The question is: For a nonzero vector v in a $2$ dimensional space $V$ over $F_q$, find the number of linear functionals $f$ from $V$ to $F_q$ such that $f(v) = 1$. q is given to be a power of prime.
I'm not quite sure how to go about this. I know how to count the number of the bases of such a space, but that's about it. Any hints will be appreciated. Thank you!
On
Consider the function
$$\begin{array}{ccccc} \phi & : & E & \to & W \\ & & f & \mapsto & \ker f \\ \end{array}$$
where $E$ is the set of linear functionals such that $f(v)=1$ and $W$ the set of linear subspaces $U \subseteq V$ of dimension $1$ of $V$ with $\mathbb F_q v \oplus U = V$. $\phi$ is a bijective map as for $U \in W$, there is a unique linear functional defined by $f(u) = 0$ for $u \in U$ and $f(v)=1$.
Therefore, we're left to compute the cardinal of $W$. There are $q^2-q$ vectors that are linearly independent with $v$. And each elements $U \in W$ contains $q-1$ non-zero vectors.
Finally the cardinal of $E$ which is equal to the one of $W$ is
$$\frac{q^2-q}{q-1} = q.$$
Fix $w\notin \text{Span}[v]$, then $\left\{v,w\right\}$ is a basis of $V$. Let $f$ and $f'$ be two functionals such that $f(v)=f'(v)=1$.
We claim that $f=f'$ if and only if $f(w)=f(w')$. Indeed, choose $x\in V$. We can write $x=\lambda v+\mu w$ for a unique $\lambda,\mu \in \mathbb{F}_q$. Hence $f(x)=\lambda+\mu f(w)$ and $f'(x)=\lambda+\mu f'(w)$. The claim follows.
To count the number of functionals $f$ satisfying $f(v)=1$ it now suffices to count the amount of $f(w)$ as this completely determines such a functional.
The "trick" here is that the choice of $w$ is independent of the number of functionals you're looking for, but it does completely determine such a functional.