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It is known that on the set $\mathbb{R}^+$ of the positive real numbers, one can introduce a linear structure $(\mathbb{R}^+,\mathbb{R}\odot, \oplus)$ such that $$ u\oplus v=u\cdot v \\ \alpha\odot u=u^\alpha, $$ where $u,v\in\mathbb{R}^+$, $\cdot$ is the usual multiplication of reals, and $\alpha\in\mathbb{R}$ is an arbitrary real.

This vector space is essentially the image $\exp(\mathbb{R})$ with the $\exp$ function being considered a linear map.


I have always found this example interesting, because the vector space and its underlying field is the same set. Or rather, the vector space is a subset of its underlying field, but is not a subspace, because the linear structures are completely different.

I have been trying to think about a vector space, where we can endow the entire set with a different linear structure, but I couldn't conjure up an actual example.

So, "question": I would be terribly appreciative of some interesting examples of sets that can be endowed with two completely different linear structures.

I assume the same construction (same as the opening example with $\mathbb{R}^+$ I mean) can be carried out on $\mathbb{C}$, where the exp function covers most of $\mathbb{C}$, but the zero complex number is also exempt from the set, so it is not a good example.

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For one, if $X$ is a vector space, you can take any bijective function $f:X\to Y$ and define $$u\oplus v = f(f^{-1}(u)+f^{-1}(v))\\ \alpha \odot v = f(\alpha\cdot f^{-1}(u))$$

and then $(Y, \oplus, \odot)$ will be a vector space (isomorhic to $X$, with the obvious isomorphism of $f$).

So, since there always exists a bijection between $\mathbb R^n$ and $\mathbb R^m$ for any pair $m,n\in\mathbb N$, you can make, using such a bijection, $\mathbb R$ into a $124$-dimensional vector space over $\mathbb R$. It's ugly, I know, but it's possible.

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    $\begingroup$ And even more, it is possible to do this while keeping the original addition (so by only changing the scalar multiplication). $\endgroup$ – Tobias Kildetoft Oct 4 '17 at 9:14
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This is just a particular case of the general answer provided by 5xum (as, by the way, your example concerning $\mathbb{R}^+$), but here it goes: consider on $\mathbb{R}$ the following operations: $x\oplus y=x+y+1$ and $\lambda\odot x=\lambda x+\lambda-1$. With these operations, $\mathbb R$ becomes (again) a real vector space.

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