Determine the image of the map

Question

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by \begin{equation*}f: \begin{pmatrix}x \\ y\end{pmatrix}\rightarrow \begin{pmatrix}u \\ v\end{pmatrix}=\begin{pmatrix}x(1-y) \\ xy\end{pmatrix}\end{equation*}

I want to determine th eimage $f(\mathbb{R}^2)$.

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We have that \begin{equation*}f\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix} \ \text{ und } \ f\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}\end{equation*}

Does this mean that $\text{im}(f)=\left \{\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0\end{pmatrix}\right \}$ ? Or do we have to do something else in this case where we don't have a linear map?

Answer 1

You can notice that the map is invertible in certain conditions.

If $(u,v)\in\mathbb R^2$ then $f(x=u+v,y=\dfrac v{u+v})=(u,v)$ except for $v=-u$.

Indeed these points cannot be reached since $x(1-y)=-xy\implies x=0$

And when $x=0$ then the whole axis is transformed to origin point : $f(0,y)=(0,0)$.

So the image $f(\mathbb R^2)$ is whole $\mathbb R^2$ except for the line $v=-u$ but still including origin.

Answer 2

$f$ is not linear. Hence, you can't assume that the image of $f$ is a subspace of $\mathbb R^2$.

Hint: Consider $$ f(x,y)=x\begin{pmatrix}1-y\\y\end{pmatrix}. $$ You see that the image of $f$ contains all lines through $\begin{pmatrix}1-y\\y\end{pmatrix}$ for some $y\in\mathbb R$.