Number of bases of an n-dimensional vector space over q-element field.

linear-algebra

Solution 1:

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 $n!$.

Related

A good Open Source book on Analytic Geometry?
Prove that a $k$-regular bipartite graph has a perfect matching
Why does the ideal $(a+bi)$ have index $a^2+b^2$ in $\mathbb{Z}[i]$? [duplicate]
Integral of $\int_0^{2\pi}\cos^n(x)\,dx$.
Bounded sequence with divergent Cesaro means
Prove $\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})$ [duplicate]
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions
Proofs of Determinants of Block matrices [duplicate]
Mean Value Theorem for complex functions?
Field Extension Notation
Reference for a tangent squared sum identity
Showing $(C[0,1], d_1)$ is not a complete metric space