Show $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined by ... is bijective Show $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined by $$
f\left(\begin{array}{l}{u} \\ {v}\end{array}\right)=\left(\begin{array}{l}{u+v} \\ {u-v}\end{array}\right)
$$ is bijective and find $f^{-1}$
My answer
My strategy is to show that $f$ is surjective and injective.
Injectivity: 
$$
f\left[\begin{array}{l}{u} \\ {v}\end{array}\right]=f\left[\begin{array}{l}{u^{\prime}} \\ {v^{\prime}}\end{array}\right]=\left[\begin{array}{c}{u+v} \\ {u-v}\end{array}\right]=\left[\begin{array}{c}{u^{\prime}+v^{\prime}} \\ {u-v^{\prime}}\end{array}\right]
 \implies
u+v=u'+v' \quad \wedge \quad u-v=u'-v'
$$ 
From this I do not think that I can conclude $(u,v)=(u',v')$. (Question 1) I do not really have a good argument for that. 
(Question 2) I do not have any ideas on how to show that it's bijective. 
 A: 
Injectivity

Building up on what you did,  add both equations $u+v + u-v = u'+v'+u'-v' $, which means $2u=2u'$, which means that $v=v'$.

Surjectivity

$\forall y \in \mathbb{R}^2$, $\exists x \in \mathbb{R}^2$ such that $f(x) = y$, which basically means that " Could you find $u,v$ such that $y_1 = u+v$ and $y_2 = u-v$ ?" Yes, just express $u,v$ in terms of $y_1,y_2$.
A: If $u+v = u'+v'$ and $u-v = u'-v'$, then solving the first in terms of $u$ gives $u = u'+v' -v$. Substituting into the second gives $(u'+v'-v) -v = u'-v'$, i.e.
$$u'+v' -2v = u'-v' \Longrightarrow -2v = -2v' \Longrightarrow v = v'.$$
This combined with one of the two original equations gives $u = u'$. I'll let you work on that.
For surjectivity, all you have to show is that for any $(u',v')$, there is $(u,v)$ such that $f(u,v) = (u',v')$.. in other words, solve the equation in reverse with $(u',v')$ as given and $(u,v)$ as the "dependent" variable so to speak.
A: injective:
You get two equations:
$$(1)\ \ u+v = u‘ + v‘$$
$$(2)\ \ u-v = u‘-v‘ \Leftrightarrow u = u‘-v‘+v$$
plug the second one into the first:
$$u‘-v‘+v + v= u‘ + v‘ \Leftrightarrow -2v‘ + 2v = 0$$
Therefore $v=v‘$. By either equation it follows that $u=u‘$. Therefore you function is injective.
surjective:
Let $(x,y)\in \mathbb{R}^2$. Now find $(u,v) \in \mathbb{R}^2$ such that
$$u+v = x$$
$$u-v = y$$
since this system of equations always has a solution, your function is surjective and the solution $(u,v)$ at the same time is your $f^{-1}(x,y)$.
Note:
This system is represented by the matrix
$$A= \begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix}$$
since $\det A = -2 \neq 0$ and $f(u,v)=A (u,v)^T$ you get bijectivity because the matrix is invertible which means you don’t need to do the previous calculations and directly get $f^{-1}$ through $A^{-1}$ by calculating
$$f^{-1} : \mathbb{R}^2 \to \mathbb{R}^2, (x,y)^T \mapsto A^{-1} (x,y)^T = \frac{1}{2} \begin{pmatrix} 1 & 1\\ 1 & -1\end{pmatrix}  (x,y)^T$$
