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 $c_1+\dots +c_n=0$.
Attempt:
$(1) $Solving $x_1+\cdots+x_n=0$ and considering $x_n$ as the dependent variable, we get the basis vectors for $W$ as $\{\alpha_1,\alpha_2,\dots,\alpha_{n-1}\}$ where $\alpha_i=\big(\underbrace{\dots}_{0\text{'s}},\underbrace{1}_{i^{\text{th}}\text{coordinate}},\underbrace{\dots}_{0\text{'s}},-1\big)$
If $f\in W^0$ is expressed as $f(x_1,\dots ,x_n)=d_1x_1+\cdots d_nx_n$ then $W^0$ is the solution space to the system $AD=0$ where $A$ is the $n-1\times n$ matrix with $\alpha_i$ as it's $i^{\text{th}}$ row and $D=\big(d_1,\cdots,d_n\big)^T.$
Since $A$ is in row-reduced row echelon form, we set $d_n=c$ and hence we get $$f(x_1, \dots , x_n) = c \sum _{j=1}^n x_j.$$
$(2)$ To find the basis $\{f_1,\dots,f_{n-1}\}$for the dual space $W^*$ we need $f_i$'s to satisfy $$f_i(\alpha_j)=\delta_{ij},1\le i,j\le n-1.$$ Looking at $\alpha_i$'s, it is natural to consider $f_i$ as $f_i(x_1,\cdots,x_n)=x_i,1\le i\le n-1$ so that $f_i(\alpha_j)=\delta_{ij},1\le i,j\le n-1.$
Thus $\{f_i\}_{i=1}^{n-1}$ is a basis for $W^*$ and if $f\in W^*$ then \begin{align} f=\sum_{i=1}^{n-1}f(\alpha_i)f_i\\ \implies f(\alpha)=\sum_{i=1}^{n-1}f(\alpha_i)x_i \end{align}
It looks like I am nowhere near in the second part of the problem. Help!
This is a related post but the answers given uses tools which are not introduced in the book yet.
