# Linear Algebra: Mapping Question

1. Jun 28, 2012

### srfriggen

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. Jun 28, 2012

### 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.