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I'm reading through a geometry proof that claims that every affine map $f:\mathbb{R}^n\to\mathbb{R}^n$ is the composition of a linear map and a translation. More precisely, $f$ is affine if and only if $f_0 =T_{-f(O)} \circ f$ is linear, where $T_{-f(O)}$ is the translation by the vector $-\overrightarrow{Of(O)}$.

One direction of the proof relies on showing that $f_0(\alpha A + \beta B)=\alpha f_0(A)+\beta f_0(B)$. However, how does this show that $f_0$ is linear? This is exactly what the definition of affine map is, so we've only shown that $f_0$ is affine, which isn't enough. I feel like we still need to show that $f_0$ transforms the origin into itself.

If anyone can recommend a good book on these topics I'd appreciate it, thanks in advance!

user401936
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1 Answers1

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That’s the definition of a linear map, not an affine map. For example, consider the 1-D function

$$f(x)=mx+b$$

for some $m,b\in\mathbb{R}$ with $b\neq0$. This function is affine, but for $x,y\in\mathbb{R}$ and $\alpha,\beta\in\mathbb{R}$ such that $\alpha+\beta\neq1$, we have

$$f(\alpha x+\beta y)=m(\alpha x+\beta y)+b\neq \alpha f(x)+\beta f(y).$$

David M.
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  • The way my teacher has defined affine maps is exactly when $\alpha + \beta = 1$. I think there's more confusion in my head than what I've asked about, I should read the definitions more carefully when I get the chance – user401936 Jun 22 '18 at 19:27
  • @AstlyDichrar Getting the definitions straight is always a good idea. If you can show that a function satisfies the property for all $\alpha,\beta\in\mathbb{R}$, then it is linear. If you can show it for all $\alpha,\beta\in\mathbb{R}$ such that $\alpha+\beta=1$, then it is affine. – David M. Jun 22 '18 at 19:29
  • Oh, maybe that is the part that I didn't notice. It's very possible that the proof didn't have the requirement that $\alpha +\beta = 1$. – user401936 Jun 22 '18 at 20:12
  • @AstlyDichrar See this old post for some further discussion – David M. Jun 22 '18 at 20:16