1. ## Quick confirmation

Hi All, this shouldn't take anyone worth their salt too long.... i'm just wondering if this is correct:

If the task is to prove that:

$\displaystyle (u,v)\rightarrow (u+v,u-v)$ is a bijection from $\displaystyle \mathbb{R}^2$onto $\displaystyle \mathbb{R}^2$...

Do we go about it this way?

To show injectivity:

Assume there exists $\displaystyle (u_1,v_1)\neq(u_2,v_2)$ such that: $\displaystyle u_1+v_1 = u_2 + v_2,$ and also that $\displaystyle u_1 - v_1 = u_2 - v_2$.

Isolating $\displaystyle u_1$ in the first equation, we can sub in to the second to get $\displaystyle u_1=u_2$ and then we can repeat the similar process, just isolating $\displaystyle v_1$ to arrive at $\displaystyle v_1=v_2$?

Then to show surjectivity:

Do we just pick any $\displaystyle (x,y)$ in $\displaystyle \mathbb{R}^2$. We set $\displaystyle (x,y)=(u+v,u-v)$ and then solve for $\displaystyle u$ and $\displaystyle v$ in terms of $\displaystyle x$ and $\displaystyle y$. I get $\displaystyle u=\frac{x+y}{2}$ and $\displaystyle v=\frac{x-y}{2}$??

2. ## Re: Quick confirmation

Originally Posted by mathswannabe
if this is correct:
If the task is to prove that:
$\displaystyle (u,v)\rightarrow (u+v,u-v)$ is a bijection from $\displaystyle \mathbb{R}^2$onto $\displaystyle \mathbb{R}^2$
Yes, that works.