Does a non-zero linear functional attain every value? I want to prove the following 

If $y$ is a non-zero linear functional on a vector space $\mathbb V$, and if $\alpha$ is an arbitrary scalar, does there neccessarily exist a vector $x \in \mathbb V$ such that $[x,y] = \alpha$ ?

EDIT: Removed my attempt at it.
 A: Maybe it will help if I remind you of some definitions. 
$y$ is a functional on $V$ means $y$ is a function with domain $V$ and codomain the reals. 
$y$ is a linear functional means if $a$ and $b$ are reals and $u$ and $v$ are in $V$ then $y(au+bv)=ay(u)+by(v)$ (or, in your notation, $[au+bv,y]=a[u,y]+b[v,y]$). 
$y$ is non-zero means there is at least one $v$ in $V$ such that $y(v)\ne0$ (in your notation, $[v,y]\ne0$). 
Now I hope you can put it all together and solve the question. 
A: Recall that the dual space $V^{\ast}$ is the vector space of all linear transformations $T$ from $V$ to $\mathbb{R}$.
So what you're asking is given a $T \in V^{\ast}$, and $a \in \mathbb{R}$ does there always exist a vector $v \in V$ such that $T(v) = a$. This is equivalent to asking if there is a solution for every right hand side, or if your map $T$ is surjective. Now if $V$ is finite dimensional, we can apply the the Rank - Nullity Theorem:


If $T : V \longrightarrow \Bbb{R}$ for $V$ finite dimensional then 
    $$\dim V = \dim \ker T + \dim \operatorname{Im} T.$$


Now if the dimension of the image is zero then this means that the dimensional of the kernel is equal to the dimension of $V$. Since $\ker T$ is a subspace of $V$, this means that $V = \ker T$ (exercise). In other words, $T$ is the zero map so we exclude this possibility.
Therefore this means that $1 \leq \dim \operatorname{Im} T \leq \dim \Bbb{R} = 1$ so that by the same reasoning as before and noting that the image of a linear transformation is always a subspace of the codomain that
$$\operatorname{Im} T = \mathbb{R}.$$
In other words, your map is surjective.
A: If $y$ is nonzero, then there is a $v_0 \in V$ such that $y(v_0)=r\neq 0$. And if $\alpha \in \mathbb R$, then $y({\alpha \over r}v_0)={\alpha \over r}y(v_0)=\alpha$. So $y$ is surjective.
