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Prove that if an affine transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ is onto, then it is one-to-one.

Here we define an affine transformation as $T:\mathbb{R}^2 \to \mathbb{R}^2$ s.t. $T(r_1A_1 + ... + r_nA_n) = r_1T(A_1) + ... + r_nT(A_n)$ for $r_1 +...+r_n =1$ and $A_1, ..., A_n \in \mathbb{R}^2$

Typically when we want to prove something is one-to-one we start by assuming that $T(A) = T(B)$ for $A,B \in \mathbb{R}^2$ and we want to conclude that $A = B$. So if we let $A = a_1A_1 + ... + a_nA_n$ and $B = b_1B_1 + ... + b_nB_n$ such that $a_1 + ... + a_n =1, b_1 +...+b_n =1$ and $A_1, ..., A_n, B_1, ..., B_n \in \mathbb{R}^2$. I'm not sure how to conclude that $A=B$.

frostyfeet
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  • Given a vector space $V$, is it true that every affine transformation $T\colon V\to V$ can be written as $T=L+R$, with $L\colon V\to V$ linear and $R\colon V\to V$ a translation ? – Rodrigo Dias Oct 05 '18 at 18:39

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Let $f$ be an affine transformation. Now, it is well known that $f$ can be written as $T+b$ with $T$ linear and $b$ a vector. Now $x \mapsto x+b$ is a bijection, so $f$ being surjective implies $T$ surjective: the latter is a linear endomorphism and so (by the dimension theorem if you will) it being an epi implies being iso. Thus $T$ is bijective and then (by composing with $x \mapsto x+b$) so is $f$.

As you can see, this works in any finite dimensional vector space because no additional assumptions have been made.

qualcuno
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