I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $1$ from page $111$.
If $W$ is a subset of a vector space $V$, we define $W^{0}=\{f\in V^{\star}|f(w)=0 \operatorname{for all}w\in W\}.$
Let $n$ be a positive integer and $\mathbb{F}$ be a field. Let $W$ be a set of all vectors $(x_{1},\ldots,x_{n})$ in $\mathbb{F}^{n}$ such that $x_{1}+\cdots+x_{n}=0$.
a) Prove that $W^{0}$ consists of all linear functionals $f$ of the form $f(x_{1},\ldots,x_{n})=c(\sum_{j=1}^{n}x_{j}).$
b) Show that the dual space $W^{\star}$ of $W$ can be "naturally" identified with the linear functionals $f(x_{1},\ldots,x_{n})=c_{1}x_{1}+\cdots+c_{n}x_{n}$ on $\mathbb{F}^{n}$ which satisfy $c_{1}+\cdots+c_{n}=0$.
My approach for part a) First, we note that $e_{i}-e_{j}\in W$ for all $i,j\in\{1,2,\ldots,n\}$. If $f\in W^{0},$ then we have $f(e_{i})-f(e_{j})=f(e_{i}-e_{j})=0.$ Therefore $f(e_{i})=f(e_{j})$ and we have $f(e_{i})=c$, for all $i\in \{1,2,\ldots,n\}$, where $c$ is a constant. Therefore, if $w=(x_{1},\ldots,x_{n})\in W$, then $f(x_{1},\ldots,x_{n})=c(\sum_{j=1}^{n}x_{j}).$ The converse is easy.
I was not able to solve the part b).