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How does one count the number of automorphisms of a vector space? If a vector space over $\mathbb F_p$ has $n$ ordered bases how many are there? I think I should be considering the mappings of a set of basis to another, but I am confused because how does/whether one (should) take into account of the identity maps between different bases, or what if a set of basis is a scaled version of another, then how does one do with them? Thanks!

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up vote 5 down vote accepted

Pick your favourite ordered basis $(v_1,v_2,\ldots,v_n)$ of $V$. For any automorphism $f$, $(f(v_1),f(v_2),\ldots,f(v_n))$ is again an ordered basis, while for any ordered basis $(w_1,w_2,\ldots,w_n)$, there's a unique automorphism that maps $v_i$ to $w_i$ for each $i$. Thus the correspondence $f \leftrightarrow (f(v_1),f(v_2),\ldots,f(v_n))$ gives a bijection between the set of automorphisms of $V$ and the set of ordered bases of $V$. So counting one set is the same as counting the other.

So you want to count ordered bases. Suppose $V$ is an $n$-dimensional space over a field with $q$ elements. Suppose further we have already picked the first $k$ elements of an ordered basis. Then these elements span a $k$-dimensional subspace, which contains $q^k$ vectors. So there are $q^n-q^k$ vectors not in the span of our partial basis, and thus available to be the $k+1$-th basis element. So the total number of ordered bases is $\prod_{k=0}^{n-1} q^n-q^k$.

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Thanks, Chris! :-) – Eddie Sep 14 '11 at 11:56

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