MHB What is the Cardinality of the Set of Ordered Bases of a Finite-Dimensional Vector Space over a Finite Field?

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The problem involves determining the cardinality of the set of ordered bases, denoted as $\mathscr{B}$, for a finite-dimensional vector space $V$ over a finite field $\mathbb{F}_q$. The cardinality is expressed as a function of the field size $q$ and the dimension of the vector space, $\mathrm{dim}(V)$. No responses were provided to the question posed, indicating a lack of engagement or clarity on the topic. The original poster has included their solution for reference. The discussion highlights the challenge of calculating the number of ordered bases in this mathematical context.
Chris L T521
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Here's this week's problem.

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Problem: Let $V$ be a finite-dimensional vector space over a finite field $\mathbb{F}_q$ of cardinality $q$. Let $\mathscr{B}$ be the set of ordered bases of $V$. Compute the cardinality of $\mathscr{B}$, as a formula involving $q$ and $\mathrm{dim}(V)$.

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No one answered this week's question. You can find my solution below.

Let $B$ be any basis of $V$ and $d:=\dim V$. First choose a non-zero vector in $V$. Since every vector in $V$ can be written as a linear combination of elements in $B$ with coefficients in $\mathbb{F}_q$, there are $q^d-1$ nonzero vectors in $V$. Now assume we have chosen linearly independent vectors $v_1,v_2,\dots, v_n$ for $n<d$. Then $\#\mathrm{Span}\{v_1,\dots,v_n\}=q^n$, and there are $q^d-q^n$ choices for vectors not in the span of $v_1,\dots, v_n$. This process must continue until we have $\dim (V)=d$ linearly independent vectors. Therefore
\[\#\mathscr{B}=\prod_{i=0}^{d-1}(q^d-q^i).\]