Results 1 to 9 of 9

Thread: Transformation

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Transformation

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

    Can you show in steps as I find this topic confusing. Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Still breathing after this morning huh ?

    $\displaystyle T \begin{pmatrix} x+y \\y \\y+z \end{pmatrix}
    = \begin{pmatrix} x+y \\x + 2y \\x +2y+z \end{pmatrix} $

    Bobak
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello !

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

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


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

    $\displaystyle x+y=m$
    $\displaystyle y=n$
    $\displaystyle y+z=p$

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


    Got it ? =)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by Moo View Post
    $\displaystyle y=n$
    How did you get $\displaystyle y=n$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Air View Post
    How did you get $\displaystyle y=n$?
    Ow ^^

    $\displaystyle
    T: \begin{pmatrix} {\color{red}m} \\ {\color{blue}n} \\ {\color{green}p} \end{pmatrix} \mapsto \begin{pmatrix} m \\ m+n \\ m+p \end{pmatrix}
    $

    $\displaystyle
    T(\textbf{v})=T \begin{pmatrix} {\color{red}x+y} \\ {\color{blue}y} \\ {\color{green}y+z} \end{pmatrix}
    $
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Can you find the 3 by 3 matrix that represents the transformation ?


    Bobak
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by bobak View Post
    Can you find the 3 by 3 matrix that represents the transformation ?


    Bobak
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    As $\displaystyle T$ is a liner transformations there exists and matrix such that $\displaystyle A \textbf{v} = T( \textbf{v} )$ to find this matrix you need to find the images of the vectors $\displaystyle i, j$ and $\displaystyle k$ under $\displaystyle T$. This all lies down the understanding how matrix multiplication works.

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

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

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

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

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

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

    to check this is correct just multiple the two.

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


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

    You following?


    Bobak
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by bobak View Post
    You following?


    Bobak

    Yes! Much better than the book too.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Transformation
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Nov 4th 2011, 06:42 AM
  2. Transformation
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Mar 8th 2011, 06:12 PM
  3. Transformation help
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Sep 27th 2010, 09:04 PM
  4. Transformation
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: Oct 31st 2009, 01:33 AM
  5. Transformation
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Nov 29th 2008, 12:42 AM

Search Tags


/mathhelpforum @mathhelpforum