Is there an automorphism of $\mathbb{R}^n$ (here it is seen as a vector space) that is not a linear mapping?
Every automorphism of $\mathbb{R}^n$ a linear mapping
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0The exponential example was to show that not every isomorphism is linear. – 2012-04-16
2 Answers
An automorphism of a vector space is, by definition, an invertible linear mapping. So no.
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1It's in the de$f$inition of automorphism that it is from the space into itself. – 2012-04-16
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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0Sorry for "overinterpreting"... In that case, your question is of a terminological nature, and you may refer to (and accept) Chris' answer above. – 2012-04-16