What is the number of basis in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$?

Question

What is the number of basis in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$ ?

I would say $(p^n -1)(p^n - p)(p^n - p^2)...(p^n - p^{n-1})$ since there are $p^n$ vectors. Do you think it is ok ?

That should be the number of ordered bases. I.e. the size of $\mathrm{GL}_n$. — jgon, Dec 19 '17 at 18:53
Yes it is the same. (fixing one basis, all function (or matrix) of $GL_n$ is bijectively determined by the choice of one other basis.) — , Dec 19 '17 at 18:53
So if you're looking for the number of ordered bases, then I agree that this is correct, otherwise divide by $n!$ if you are looking for the number of unordered bases. — jgon, Dec 19 '17 at 18:55
Not particularly, but I usually think of a basis for a vector space as being an (unordered) set rather than a list, so I just wanted to make sure whether you intended to count ordered bases or unordered bases. — jgon, Dec 19 '17 at 19:02

score 1 · Accepted Answer · answered Dec 19 '17 at 19:00

I'll begin the process of counting, you finish it...and I will count the elements of $\;GL_n(\Bbb F_p)\;$ :

For the first column (of some matrix there) we can take any element in $\; \left(\Bbb F_p\right)^n\;$, except the vector $\;(0,0,...,0)\;$ , thus: there are $\;p^n-1\;$ options.

For the second column we can take any element in $\; \left(\Bbb F_p\right)^n\;$ except the scalar multiples of the first chosen column, and thus there are $\;p^n-p\;$ options ...etc.

You finish the above counting

Indeed, that was what I thought. Thanks. – Dec 19 '17 at 19:01 — , Dec 19 '17 at 19:01

What is the number of basis in a $\mathbb Z/p\mathbb Z$ - vector space of dimension $n$?

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