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'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.

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  • $\begingroup$ If the underlying field ils infinite, there is an infinite number of such basis : if you have a basis $(e_1,...,e_n)$, then $(\lambda e_1,...,\lambda e_n)$ is also a basis for every $\lambda\neq 0$. $\endgroup$ – Tlön Uqbar Orbis Tertius Nov 2 '15 at 7:16
  • $\begingroup$ There are infinite number of bases even for a $2$ dimensional space. $\endgroup$ – cr001 Nov 2 '15 at 7:17
  • $\begingroup$ The cardinal of the set of basis depends on the cardinal of the underlying field and on the dimension of the vector space. If the underlying field is finite the number of basis can be finite. $\endgroup$ – mathcounterexamples.net Nov 2 '15 at 7:21
  • $\begingroup$ Since it impacts the answer over finite fields, are your bases ordered? $\endgroup$ – alex.jordan Nov 2 '15 at 8:19
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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.

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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.

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