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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.

Guerlando OCs
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4 Answers4

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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.

Ivo Terek
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If $(r, s)=(2x+y,x+4y)$ solve $(x, y)$ in terms of $(r, s)$.

jimbo
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$$ (2x_1+y_1, x_1+4y_1) = (2x_2+y_2, x_2+4y_2)\implies\\2x_1+y_1 = 2x_2+y_2\\x_1+4y_1 = x_2+4y_2 $$ from the second equation you get $x_1 = x_2+4y_2 - 4y_1$ substitute this in the first equation to get: $$ 2x_2 + 8y_2-8y_1 + y1 = 2x_2 + y_2 \implies y_1=y_2 $$

benji
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There are a few methods here:

First, is the continuation of yours: As you've observed, $F(x_1, y_1) = F(x_2, y_2)$ implies \begin{align} 2 x_1 + y_1 &= 2 x_2 + y_2 \\ x_1 + 4 y_1 &= x_2 + 4 y_2 . \end{align} Respective subtracting twice each side of the second equation from each side of the first give $$- 3 y_1 = - 3 y_2,$$ and multiplying both sides by $-\frac{1}{3}$ gives $$y_1 = y_2$$ as desired. An analogous pair of operations gives $y_1 = y_2$.

Alternatively, we can simply look for an inverse: That is solve $(u, v) = F(x, y) = (2 x + y, x + 4y)$ for $x, y$ in terms of $(u, v)$. If we can do so, $F$ is invertible, with inverse $F^{-1}$ given by this expression.

Finally, since $F$ is linear, we may write it in matrix form: $$F\left(\begin{pmatrix}x\\y\end{pmatrix}\right) = \begin{pmatrix}2 & 1\\1 & 4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}.$$ If the $2 \times 2$ matrix is invertible, then we can solve for $F(x, y)$ in terms of $(x, y)$, which shows that $F$ is invertible: $$\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2 & 1\\1 & 4\end{pmatrix}^{-1}F\left(\begin{pmatrix}x\\y\end{pmatrix}\right).$$

To show that this matrix is invertible we need not compute its inverse directly---a matrix is invertible iff its determinant is nonzero, which is the case for this matrix: $$\det \begin{pmatrix}2 & 1\\1 & 4\end{pmatrix} = (2)(4) - (1)(1) = 7 \neq 0.$$

Travis Willse
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