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If I have an n-dimensional vector space over a field with q elements, how can I find the number of bases of this vector space?

käyrätorvi
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1 Answers1

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There are $q^n-1$ ways of choosing the first element, since we can't choose zero. The subspace generated by this element has $q$ elements, so there are $q^n-q$ ways of choosing the second element. Repeating this process, we have $$(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$$ for the number of ordered bases. If you want unordered bases, divide this by $n!$.

wj32
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    I was wondering if this problem can be solved using matrices. As I see it, the number of bases in an n-dimensional vector space over the field q is equal to the number of non-singular nxn matrices over the same field.

    Is this correct? And if so then how can I get an equivalent formula using matrices?

     – käyrätorvi Apr 15 '13 at 12:31
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    @Faz3r: Your observation is correct, but then to compute the number of non-singular matrices, the easiest thing to do is to compute the number of possible first columns, $q^n - 1$, and then given a choice for the first column to compute the number of possible second columns, $q^n - q$, and so on. Since the matrix is invertible, the first $k$ columns must have rank $k$ which is equivalent to saying that the $k$th column must not be in the span of the first $k-1$ columns. – Michael Joyce Apr 15 '13 at 12:48
  • Okay, I think i get it now. Thanks! – käyrätorvi Apr 15 '13 at 13:16