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Math Help - Transformation

  1. #1
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    Transformation

    If T: \begin{pmatrix} x \\y \\z \end{pmatrix} ->\begin{pmatrix} x \\x+y \\x+z \end{pmatrix} then what does T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix} equal?

    Can you show in steps as I find this topic confusing. Thanks in advance.
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  2. #2
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    Still breathing after this morning huh ?

    T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}<br />
 =   \begin{pmatrix} x+y \\x + 2y \\x +2y+z \end{pmatrix}

    Bobak
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  3. #3
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    Hello !

    Quote Originally Posted by Air View Post
    If T: \begin{pmatrix} x \\y \\z \end{pmatrix} ->\begin{pmatrix} x \\x+y \\x+z \end{pmatrix} then what does T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix} equal?

    Can you show in steps as I find this topic confusing. Thanks in advance.
    I'd suggest you to rewrite that because using the same x & y & z stuff makes things complicated !

    T: \begin{pmatrix} m \\ n \\ p \end{pmatrix} \mapsto \begin{pmatrix} m \\ m+n \\ m+p \end{pmatrix}


    T(\textbf{v})=T \begin{pmatrix} x+y \\ y \\ y+z \end{pmatrix}

    x+y=m
    y=n
    y+z=p

    ---> m=x+y
    m+n=(x+y)+y=x+2y
    m+p=(x+y)+(y+z)=x+2y+z


    Got it ? =)
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  4. #4
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    Quote Originally Posted by Moo View Post
    y=n
    How did you get y=n?
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  5. #5
    Moo
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    Quote Originally Posted by Air View Post
    How did you get y=n?
    Ow ^^

    <br />
T: \begin{pmatrix} {\color{red}m} \\ {\color{blue}n} \\ {\color{green}p} \end{pmatrix} \mapsto \begin{pmatrix} m \\ m+n \\ m+p \end{pmatrix}<br />

    <br />
T(\textbf{v})=T \begin{pmatrix} {\color{red}x+y} \\ {\color{blue}y} \\ {\color{green}y+z} \end{pmatrix}<br />
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  6. #6
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    Can you find the 3 by 3 matrix that represents the transformation ?


    Bobak
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  7. #7
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    Quote Originally Posted by bobak View Post
    Can you find the 3 by 3 matrix that represents the transformation ?


    Bobak
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  8. #8
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    As T is a liner transformations there exists and matrix such that A \textbf{v} = T( \textbf{v} ) to find this matrix you need to find the images of the vectors  i, j and k under T. This all lies down the understanding how matrix multiplication works.

    I do this row by row, you see that the i vector is preserved under the transformation, so the first row is one i combined linearly with no j's or k's

    so the first row is \left(\begin{array}{ccc}1 & 0 & 0 \\? & ? & ? \\? & ? & ?\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)

    for the next row you combining one i with one j and no k's

    so we have \left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\? & ? & ?\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)

    Then similarly for the next row your combining 1 i with 1 k.

    giving \left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right)

    to check this is correct just multiple the two.

    \left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right)\left(\begin{array}{c}x \\y \\z\end{array}\right) = \begin{pmatrix} x \\x+y \\x+z \end{pmatrix}


    so T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix} is given by \left(\begin{array}{ccc}1 & 0 & 0 \\1 & 1 & 0 \\1 & 0 & 1\end{array}\right) \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}

    You following?


    Bobak
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  9. #9
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    Quote Originally Posted by bobak View Post
    You following?


    Bobak

    Yes! Much better than the book too.

    Thanks!
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