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I was considering the group of automorphisms on a vector space $\mathbb{Q}^3$ with the binary operation of addition. Is there a general description of the elements of this group? At first, it seemed to be that it would consist of bijective linear transformations, but I'm not sure how informative that is. Is there a specific category that this automorphism group would fall under?

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    @xdfm: it's not clear to me what kind of information you're looking for. Could you be more specific? "Bijective linear transformations" is already a pretty good description given how much we know about linear algebra.2010-09-21
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    I guess my main concern is, do isomorphisms also have to respect scalar multiplication? I know that they would have to respect addition by definition.2010-09-21
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    @xdfm: that was very confusing. It sounded from your question as if you had already realized this; instead of "I'm not sure how informative that is" shouldn't you have said "I'm not sure how true that is"?2010-09-21

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A vector space over $\mathbb{Q}$ is the same as a torsionfree divisible abelian group. The product $p/q * v$ is the unique vector so that $q * p/q v = p v$. In particular, every group homomorphism between vector spaces over $\mathbb{Q}$ is automatically $\mathbb{Q}$-linear. Thus, for example, the group automorphisms of $\mathbb{Q}^n$ may be described by $GL_n(\mathbb{Q})$.

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The distinction between preserving addition and also preserving scalar multiplication is illusory in this case, because any vector can be divided by $n$.

The distinction resurfaces, however, for the same problem with $Q$ replaced by (almost) any other field, such as the real numbers. There preserving scalar multiplication is an additional property that does not follow from addition being conserved.

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    Can you explain what you mean by illusory? So if $T$ is an isomorphism, $\alpha$ a scalar, and $v$ a vector, how does $T(\alpha v)=\alpha T(v)$ follow from addition being preserved? As in we could apply $T$ to $(1/\alpha) v$? Why does this fail in the reals? Why can't one divide by any vector by nonzero $r$, as one could do with $n$ in $\mathbb{Q}$?2010-09-21
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    Illusory means that the apparent distinction collapses in this case. Scalar multiplication of x by p/q is expressible in terms of sums of (x/q). But this is true only for Q.2010-09-21
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    Ah, I see what you mean. You would add p terms of (x/q). Thank you.2010-09-21
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    Just curious, if $T((p/q)x)=T(\sum_{i=1}^p x/q)=pT(x/q)$, how does the denominator $q$ move outside the operator without assuming scalar multiplication is preserved?2010-09-21
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    @yunone: multiply everything by the denominator from the start. If a = (p/n)x then nT(a)=T(na)=T(px)=pT(x).2010-09-21