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

  • Thread starter Thread starter Chris L T521
  • Start date Start date
Click For Summary
SUMMARY

The cardinality of the set of ordered bases, denoted as $\mathscr{B}$, for a finite-dimensional vector space $V$ over a finite field $\mathbb{F}_q$ can be computed using the formula $q^{\mathrm{dim}(V)}(q^{\mathrm{dim}(V)} - 1)(q^{\mathrm{dim}(V)} - q)(q^{\mathrm{dim}(V)} - q^2)...(q^{\mathrm{dim}(V)} - q^{\mathrm{dim}(V)-1})$. This formula accounts for the number of choices available for each vector in the basis, considering the constraints imposed by linear independence. Understanding this cardinality is crucial for applications in linear algebra and coding theory.

PREREQUISITES
  • Finite fields and their properties
  • Concept of vector spaces and bases
  • Linear independence in vector spaces
  • Cardinality and combinatorial counting principles
NEXT STEPS
  • Study the properties of finite fields, specifically $\mathbb{F}_q$
  • Explore linear algebra concepts related to vector spaces and bases
  • Investigate applications of cardinality in coding theory
  • Learn about combinatorial counting techniques in linear algebra
USEFUL FOR

Mathematicians, computer scientists, and students studying linear algebra, particularly those interested in finite fields and their applications in coding theory and combinatorial mathematics.

Chris L T521
Gold Member
MHB
Messages
913
Reaction score
0
Here's this week's problem.

-----

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 $\mathrm{dim}(V)$.

-----

 
Physics news on Phys.org
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
\[\#\mathscr{B}=\prod_{i=0}^{d-1}(q^d-q^i).\]
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
Replies
1
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
3K