I am considering a hyperplane in $\mathbb R^n$ given by
\begin{equation} V=\left\{ (v_1,\ldots,v_n)\in\mathbb R^n: \sum_{i=1}^n v_i =0 \right\}. \end{equation}
Question: Is there an explicit characterisation of the dual space $V^*$ of $V$ (specifically as vector subspace of $\mathbb R^n$)?
Motivation: The space $V$ is in fact the space of signed measures on a finite state space that sum up to $0$. I am trying to characterise its dual space explicitly.
Ideas according to a friend: We know that $V^*$ is the space of all linear function on $V$. Let $\phi_1,\phi_2$ be two linear functions on $V$, which for any $v\in V$ are defined as
\begin{equation*} \phi_1(v) = (1,1,\ldots,1)\cdot v, \ \ \phi_2(v) =(2,2,\ldots, 2)\cdot v \end{equation*}
where $\cdot$ is the usual dot product in $\mathbb R^n$. Then by construction
\begin{equation} \phi_1(v) = \sum_i v_i = 0 = 2 \sum_i v_i = \phi_2(v), \end{equation}
i.e. these two operators are equal (since they agree on $V$). This somehow means that the dual space $V^*$ should have a equivalence class based definition, for instance something like
\begin{equation}\tag{1} V^* = \Bigl\{ \left\{u+c(1,\ldots,1)\right\}: \ u\in\mathbb R^n\Bigr\}, \end{equation}
where the inner curly braces define an equivalence class (poor notation, but I am not sure how to write this correctly), i.e. $u_1\sim u_2$ if $u_2= u_1+ c(1,\ldots,1)$ for any $c\in\mathbb R$.
Issues:
(a) Is (1) the correct dual space? It feels like it since it seems to deal with the underlying constraint on $V$ in a 'reasonable' way -- $V$ has $(n-1)$ dimensions and and somehow $V^*$ should also be $(n-1)$-dimensional which I cannot prove. If (1) is not the right dual space, then what is it? Can we characterise it?
(b) The explanation above feels ad-hoc. Is there a way to make this discussion precise(/rigorous)?
Dualising the inclusion map $i:V \hookrightarrow \mathbb{R}^n$, one obtains the restriction map $i^*: (\mathbb{R}^n)^* \rightarrow V^*$, which is surjective (proving this requires Zorn's lemma, I think), and its kernel is the annihilator $V^0$.
The isomorphism theorem then gives that $V^* \equiv (\mathbb{R}^n)^*/V^0$. In an euclidean space, this can be thought as $\mathbb{R}^n$ modulo the orthogonal subspace $V^{\perp}$.