Concerning $f(x_1, \dots , x_n)$ I am not getting even an intuition as how to solve this problem. Please help me with a solution.
Let $n$ be a positive integer and $F$ a field. Let $W$ be the set of all vectors $(x_1, \dots , x_n)$ in $F^n$ such that $x_1+\dots +x_n =0$.
$1)$ Prove that $W^0$ (annihilator of $W$) consist of all linear functionals $f$ of the form $$f(x_1, \dots , x_n) = c \sum _{j=1}^n x_j.$$
$2)$ Show that the dual space $W^*$ of $W$ can be naturally identified with the linear functionals of the form $$f(x_1, \dots , x_n) =c_1x_1 +\dots +c_nx_n$$ on $F^n$ which  satisfy $c_1+\dots +c_n=0$.
 A: Hint. First prove that
$$
\dim W^0=\dim F^n-\dim W
$$
Then prove that $\dim W=n-1$ so
$$
\dim W^0=1
$$
Do you see how this proves (1)?
A: Put $V = F^n$. By definition, $W^0 = \{f\in V^*:\, f\vert W = 0\}$. The inclusion $W \to W\oplus W^\perp = V$ induces an isomorphism $W^0 \oplus W^* \to V^*$, given by $(f, g) \to f + g$. Hence 
$$\dim W^0 = \dim V - \dim W^* = \dim V - \dim W = 1.$$
The function $f(x) = x_1 + \cdots + x_n$ clearly lies in $W^0$, so it spans it. Part (2) is just the definition of dual space.
A: Let's look like this. 
1) A functional $f$ is in $W^\circ$ if and only if $f(x)=0$ whenever $x_1+\cdots+x_n=0$. 
Define a functional $G$ by 
$$g(x)=x_1+\cdots+x_n.$$ Then the kernel of the functional $g$ is contained in the kernel of the functional $f$. With symbols this means $\ker g\subseteq \ker f$.
We claim that there exists a scalar $c$ such that $g=cf$ and the proof will be finished. 
Write $F^n=\ker f\oplus Fe$ where $f(e)=1.$ Then, if $x=y+\lambda e$ for some $y\in \ker f$ and $\lambda \in F$, we have
$$g(x)= g(y)+\lambda f(e)=\lambda g(e)=g(e) \cdot f(x).$$
Now you have $c=g(e)$. 
2) If you write 
$x\in W$ as $x=x_1e_1+\cdots+x_ne_n$, then apply $x_n=-x-\ldots-x_{n-1}$ to get 
$$x=x_1(e_1-e_n)+\cdots+x_{n-1}(e_{n-1}-e_n).$$ If $f\in W^*$, then 
$$f(x)=x_1 c_1+\cdots x_{n-1}c_{n-1}$$ where $c_{i}=f(e_i-e_n).$
As you see the dimension of $W^*$ is $n-1$, so that the basis of the dual of $F^n$ consists of basis for $W^*$ and an additional functional $g$ defined by 
$g(x_1,\ldots,x_n)=-(c_1+\cdots+c_{n-1})x_n.$ Since $(F^n)^*=W^*\oplus Fg$ by the decomposition above every functional $f\in W^*$ is of desired form. 
I hope this helps. 
