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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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    $\begingroup$ What do you mean by well-defined? There is no choice involved in the definition which might change the value of the function. $\endgroup$ – Tobias Kildetoft May 5 '16 at 8:37
  • $\begingroup$ It means $t_{A,v}=t_{A',v'}$. @TobiasKildetoft $\endgroup$ – Mathematicing May 5 '16 at 9:08
  • $\begingroup$ @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? $\endgroup$ – Travis Willse May 5 '16 at 10:00
  • $\begingroup$ 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. $\endgroup$ – Mathematicing May 5 '16 at 10:05
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    $\begingroup$ 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. $\endgroup$ – Travis Willse May 5 '16 at 10:26

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