How to prove that this function is injective and find its inverse

Question

If we define $f:R^2\to R^2$ by $f(x,y)=(3x+2y, -x+5y)$ how can I show that this function is injective, and go on to find its inverse? I think it can be done with matrices but I would rather not as we haven't really covered them yet.

Thanks

Answer 1

You could prove it is injective by exhibiting its inverse. In a case like this, where you feel you understand the transformation because it's linear, and you have good reason to expect its invertible, this is a reasonable idea.

And if you know matrices, whether you like it or not you'll be computing the inverse transformation.

You will be done if you can solve $\begin{bmatrix}a\\ b \end{bmatrix}=\begin{bmatrix}3x+2y\\ -x+5y \end{bmatrix}$ in terms of $x$ and $y$.

At once you find that $a+3b=17y$. Then $a=3x+2\frac{a+3b}{17}$.

Using both expressions, you have $x$ and $y$ in terms of $a$ and $b$, and you have, in the course of solving, found the inverse matrix of the transformation.

If you just had to prove that it's injective and you didn't care about the inverse, another approach might be to prove it is a linear transformation, and then show that the kernel is zero. But this doesn't help to compute the inverse function.

Answer 2

An injective function is a function such that $f(x_1, y_1) = f(x_2, y_2) \implies x_1 = x_2, y_1 = y_2$. Let us take the coordinates $(x_0, y_0)$ and $(x_0 + a, y_0 + b)$. To show that our function is injective, we will show that for all points, $f(x_0, y_0) = f(x_0 + a, y_0 + b) \implies a, b = 0$.

Assume for the sake of contradiction that $f(x_0, y_0) = f(x_0 + a, y_0 + b)$ and $ a,b \neq 0$. Note this implies $(3x_0+2y_0, -x_0+5y_0) = (3x_0 + 3a +2y_0 + 2b, -x_0 - a+5y_0 + 5 b)$, so we form the system of equations $3x_0+2y_0 = 3x_0 + 3a +2y_0 + 2b$ and $ -x_0+5y_0 = -x_0 - a+5y_0 + 5 b$. Simplifying the first equation yields $3a + 2b = 0$, and the latter equation yields $-a + 5b = 0$. Solving for $a, b$ shows that $a = b = 0$. This is a contradiction, so we see $f$ is injective.

As for finding the inverse, I believe you will have to use matrices here.

Hope this helps!

Answer 3

Another way

$f(1,0)=(3,-1)$ and $f(0,1)=(2,5)$, set $A=\left(\begin{matrix}3&2\\-1&5\end{matrix}\right)$ and $\text{x}=\left(\begin{matrix}x\\y\end{matrix}\right)$ thus $$f(x,y)=A\text{x}=\left(\begin{matrix}3&2\\-1&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)$$ $f$ is injective because $\text{ker}_f=\{(x,y)\in\mathbb{R}^2|f(x,y)=(0,0)\}=\{(0,0)\}$ and $$f^{-1}(x,y)=\frac{1}{17}\left(\begin{matrix}5&-2\\1&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)$$

Answer 4

Without using the inverse matrix concept: $$f(x,y)=f(z,w)\Rightarrow (3x+2y, -x+5y)=(3z+2w, -z+5w)\Rightarrow \begin{cases} 3x+2y=3z+2w\\ -x+5y=-z+5w\end{cases}$$ $$\Rightarrow \begin{cases} 3(x-z)+2(y-w)=0\\ -(x-z)+5(y-w)=0\end{cases}\underbrace{\Rightarrow}_{\text{solv. system}} x-z=0\wedge y-w=0\Rightarrow x=z\wedge y=w$$ $$\Rightarrow (x,y)=(z,w)\Rightarrow f\text{ is } \text{injective}.$$ To see that there exists $f^{-1}$ you have to proof that $f$ is also surjective. Consider $(x',y')\in \mathbb{R}^2$ and prove that the equation $f(x,y)=(x',y')$ has solution for every $(x',y').$ Solving it, you´ll inmediatly have the expression of $f^{-1}(x',y')=(x,y)=\ldots$

Answer 5

Considere the point $(x_1,y_1)$ and $(x_2,y_2)$. By definition f(x,y) is injective if $f(x_1,y_1)=f(x_2,y_2)$ if and only if $x_1=x_2$ and $y_1=y_2$. We can write: \begin{equation} f(x_1,y_1)=(3 x_1+2 y_1,-x_1+5 y_1) \end{equation} And \begin{equation} f(x_2,y_2)=(3 x_2+2 y_2,-x_2+5 y_2) \end{equation}

Then if $f(x_1,y_1)=f(x_2,y_2)$ we have

\begin{equation} 3 x_1+2 y_1=3 x_2+2 y_2,\\ -x_1+5 y_1 =-x_2+5 y_2 \end{equation} By multiplying the second equation by 3 and then adding the first equation the the previous one we have $17 y_1=17 y_2$, then $y_1=y_2$ and substituting back $x_1=x_2$. In conclusion we proved that when $f(x_1,y_1)=f(x_2,y_2)$ we have $y_1=y_2$ and $x_1=x_2$. The second part of the proof is easier. As a matter of fact when $x_1=x_2$ and $y_1=y_2$ is trivial to show that $f(x_1,y_1)=f(x_2,y_2)$. To see that there is an inverse you've to prove that $f(x,y)$ is bijectiove i.e. iniective and surjective (or onto). To show that $f(x,y):U \rightarrow V$ is surjective, we suppore there exists $a_y \in V$ and we prove that exists $a_x \in U$ for which $f(a_x)=a_y$