Proving range of transformation

[jonroberts74]

Homework Statement

I haven't learned kernel yet so if that's of use here I don't know it yet

let $T: \mathbb{R^3} \rightarrow \mathbb{R^2}$ where $T<x,y,z>=<2y,x+y+z>$[/B]

prove that the range is $\mathbb{R^2}$

The Attempt at a Solution

I know that T is not one-to-one, I checked that for a previous question. I did read if the range is $\mathbb{R^n}$ then T is onto. but I am not really sure how to prove the range is $\mathbb{R^2}$

 

[RUber]

A good way to do this is to choose any (a,b) in $\mathbb{R}^2$. Since they are arbitrary points in the range, if you can find a general form for them which is in the domain (in this case $\mathbb{R}^3$), then the range is the entirety of $\mathbb{R}^2$.

 

[jonroberts74]

so it goes both way "if $T$ is onto $\Leftarrow\Rightarrow$ the range is $\mathbb{R^n}$"

thanks

I think I was misunderstanding what range meant in this sense.

 

[Mark44]

so it goes both way "if $T$ is onto $\Leftarrow\Rightarrow$ the range is $\mathbb{R^n}$"

Since T is a transformation to $\mathbb{R^2}$ what you said should refer to that space, not $\mathbb{R^n}$.

thanks

I think I was misunderstanding what range meant in this sense.