I don't understand the task from (b). Is it equivalent to : for every linear functional $f(x_{1}, ..., x_{n})=c_{1}x_{1}+...+c_{n}x_{n}$ on $F^{n}$ which satisfy $c_{1}+...+c_{n}=0$ there exists exactly one and unique functional that belongs to the dual space of $W$?
If it is so, could someone please give me a hint?
Yes. That is, for each $(c_1,\ldots,c_n)\in W$, define $$f_{(c_1,\ldots,c_n)}(x_1,\ldots,x_n) = c_1x_1 + \cdots + c_nx_n$$ and let $W'= \{f_{(c_1,\ldots,c_n)} : (c_1,\ldots,c_n)\in W\}$. Then $W'$ is a subspace of $(F^n)^\ast$.
Part (b) asks to prove that there exist an isomorphism between $W^\ast$ and $W'$.
Clearly $W$ and $W'$ are isomorphic. Can you take it from here?
I worked out the details from leo's hint.
Consider the sequence of functions $$W\rightarrow (F^n)^{*} \rightarrow W^{*}$$ where the first function is $$(c_1,\dots,c_n)\mapsto f_{c_1,\dots,c_n}$$ where $f_{c_1,\dots,c_n}(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ and the second function is restriction from $F^n$ to $W$
We know both $W$ and $W^{*}$ have the same dimension. Thus if we show the composition of these two functions is one-to-one then it must be an isomorphism. Suppose $(c_1,\dots,c_n)\in W\mapsto f_{c_1,\dots,c_n}=0\in W^{*}$.
Then $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)\in W$.
In other words $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)$ such that $\sum x_i=0$.
Let $\{\alpha_1,\dots,\alpha_{n-1}\}$ be the basis for $W$ from part (a) ($\alpha_i$ has one in the first component and $-1$ in the $i$-th component). Then $f_{c_1,\dots,c_n}(\alpha_i)=0$ $\forall$ $ i=1,\dots,n-1$; which implies $c_1=c_i$ $\forall$ $i=2,\dots,n$. Thus $\sum c_i=nc_1$. But $\sum c_i=0$, thus $c_1=0$. Thus $f_{c_1,\dots,c_n}$ is the zero function.
Thus the mapping $W\rightarrow W^{*}$ is a natural isomorphism. We therefore naturally identify each element in $W^{*}$ with a linear functional $f(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ where $\sum c_i=0$.
I am just reading the book and was confused by this question as well. Thanks Leo and Gregory for all the good answers. After reading those answers, I think the part I'm still confusing is that why $f(x) = c_1x_1 + ... + c_n x_n$ where $c_1 + ... + c_n = 0$ spans the entire space of $W^*$. I want to write my thought here.
Clearly a basis for $W$ is the ordered basis $\alpha_1, ..., \alpha_{n-1}$ where $\alpha_i = (1, ..., -1, ...)$ that the first entry is $1$ and the $i+1$ th entry is $-1$.
Then the dual basis of this ordered basis is $f_1 = -x_2$, $f_i = -x_{i+1}$ for $i$ from $1 \to n-1$. This dual basis maps has $f_i(\alpha_j) = \delta_{ij}$ where $\delta_ij$ is the Kronecker delta.
Now for any linear functional in this $W^*$ we have $f = \sum_i^{n-1} k_i f_i$ for some $k_1, k_2,..., k_{n-1}$. This means $f(x) = \sum_i^{n-1} - k_i x_{i+1}$ for $x \in W$. When extend this f from $x \in W$ to $x \in F^n$, one way to do that is to add $\lambda (x_1 + x_2 +... +x_n)$, so $f^\prime(x) = \sum_i^{n-1} - k_i x_{i+1} + \lambda (x_1 + x_2 +... +x_n)$. This ensures that $f^\prime(x) = f(x)$ for $x \in W$.
Rearranging coefficients we have $f^\prime(x) = \lambda x_1 + (\lambda - k_1) x_2 + (\lambda - k_2) x_3 + ... + (\lambda -k_{n-1}) x_n$. Since $k_1$ to $k_{n-1}$ are already given when forming any linear functional $f \in W^*$, we can find a unique lambda such that the coefficients of $f^\prime$ add to $0$. This ensures that each linear functional in $W^*$ can be extended to $F^n$ such that $c_1 + c_2 + ... + c_n = 0$.
Finally, since the dimension of $W^*$ is $n-1$ and the dimension of linear functional such that $c_1 + c_2 + ... + c_n = 0$ is also $n-1$. We get what the author said each linear functional in $W^*$ can be "naturally" identified by a linear functional such that $c_1 + c_2 + ... + c_n = 0$.
I think what's missing here is that the author did not mention anything about extending linear functional from a subspace to the original space (maybe I missed... I did not double check). So I did not know we can have multiple different representations in $F^n$ for each linear functional from $W^*$.
