Let F3 be the field of 3 elements and let F3×F3 be the vector space over F3. What is the number of distinct linearly independent sets of the form {u, v}, where u, v ∈ F3×F3 \{(0,0)... Let F3 be the field of 3 elements and let F3×F3 be the vector space over F3. What is the number of distinct linearly independent sets of the form {u, v}, where u, v ∈ F3×F3 \{(0,0)} and u ≠ v?
Understand the Problem
The question is asking about the number of distinct linearly independent sets of the form {u, v} in the vector space F3×F3, where u and v are non-zero vectors and are not equal to each other. This involves concepts from linear algebra and finite fields.
Answer
$$ \frac{(q^3 - 1)(q^3 - q)}{2} $$
Answer for screen readers
The number of distinct linearly independent sets of the form ${u, v}$ in the vector space $F^3$ where $u$ and $v$ are non-zero and not equal is given by: $$ \text{Distinct sets} = \frac{(q^3 - 1)(q^3 - q)}{2} $$
Steps to Solve
- Determine the vector space dimensions
The vector space $F^3$ consists of all 3-dimensional vectors over the field $F$, which we can assume has $q$ elements. Thus, $\dim(F^3) = 3$.
- Count non-zero vectors
In $F^3$, there are a total of $q^3$ vectors, and since we are interested in non-zero vectors, we have $q^3 - 1$ choices for the first vector $u$.
- Selecting the second vector
After choosing $u$, we need to select $v$. The vector $v$ must be different from $u$, and since $u$ is already chosen (and cannot be zero), we still have $q^3 - 1$ vectors to choose from for $v$.
- Linearly independence condition
The vectors $u$ and $v$ must be linearly independent, which means they cannot be scalar multiples of each other. For a fixed $u$, there are $q - 1$ choices for $v$ that maintain linear independence (because for each non-zero scalar in the field, $au$ (with $a \in F$) corresponds to the scalar multiples).
- Calculate total combinations respecting independence
To sum it all up, the total number of distinct pairs ${u, v}$ can be calculated as: $$ (q^3 - 1)(q^3 - 1) - (q - 1)(q^3 - 1) $$
This accounts for choosing $u$, then choosing unrestricted $v$, minus the $q - 1$ scalar multiples of $u$.
- Final formula simplification
To get the final count of distinct linearly independent pairs, we use the formula: $$ \text{Total pairs} = (q^3 - 1)(q^3 - 1) - (q - 1)(q^3 - 1) $$
This simplifies to: $$ (q^3 - 1)(q^3 - q) $$
Since we should only count each pair once (order doesn't matter), we divide by 2: $$ \text{Distinct sets} = \frac{(q^3 - 1)(q^3 - q)}{2} $$
The number of distinct linearly independent sets of the form ${u, v}$ in the vector space $F^3$ where $u$ and $v$ are non-zero and not equal is given by: $$ \text{Distinct sets} = \frac{(q^3 - 1)(q^3 - q)}{2} $$
More Information
This formula counts the number of ways to select two linearly independent vectors in a finite-dimensional vector space, considering only non-zero unique sets. This is a fundamental result in linear algebra when dealing with vector spaces over finite fields.
Tips
One common mistake is forgetting to account for the linear independence condition, leading to the overcounting of pairs that are scalar multiples of each other. Always ensure to subtract these cases appropriately.
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