# Showing a function is bijective and finding its inverse

The function $$f: \mathbb{R}^2 \rightarrow\mathbb{R}^2$$ is defined by $$f(x,y)=(2x+3y,x+2y)$$. Show that $$f$$ is bijective and find its inverse.

I've got so far: Bijective = $$1-1$$ and onto.

$$1-1$$ if $$(2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$$

Then $$2x_1+3y_1=2x_2+3y_2 \qquad (1)$$ $$x_1+2y_1=x_2+2y_2 \qquad (2)$$

$$(1)-(2)$$

$$x_1+y_1=x_2+y_2$$

$$x_1=x_2+y_2-y_1$$

Substituting into equation 1:

$$2(x_2+y_2-y_1)+3y_1=2x_2+3y_2$$

$$y_1=3y_2-2y_2$$

$$y_1=y_2$$

Substituting into equation 2:

$$x_1+2y_1=x_2+2y_1$$

$$x_1=x_2$$

$$(2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$$

Thus 1-1

Onto

if $$(u,v) \in \mathbb{R}^2$$ (codomain) we want $$(x,y)$$ with $$f(x,y)=(u,v)$$

$$(2x+3y,x+2y)=(u,v)$$

$$2x+3y=u$$

$$x+2y=v$$

Eliminating x:

$$y=2v-u$$

Substituting:

$$2x+3(2v-u)=u$$

$$2x+6v-3u=u$$

$$2x=4u-6v$$

$$x=2u-3v$$

Therefore:

$$f(2u-3v,2v-u)=(u,v)$$

Now how do you find it’s inverse? And is that correct what I have done?

• Do you know about matrices? Commented Nov 2, 2013 at 19:24
• Looks right. And you have found the inverse transformation. It sends $(u,v)$ to $(2u-3v, 2v-u)$. If you like, you can change the variables to $x$ and $y$. Commented Nov 2, 2013 at 19:28
• So u to be x and v to be y? Commented Nov 2, 2013 at 19:29
• Yes, if you want to use those names for variables. To me, $u$ and $v$ are just as good, but they may be expecting you to use $x$ and $y$. Commented Nov 2, 2013 at 19:31
• so the inverse is literally: (2-3y,2y-x)? Commented Nov 2, 2013 at 19:41

Superhint: If you had started with the second part ("and find its inverse") you could have "coimpletely" avoided the first part calculations.

You have $f(x,y)=(2x+3y,x+2y)$. This function is what we call linear, because it satisfies the following property: $f(\lambda v+w) = \lambda f(v)+f(w)$ for $v,w\in \Bbb R^2$ as you can check.

Now, linear functions are much easier do deal with than general functions. Why? Because they can be (at least in spaces like $\mathbb{R}^2$ that are what we call finite dimensional) be expressed in terms of matrices. Indeed, look that you have the following:

$$f(x,y)=xf(1,0)+yf(0,1)$$

Exactly because of that property of linearity. In that case, if you know about matrix multiplication, you'll see that this is equal to saying that

$$f(x,y)=\begin{pmatrix}f(1,0) & f(0,1)\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix},$$

where we consider elements of $\mathbb{R}^2$ as columns. In that case, you have

$$f(x,y)=\begin{pmatrix}2 & 3\\ 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix},$$

if we call that matrix $A$, then $f(v)=A v$ where $v=(x,y)$. Now, Look that because of that, things get easier. Indeed, to show that $f$ is injective, is equivalent to showing that $f(v)=f(w)$ implies $v=w$, but $f(v)=f(w)$ is the same as $Av=Aw$, or $A(v-w)=0$. This implies $v=w$, if and only if the homegenous system of linear equations $Av=0$ has only the trivial solution. That is, if and only if $\det A \neq 0$. But you can check that $\det A = 1$, so that $A$ is invertible, hence, $A(v-w)=0$ implies $v=w$, and hence $f$ is injective.

To show surjectivity, notice that $A$ being invertible, already implies that, because you can multiply the following

$$f(x,y)=A\begin{pmatrix}x\\y\end{pmatrix}$$

on the left by $A^{-1}$ to find $(x,y)$. It is easy, because

$$A^{-1}=\begin{pmatrix}-2 & \phantom{-}1 \\ \phantom{-}3 & -2\end{pmatrix},$$

and so we can take the inverse to be $g(v)=A^{-1}v$. In that case, $f(g(v))=A(A^{-1}v)=v$ and similarly $g(f(v))=A^{-1}(A(v))=v$, so $g$ is inverse to $f$ and hence $f$ is bijective.

Hint: You want a function $g(x,y)$ such that $g(2x+3y,x+2y) = (x,y)$. In the work you've already shown, you set $u = 2x+3y$ and $v = x+2y$ and then solved for $x$ and $y$ in terms of $u$ and $v$. These expressions for $u$ and $v$ are somehow related to $g$.

That $f(2u-3v,2v-u)=(u,v)$ means that with $g(u,v)=(2u-v,2v-u)$ you have $f\circ g=\mathrm{id}$.

Had you not already shown that $f$ is injective, you could do so by checking that $g\circ f=\mathrm{id}$. (Indeed linear algebra tells us that “for dimension reasons” this follows automatically in this case.)