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Let V be an n-dim vector space over the field $\mathbb{F}.$

$A \in GL\left ( n,\mathbb{F} \right )$ and $v \in V$

Define the affine transformation $t_{A,v}$:

$V\rightarrow V$

$x \mapsto xA+v$

Showing that this map is a bijection from V to V is simple. Indeed, by definition, $t_{A,v}$ is a permutation. I'd like to show that this map is well-defined but unsure how to do so. The link below provides the lemma to do so.

helpful link from Mathstack

With the lemma, I am unsure how to begin. Looking only for hints.

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Mathematicing
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    What do you mean by well-defined? There is no choice involved in the definition which might change the value of the function. – Tobias Kildetoft May 05 '16 at 08:37
  • It means $t_{A,v}=t_{A',v'}$. @TobiasKildetoft – Mathematicing May 05 '16 at 09:08
  • @Mathematicing It's still not clear what you mean. Do you mean that you want to show that $t_{A, v} = t_{A', v'}$ implies that $(A, v) = (A', v')$? This says that map $(A, v) \mapsto t_{A, v}$ is injective. Or do you want that the map is surjective (that is, that every affine map can be written this way)? Both? – Travis Willse May 05 '16 at 10:00
  • Hi @Travis I apologise for the vagary of my question. On my lecture slides, it says that if $A\neq A'$ and $v\neq v'$ then $t_{A,v}\neq t_{A',v'}$ and this can be verified. So my interpretation of the wording is that the map can be checked if it is well-defined. – Mathematicing May 05 '16 at 10:05
  • As an aside, since $t_{A,v}$ is a permutation, is there a reason for which the notation is written as $(x)^{t_{A,v}}$ rather than as $(x)t_{A,v}$? The exponential notation in the former is usually used to denote group action as I am aware. – Mathematicing May 05 '16 at 10:08
  • @Travis I think your former question is what I am looking for since it fits the contraposition of what is written on my lecture slides. – Mathematicing May 05 '16 at 10:15
  • No, what you wrote there means that the map is injective, not that it is well-defined (it is trivially well-defined). – Tobias Kildetoft May 05 '16 at 10:15
  • What Tobias said---well-definedness is something completely separate that isn't an issue here. Note that injectivity of this map is the same thing as faithfulness of the group action. (By the way, my previous comment should say "...every invertible affine map can be written this way...".) – Travis Willse May 05 '16 at 10:24
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    As for the notation, I'd find the superscript notation probably unwieldy in this case, and the notation $(x) t_{A, v}$ simply misleading, because the action $x \mapsto A x + v$ is a left action, not a right one. – Travis Willse May 05 '16 at 10:26
  • @Travis I have more question pertaining to this question. Do I construct a thread? – Mathematicing May 05 '16 at 10:40
  • If it's a distinct question from this yes, in short, yes. – Travis Willse May 05 '16 at 10:43

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