Linear Algebra: Mapping Question

[srfriggen]

Homework Statement

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.

Homework Equations

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.

 

[christoff]

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 \mathbb{R}^2, 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.