We know the homeomorphism $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ maps straight lines to straight lines and the zero vector to the zero vector. Is it Linear?? If so, how can we prove it?
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1I've changed the tags a bit. "soft-question" and "recreational-mathematics" don't seem to be a good fit. In turn, I've added the "real-analysis" tag. I've also added a more informative title. The old title, while attention grabbing, does not provide problem context. – EuYu Aug 31 '13 at 04:01
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When you mean straight lines to straight lines, do you mean any line in $\mathbb{R}^n$ or only lines passing through the origin? – KReiser Aug 31 '13 at 04:05
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2I checked mathoverflow. Your question was discussed and answered there: mathoverflow.net/questions/123356 – Moishe Kohan Aug 31 '13 at 06:02
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Are you asuming $n>1$? If not a good counter-example could be the function $\phi(x)=x^3$ ( $x\in\mathbb{R}$) . It's a homeomorphism, maps $\mathbb{R}$ to $\mathbb{R}$ (the only line in $\mathbb{R}$) and obviously it maps $0$ to $0$
JDPoincare'S
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The same way, it could be done in any dimension: $φ(x) = ||x||^2 x$, $x ∈ \mathbb{R}^n$. – user87690 Aug 31 '13 at 08:36
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1@user87690 This preserves straight lines through the origin, but not all straight lines in $\mathbb R^n$. Consider, e.g., the points $(0,1),(1,1),(1,0)\in\mathbb R^2$. – WillG Dec 17 '22 at 20:51
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This is not enough to suggest linearity w/o it being a homeomorphism... It has satisfy $\phi(a*v_1 + b*v_2) = a\phi(v_1)+b\phi(v_2)$. So lines to lines does not promise linearity, though it is a requirement as lines are a linear combination of vectors in $\mathbb{R}^n$. Assuming you are looking at $\mathbb{R}^n$ as a combination of vectors in the typical sense.
The homeomorphism takes away the discontinuous cases and the situations with a kernel. My bad I did not read the problem correctly initially.
Lines can be expressed as a linear combination of elements in a vector space with an indeterminate to the first power. Just express a matrix with such a basis as the mapping and you have you linearity in the form of a matrix.
wfw
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Why is mapping lines to lines an insufficient requirement for linearity? Did you have a specific counter-example in mind? As it stands, your answer right now simply states the definition of linearity. – EuYu Aug 31 '13 at 04:11