Let $S$ be a set of $n$ nonzero vectors in $\mathbb{R}^2$ such that $S$ spans the whole $\mathbb{R}^2$ and let $q$ be the number of pairs of linearly independent vectors from $S$. What is the smallest and the largest possible value of $q$?
My attempt: Since we are in $\mathbb{R}^2$, $q$ cannot be larger than $2$. Since $S$ does not contain zero vector, minimal $q$ is $1$?
Is this correct?
If $n \geq 2$ then $n-1 \leq q \leq \binom{n}{2}$. There are two linearly independent vectors in $S$ let us choose two $v_0, v_1$, then in the worst case all the other vectors are scalar multiples of one of these. But if $0 \neq v \in S$ is a scalar multiple of $v_0$ then $v_1, v$ is is a linearly independent pair. So the smallest possible $q $ is $n-2+1$. And for the largest, any two pairs could be linearly independent in $S$ where $S$ is a collection of nonzero vectors chosen from $n$ distinct lines passing through the origin.
Based on how you have defined $q:=$ number of pairs of linearly independent vectors. The answer for maximum value of $q$ should be $\binom{n}{2}$. For example, if $$S=\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\2\end{bmatrix}, \ldots , \begin{bmatrix}1\\n-1\end{bmatrix}\right\}.$$ Then $|S|=n$, $S$ definitely spans $\mathbb{R}^2$ and if you choose any $2$ vectors from $S$, they will form a basis of $\mathbb{R}^2$, hence linearly independent.
