Can all vectors be expressed as the product of a symmetric matrix with non-zero off-diagonal elements and the vector of ones?

Question

\begin{equation} \forall \ (\mathbf{x} \in \mathbb{R}^n) \ \exists \ (A \in \mathbb{A}) \ : \ A \mathbf{1} = \mathbf{x} \end{equation}

Hi, I'm writing a paper about certain difference equations on undirected graphs. I'm trying to prove that any steady-state solution is possible on the complete graph. Here $\mathbb{A}$ is the set of symmetric matrices which have non-zero off-diagonal elements (all weighted adjacency matrices for the complete graph), and $\mathbf{1} $ is the vector of ones.

I'm an undergraduate, and I don't have a clue as to where to start with this. Is is trivially true because we always have more elements of $A$ than equations to solve them with? If not, what are the possible $\mathbf{x}$? Which $A$ satisfy $A \mathbf{1}=\mathbf{x}$ for a given $\mathbf{x}$, and is there an algorithm for finding the elements of $A$?

Update: I think I've made some headway, but not much. For the $2 \times 2$ case the following holds:

\begin{bmatrix} c_1 x_1+c_2 x_2 & (1-c_1)x_1-c_2 x_2\\ (1-c_1)x_1-c_2 x_2 & -(1-c_1)x_1 + (1+c_2)x_2\\ \end{bmatrix}

Answer 1

Let $\ b\ $ be any non-zero number, $\ a_i= x_i-(n-1)b\ $ for all $i$, and $$ A =\pmatrix{a_1&b&b&\dots&b&b\\ b & a_2& b&\dots&b&b\\ b & b& a_3 &\dots &b&b\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ b&b&b&\dots&a_{n-1}& b\\ b&b&b&\dots&b& a_n }\ . $$

Answer 2

Define $A_{ij} = x_i / n$. Would that suffice?

EDIT: duh, this is not symmetric. See lonza's answer instead.