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Thread: Linear transformation

  1. #1
    Senior Member
    Jan 2009

    Linear transformation

    I don't understand this definition.

    In a linear transformation, 0,0) \to (0,0)" alt="f0,0) \to (0,0)" />. If 0,0)" alt="f0,0)" /> (the "to" symbol with the "not" diagonal cross in it) (0,0), the mapping is not a linear transformation.

    Now I have to determine if each mapping that moves (x,y) to the given point is a linear transformation. I don't get how it works.

    Why is moving (x,y) to (0,x) is a linear transformation and moving it to (x,-1) is not?

    Another question I don't get.

    Given the linear transformation \left( \begin{array}{lcr} x' \\y' \end{array} \right) = \left( \begin{array}{lcr} 2&1 \\ 3&4 \end{array} \right)=\left( \begin{array}{lcr} x \\ y \end{array} \right), determine the matrix of the linear transformation which moves points (1,0) and (0,1) to points (1,2) and (3,4), respectively.

    How do you get the answer \left( \begin{array}{lcr} 1&3 \\ 2&4 \end{array} \right)?
    Last edited by chengbin; Aug 16th 2009 at 07:10 PM.
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  2. #2
    ynj is offline
    Senior Member
    Jul 2009
    fis linear transformation iff
    f(\alpha x)=\alpha f(x)
    1: f((x,y)+(x_1,y_1))=f((x+x_1,y+y_1))=(x+x_1,0)=f((x  ,y))+f((x_1+y_1))
    f(\alpha(x,y))=f((\alpha x,\alpha y))=\alpha x=\alpha f((x,y))
    2: f(\alpha(x,y))=f((\alpha x,\alpha y))=(\alpha x,-1)
    but \alpha f(x,y)=(\alpha x,-\alpha)

    f\left(\begin{array}{cc}x\\y\end{array}\right)=xf\  left(\begin{array}{cc}1\\0\end{array}\right)+yf\le  ft(\begin{array}{cc}0\\1\end{array}\right)=x\left(  \begin{array}{cc}1\\2\end{array}\right)+y\left(\be  gin{array}{cc}3\\4\end{array}\right)=\left(\begin{  array}{cc}1&3\\2&4\end{array}\right)\left(\begin{a  rray}{cc}x\\y\end{array}\right)
    Last edited by ynj; Aug 16th 2009 at 07:39 PM.
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