Can duality be used to find the dimension of this space of linear functionals? Let $F$ be any subfield of the field of complex numbers.
The subspace $W$ of linear functionals: $f(x_1 , \cdots, x_n) = c_1x_1 + \cdots + c_nx_n$ on $F^n$ which satisfies $c_1 + \cdots + c_n = 0$. what is the dimension of this space? I can guess the dimension is $n-1$. Also I can prove this by exhibiting a basis on this space.
I know subspace $V$ of all vectors $(x_1 , \cdots, x_n)$ in $F^n$ such that $x_1 + \cdots + x_n = 0$ also has the dimension $n-1$ by  exhibiting a basis on this space.
Is there a natural way to relate these two facts by duality?
Also, how to prove that $W$ is of dimension $n-1$ using this relation?
EDIT:
Based on the accepted answer(@Mordeus Morgenstern's) I have explicitly constructed an isomorphism between the space $W$ and space $V$, which shows that $W$ has the same dimension as $V$ does, that is $n-1$.
Given any tuple $(x_1, \cdots, x_n)$ in $F^n$, and fix the standard basis on $F^n$, we can just define directly by
$\phi(f) = (f(e_1), \cdots, f(e_n))$, where $\sum_1^nf(e_i) =0$, one can show this is an isomorphism from $W$ onto $V$. This is a restriction of the isomorphism between $(F^n)'$ to $F^n$ provided in the accepted answer.
 A: Let $F$ be any field. Let $f: F^n \to  F$ be a linear functional and let $(e_i)_{i=1}^{n}$ be the canonical basis of $F^n$. Then, you can write:
$$f(x_1,\ldots,x_n) = \sum_{i=1}^{n} x_i f(e_i)$$
Now, $f$ is entirely determined by its action on the canonical basis. In particular, you can define a map:
$$\phi: (F^n)' \to F^n \ , \ f \mapsto (f(e_i))_{i=1}^{n}$$
You can show that this is an isomorphism of vector spaces. Now, in particular, any linear functional $f: F^n \to F$ is just described entirely by an associated vector in $F^n$. In particular, the functional you described can be referenced entirely by the $n$-tuple $(c_1,\ldots,c_n)$. So, they can definitely both be associated by duality.
Edit:
Define the set:
$$S = \{(c_1,\ldots,c_n) \in F^n:c_1+\ldots,c_n = 0\}$$
This is a linear subspace of $S$. We'll show that this has dimension $n-1$. Indeed:
$$c_n = -c_1-c_2-\ldots-c_{n-1}$$
So, it follows that:
$$(c_1,\ldots,c_n) = (c_1,\ldots,-c_1-c_2-\ldots-c_{n-1}) = c_1(1,0,\ldots,-1) + c_2(0,1,\ldots,-1) + \ldots + c_{n-1}(0,0,\ldots,1,-1)$$
Now, I'll leave it to you to show that the vectors which have been listed as part of the expansion above are actually linearly independent.
