I need to prove that
$$F(x,y) =(2x+y, x+4y)$$
is injective and surjective. I started by assuming $$F(x_1,y_1) = F(x_2,y_2)$$
or
$$(2x_1+y_1, x_1+4y_1) = (2x_2+y_2, x_2+4y_2)\implies\\2x_1+y_1 = 2x_2+y_2\\x_1+4y_1 = x_2+4y_2$$
But I don't know how to show that $x_1 = x_2$ and $y_1 = y_2$.
Also, for surjectivity, I don't know how to prove that $(2x+y, x+4y)$ will always have a vector to get mapped to. I know it has, it's obvious, but I don't know how to show it.
Best Answer
Your $F$ is a linear mapping. Take $(x,y) \in \ker F$. So : $$\begin{cases}2x+y=0 \\x+4y = 0\end{cases}$$ We have that $(0,0)$ is a solution, and it is the only one, because $$\begin{vmatrix}2 & 1 \\ 1 & 4\end{vmatrix} = 7 \neq 0.$$ Conclusion: $\ker F = \{(0,0)\}$ and $F$ is injective. On the other hand, by the rank-nullity theorem, we have that $$2 = \dim \Bbb R^2 = \dim \ker F + \dim {\rm Im}\ F = \dim {\rm Im}\ F,$$ and hence ${\rm Im} \ F = \Bbb R^2.$ So $F$ is surjective.
A more elementary approach:
For injectivity, suppose that $F(x_1,y_1) = F(x_2,y_2)$. Then we have: $$\begin{cases}2x_1+y_1 = 2x_2+y_2 \\ x_1+4y_1 = x_2+4y_2\end{cases} \iff \begin{cases} 2(x_1-x_2)+(y_1-y_2) =0 \\ (x_1-x_2)+4(y_1-y_2) = 0\end{cases} \iff \begin{cases}2\Delta x+\Delta y=0 \\ \Delta x+4\Delta y = 0\end{cases},$$ where $\Delta x = x_1-x_2$ and $\Delta y = y_1 - y_2$. But I've already told you what is the only solution of this system, which is linear in $\Delta x$ and $\Delta y$. This process of thought is the same used in proving that $F$ injective $\iff \ker F = \{{\bf 0}\}$.
For surjective, let $(r,s)$ in the image of $F$. Then, we have for some pair $(x,y)$, the relation $(r,s) = F(x,y)$, that is: $$\begin{cases}2x+y=r \\x+4y = s\end{cases}.$$ Can you find $r$ and $s$ in terms of $x$ and $y$? This proves surjective.