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Linear Algebra: Mapping Question

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data

    From Serge Lang's "Linear Algebra, 3rd Edition", pg 51 exercise 9.

    Prove that the image is equal to a certain set S by proving that the image is contained in S, and also that every element of S in in the image.

    9. Let F:R2→R2 be the mapping defined by F(x,y)=(xy,y). Describe the image under F of the straight line x=2.



    2. Relevant equations



    3. The attempt at a solution


    I first simply drew out a vertical line of x=2 and applied F to (2,y) for various values of y.
    F(2,0)=(0,0)
    F(2,1)=(2,1)
    F(2,2)=(4,2).

    I noticed it seemed to form a line which could be described by y=(1/2)x .

    I got a little stuck and looked at the answer and to my surprise the image was a line with slope 2.


    Below is the answer, and I just don't understand it...

    SOLUTION The image of F is the line whose equation is y=2x. Indeed, if (2,y) belongs to the line x=2, then F(2,y)=(2y,y), and clearly (2y,y) belongs to the line y=2x. Conversely, suppose v=2u; then F(2,v/2)=(v,v/2)=(v,u).



    I don't see why "clearly (2y,y) belongs to the line y=2x". If x is twice y in that tuple, wouldn't the equation be x=2y, or equivalently y=x/2 ?


    The other part I partly understand. I see why F(2,v/2)=(v,u) after applying the transformation, but I don't understand the significance of using v and u and setting up the equation v=2u.
     
    Last edited by a moderator: Jun 28, 2012
  2. jcsd
  3. Jun 28, 2012 #2
    I think you are right. It appears to be an error in the solution, or at the very least an ambiguous use of notation. By convention, when you're dealing with [itex] \mathbb{R}^2 [/itex], you always consider the first coordinate to be the ''horizontal'' coordinate. I would agree with you when you say that the set of points (2y,y), for all values of y, belongs to the line y=2x, where y is the vertical axis and x is the horizontal axis.
     
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