Let $K$ be a field with $|K|=q$ elements and let $V$ be a $K$ vector space.
If $v_1,...,v_d$ linearly independent in $V$. How many vectors $w\in V$ are there such that $v_1,...,v_d,w$ are linearly independent?
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1$\begingroup$ I think you typed $v$ where you meant $w$, and what's the dimension of $V$? $\endgroup$– J. W. TannerCommented May 24, 2020 at 19:21
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$\begingroup$ yes. you're right. $\endgroup$– AnalysisCommented May 24, 2020 at 19:39
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2$\begingroup$ Given that the vectors $\{v_1,...,v_d\}$ are linearly independent, we have that two different linear combinations of those vectors are different. We have that the possible linear combinations of the vectors $\{v_1,...,v_d\}$ is $q^d$. As the dimension of the vector space $V$ is not given, it could be that the dimension is just any natural number $n$ that is larger than $d$. The number of distinct vectors in the vector space $V$ is therefore $q^n$. Hence the number of vectors that is not in the spanning set of $\{v_!,...,v_d\}$ is just $q^n-q^d$. I hope this is correct. $\endgroup$– user614287Commented May 24, 2020 at 19:57
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$\begingroup$ The reason that the possible linear combinations of the vectors $\{v_1,...,v_d\}$ is $q^d$ is because a linear combination of the vectors $\{v_1,...,v_d\}$ is of the form $a_1v_1+...a_dv_d, a_i \in K$. So the possible value that $a_1$ can take is $q$, the possible values that $a_2$ can take is $q$ and so on. so the total possible values that each of the $a_i$ can take is just $q^d$. We give similar reasoning for counting the number of vectors in the vector space $V$ of dimension $n$. As a vector space of dimension $n$ is spanned by $n$ linearly independent vectors. $\endgroup$– user614287Commented May 24, 2020 at 19:59
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