other ways to solve this problem, I just happen to like the approach that uses linear combinations. I looks like that approach uses least amount of computation since these are just 2\times 2 matrices.
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  5. #5
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    I understand that. Thank you! The way I did it is still correct then, just more work?
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  6. #6
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    Quote Originally Posted by antman View Post
    The way I did it is still correct then, just more work?
    Yes, setting up the matrix and reducing it by Gauss-Jordan elimination to see if it gets solved will tell you how to find a v so that L(v) = w. If you are more comfortable doing it that way then do it that way.
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