In physics, multicomponent-diffusion can be modeled with the following system of equations
$$\sum_j A_{ij}(v_j-v_i)+B_{i}=0 \tag{1}$$.
where $A_{ij}=A_{ji}$, and $0\leq i,j\leq n$. I would like to solve it and find the vector $\mathbf v$.
So I thought this could be rewritten : $(\mathbf V)_{ij}=v_j-v_i$, and then we can rewrite $(1)$ : $(\mathbf A \mathbf V)_{ii} + (\mathbf B)_{ii}=0$. Meaning that we would interpret the $B_i$ as $B_{ii}$, and all the $B_{ij}$ with $i\neq j $ would be random numbers since they don't enter directly the calculation. We could say to simplify that they are $0$. My idea is to interpret the vector $\mathbf B$ as the diagonal of a matrix $\mathbf B$.
If one could write $\mathbf A \mathbf V + \mathbf B=0$, the game would be almost won, since $\mathbf V +\mathbf A^{-1}\mathbf B=0$.
So my questions are : is there a way to extend $(\mathbf A \mathbf V)_{ii} + (\mathbf B)_{ii}=0$ to something like $\mathbf A \mathbf V + \mathbf B=0$ ? And once $\mathbf V =-\mathbf A^{-1}\mathbf B$, how could I get the $v_i$ ?
Edit 1 : adding a constraint.
As noticed in the comments, the system being underdetermined, let's add another constraint : $\sum_i \alpha_i v_i=0$. Or $\sum_{i,j} \beta_i (v_j-v_i)=0$.
Edit 2 : To be more concret.
As an example, let's take 3 equations.
$ A_{12}(v_2-v_1)+A_{13}(v_3-v_1)+B_{1}=0 \\A_{12}(v_1-v_2)+A_{23}(v_3-v_2)+B_{2}=0 \\ A_{13}(v_1-v_3)+A_{23}(v_2-v_3)+B_{3}=0 $.
The matrix $\mathbf A= \begin{bmatrix} A_{11} &A_{12}&A_{13}\\A_{12} &A_{22}&A_{23}\\A_{31}&A_{32}&A_{33} \end{bmatrix}$. For simplicity, since the diagonal doesn't appear in the equation, one could choose: $\mathbf A= \begin{bmatrix}0 &A_{12}&A_{13}\\A_{12} &0&A_{23}\\A_{31}&A_{32}&0 \end{bmatrix}$.
The matrix $\mathbf B=\begin{bmatrix} B_{1}&0&0 \\0&B_{2}&0\\0&0&B_3 \end{bmatrix}$, if we replace all the non diagonal term by $0$.
And the matrix $\mathbf V=\begin{bmatrix} 0&v_1-v_2&v_1-v_3 \\v_2-v_1&0&v_2-v_3\\v_3-v_1&v_3-v_2&0 \end{bmatrix}$.
This case with only three equations can be solved by hand. What's more intersting are the cases with a lot of components. There, having a generalized formula can be convenient.
Suppose we have a vector $\mathrm x \in \mathbb R^n$ of unknowns to be determined. Suppose we are interested in $m$ differences of unknowns. Introducing a signed incidence matrix $\mathrm C \in \mathbb R^{m \times n}$, the vector of differences is $\rm C x$. Suppose now that we have $m$ linear equations in these $m$ differences. Hence, we have
$$\rm A C x = b$$
where $\mathrm A \in \mathbb R^{m \times m}$ and $\mathrm b \in \mathbb R^m$. Note that $\mathrm C 1_n = 0_m$, which is why electrical circuits have a ground node.
Suppose $m = n-1$. Hence, matrix $\rm A C$ is fat and the linear system is underdetermined. Given that there are infinitely many solutions, we can consider a special one, such as the least-norm solution
$$\begin{array}{ll} \text{minimize} & \|\mathrm x\|_2^2\\ \text{subject to} & \mathrm A \mathrm C \mathrm x = \mathrm b\end{array}$$