The number of possible basis in $n$ dimension.

Question

'Basis' is the set of linearly independent vectors to span the whole vector space, and Basis is not unique for each dimension space.

Then, my question is'how many possible basis for 'n dimension' vector space'

Thanks.

Answer 1

Every invertible matrix gives a basis, which is its columns. These form an $n^2$ dimensional subset of the set of $n\times n$ matrices. That is, almost all matrices are invertible.
On the other hand, if the underlying field is finite, and has $q$ elements, I imagine the number of bases is $(q^n-1)(q^n-q)(q^n-q^2)...(q^n-q^{n-1})/n!$, though come to think of it, the base prime $p$, ($q=p^k$) may cause a problem.

Answer 2

Here are some intuitive answers that suggest the answer is infinite when the underlying field is infinite.

If we consider $\mathbb{R}$ as a one dimensional vector space over itself then it is clear that a basis is simply a choice of non-zero number. There are infinitely many choices.

If we consider $\mathbb{R}^2$ as a two dimensional vector space over $\mathbb{R}$ then a basis is given by a pair of non-zero vectors not lying on the same line. Again you can imagine there being infinitely many choices.