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An Affine transformation is a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(v) = Av+b $, where $\det A \neq 0 $ and $b \in \mathbb{R}^2.$

My professor give me a task, I need to prove that a function $f$ is Affine transformation iff $f$ is one-one.

It's a little confuse, I have proved that if $f$ is affine then $f$ is one - one, but the other hand I don't know and I suspect it's not true. Of course, if I take $f: \mathbb{R}_{+} \rightarrow \mathbb{R},$ where $f(x) = x^2,$ then this function is one- one and not affine, but I need some example in $\mathbb{R}^2$ with the form $f(v)=Av + b.$

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I think you have mis-interpreted the question. What is true is if $f(v)=Av+b$ for some square matrix $A$ and some vector $b$ then $f$ is one-to-one iff $\det(A) \neq 0$.

Proof: If $\det(A) \neq 0$ and $Av+b=Aw+b$ then $A(v-w)=0$ and this implies $v=w$ because $A$ is non-singular (and its kernel is $\{0\}$).

If $\det(A)=0$ the there exists a nonzero vecto $v$ such that $Av=0$. This gives $f(v)=f(2v)$ so $f$ is not one-to-one.

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  • $\begingroup$ @user638057, also note that your question in this form has a counter example$$f(x)=||x||^2x$$where $x, f(x) \in \Bbb R^2$ $\endgroup$
    – Mostafa Ayaz
    Commented Aug 22, 2019 at 9:57
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$f$ can be one-one but not affine.

For example, $f(x,y) = (x^3,y)$ This function is one-one, but not affine.

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  • $\begingroup$ yeh, I know and I take a appeared example, but I want $f$ in the way $f(v) = Av+b$, where $A$ is a matrix with $\det = 0$. $\endgroup$
    – Joãonani
    Commented Aug 22, 2019 at 4:18

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