# Question about affine transformation and one-one function.

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 $$f(v)=Av + b.$$

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.

• @user638057, also note that your question in this form has a counter example$$f(x)=||x||^2x$$where $x, f(x) \in \Bbb R^2$ Commented Aug 22, 2019 at 9:57

$$f$$ can be one-one but not affine.

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

• 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 $\det = 0$. Commented Aug 22, 2019 at 4:18