Every automorphism of $\mathbb{R}^n$ a linear mapping

Is there an automorphism of $\mathbb{R}^n$ (here it is seen as a vector space) that is not a linear mapping?

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The answer is "no", but the way to prove that depends on your definition of automorphism, $\mathbb{R}^n$ and "linear mapping". –  akkkk Apr 16 '12 at 13:57
You need to explain this better. Do you mean that "every automorphism of the topological group $\mathbb{R}$ is a linear mapping"? An automorphism in the category of vector spaces and linear mappings is by definition a linear mapping. –  Xabier Domínguez Apr 16 '12 at 13:57
@XabierDomínguez, I can't even understand what you said. My qustion was more in the sense of "The exponential function is an isomorphism between $(\mathbb{R},+)$ and $(\mathbb{R}^+,\times )$, but if I add the hypotheses that this isomorphism is an automorphism and the spaces are a vector space with usual sum, will those automorphisms be necessarily linear?". –  Marra Apr 16 '12 at 14:07
The exponential example was to show that not every isomorphism is linear. –  Marra Apr 16 '12 at 14:10

An automorphism of a vector space is, by definition, an invertible linear mapping. So no.

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"...invertible linear mapping from the space to itself" perhaps? –  Ragib Zaman Apr 16 '12 at 13:57
It's in the definition of automorphism that it is from the space into itself. –  Marra Apr 16 '12 at 14:08

This is an answer to the OP's question interpreted as "Is there an automorphism of the group $(\mathbb{R}^n,+)$ that is not $\mathbb{R}$-linear?"

The answer is yes. Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}.$ Note that as a $\mathbb{R}$-vector space, $\mathbb{R}$ has dimension 1, but as a $\mathbb{Q}$-vector space its dimension is infinite (actually it is the continuum). Let $\{e_i\}$ and $\{v_i\}$ be two different basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. Define $f:\mathbb{R}\to \mathbb{R}$ as follows: $f(\sum r_i e_i)=\sum r_i v_i.$ This is an isomorphism of the vector space $\mathbb{R}$ (over $\mathbb{Q}$) onto itself because it maps one basis onto another. In particular it is an automorphism of the additive group $\mathbb{R}.$ But most of these maps are not linear over $\mathbb{R}.$ For example, you can choose $e_1=1,$ $e_2=\pi,$ and complete the basis $\{e_i\}$ from there (note that these two are independent over $\mathbb{Q}$). Now choose $v_1=f(e_1)=1,$ and as $v_2=f(e_2)$ pick any irrational number different from $\pi,$ and complete the basis $\{v_i\}$ from there. This $f$ cannot be of the form $x\mapsto \lambda x,$ i. e. it cannot be $\mathbb{R}$-linear.

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Can you give me an example? –  Marra Apr 16 '12 at 14:13
I can't because these maps are not constructible. They depend on the Axiom of Choice. –  Xabier Domínguez Apr 16 '12 at 14:14
Are you talking about finite-dimension vector spaces? I think I should have said that. –  Marra Apr 16 '12 at 14:15
I'll try to ellaborate my answer a bit further. –  Xabier Domínguez Apr 16 '12 at 14:16
Ok, I'll make my question again: Consider $\mathbb{R}^n$ as the classical vector space over $\mathbb{R}$ (which was my obvious intent since the beggining). Then this must not hold. –  Marra Apr 16 '12 at 16:04