I considered the basis $$\{e_1-e_2, ... , e_1-e_n\}$$ for $W$, and so found the dual basis $\{f_1, ..., f_n\}$, where $$f_i =-x_i+c \sum_{k=1}^n x_k .$$
However, it is asked to show that $W^*$ can be naturally identified with the linear functionals $$f(x_1, ..., x_n) = c_1x_1+...+c_nx_n$$ with $c_1+...+c_n=0$.
Can someone give me some hint on how to show this.
$V=\mathbb{R}^n$ Then $V^\ast =\{ f_x|x\in V\},\ f_x(y)=(x,y)$.
Hence $W^\ast =\{f_x| x\in W\}$. Hence $ f_x(y)=\sum_{i=1}^n x_iy_i$
[Another] $f(x)=\sum_i\ c_ix_i$ is linear map. Hence $f\in W^\ast$.
If $$g(x)=\sum_i\ c_ix_i -\sum_i\ \frac{ (\sum_i\ c_i) }{n} x_i,$$ then $f=g$. That is, $W^\ast \subset \{ f|f(x)=\sum_i\ c_ix_i,\ \sum_i\ c_i=0\}$. By considering a dimension, they are equal.