Results 1 to 3 of 3

Math Help - Linear Transformation: finding the image.

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    10

    Linear Transformation: finding the image.

    Hi there.

    I've a doubt in getting the image of a linear transformmation.

    given this LT: R^2 \rightarrow R^3\ T(x,y)=(x-2y,y-x,3x)

    the book I'm using always attempts to do as following:

    (x-2y,y-x,3x)=(a,b,c) and solve the system \begin{pmatrix} +1-2=a \\-1+1=b \\+3+0=c\end{pmatrix}

    \Rightarrow \\ c-3a+6a+6b = 0
     c+3a+6b = 0\\ \Rightarrow base=\{(1,0,-3);(0,1,-3)\}


    I wonder if I get an equivalent result if I do it the following manner:

    (x-2y,y-x,3x) = x(1,-1,3)+y(-2,1,0) \Rightarrow base=\{(1,-1,3);(-2,1,0)\}


    of course, in this case I'll need to check if are them linearly independent if I want to get a base.

    thanks in advice
    VIKKO
    Last edited by viko; October 4th 2009 at 09:36 AM. Reason: latex mistake
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,441
    Thanks
    1862
    Quote Originally Posted by viko View Post
    Hi there.

    I've a doubt in getting the image of a linear transformmation.

    given this LT: R^2 \rightarrow R^3\ T(x,y)=(x-2y,y-x,3x)

    the book I'm using always attempts to do as following:

    (x-2y,y-x,3x)=(a,b,c) and solve the system \begin{pmatrix} +1-2=a \\-1+1=b \\+3+0=c\end{pmatrix}

    \Rightarrow \\ c-3a+6a+6b = 0
     c+3a+6b = 0\\ \Rightarrow base=\{(1,0,-3);(0,1,-3)\}
    Actually, there is an error here: with a= 0, b= 1, c= -6, not -3. A basis for the image is {(1, 0, -3); (0, 1, -6)}


    I wonder if I get an equivalent result if I do it the following manner:

    (x-2y,y-x,3x) = x(1,-1,3)+y(-2,1,0) \Rightarrow base=\{(1,-1,3);(-2,1,0)\}


    of course, in this case I'll need to check if are them linearly independent if I want to get a base.

    thanks in advice
    VIKKO
    Yes, that is perfectly valid. Any you can check that a= 1, b= -1, c= 3 and a= -2, b= 1, c= 0 also satisfy c+3a+6b = 0.

    (If they were not independent that linear transformation would be singular and map [itex]R^2[/itex] into a one dimensional subspace of [itex]R^3[/itex].)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    10
    thank you very much

    yoour answer was very useful for me
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 30th 2011, 04:36 PM
  2. Kernel/Image of a repeated linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 5th 2011, 03:33 PM
  3. Image of a linear transformation matrix...
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 18th 2010, 11:34 AM
  4. A question about rank/image of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: February 11th 2009, 05:54 PM
  5. Kernal and Image of a Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 15th 2008, 06:04 PM

Search Tags


/mathhelpforum @mathhelpforum