How the functions in a dual space look like? Let $n$ be a positive integer and $F$ a field. Let $W$ be the set of all vectors $(x_1,....,x_n)$ in $F^n$ such that $x_1+.....+x_2=0.$
Show that the dual space $W^*$ of $W$ can be 'naturally' identified with the linear functionals
 $f(x_1,....,x_2)=c_1x_1+...+c_nx_n$ on $F^n$ which satisfy $c_1+...+c_n=0.$
I have found out the basis for $W$ which is $B_W=\{(-1,1,0...,0),(-1,0,1,0...,0),(-1,0,0,1,0...,0),.....,(-1,0,....,0,1)\}$ and 
$dim$ $W=n-1.$ After that I am not able to proceed.
Thanks!
 A: Write $F^n=:V$. The set
$$W:=\bigl\{(x_1,\ldots, x_n)\in V\>\bigm|\> x_1+x_2+\ldots+x_n=0\bigr\}\subset V$$
is a subspace of dimension $n-1$.
On the other hand, for any functional $\phi\in V^*$ there are uniquely determined coefficients $c_i$ such that
$$\phi(x)=\sum_{i=1}^n c_ix_i\qquad(x\in V)\ .$$
Let $e:=(1,1,\ldots,1)\in V$. Then $\phi(e)=\sum_{i=1}^n c_i$. Regarding the question we can therefore say that we are considering the subspace $X\subset V^*$ defined by
$$X:=\bigl\{\phi\in V^*\>\bigm|\>\phi(e)=0\bigr\}\ .$$
It is claimed that the map
$$T:\quad X\to W^*,\qquad \phi\ \mapsto\ \  \phi\restriction W$$
is an isomorphism. 
Proof. It is obvious that the restriction of a $\phi\in X\subset V^*$to $W$ is  an element of $W^*$, and that $T$ is linear. Since both $X$ and $W^*$ have the same dimension $n-1$ it is therefore sufficient to prove that ${\rm ker}(T)=0$. To this end consider a $\phi\in X$ with $T(\phi)=\phi\restriction W=0$. Then $\phi(e)=0$, and $\phi(y)=0$ for all $y\in W$. Consider now an arbitrary $x\in V$, and put $${1\over n}\sum_{i=1}^n x_i=:\mu\in F\ .$$ Then $y:=x-\mu e\in W$, and therefore
$\phi(x)=\phi(y)+\mu\phi(e)=0$. Since $x\in V$ was arbitrary this shows that $\phi=0\in V^*$.$\qquad\square$
