Coefficients of capacitance In my syllabus about electromagnetism, they introduce the coefficients of capacitance by stating that, if we have $n$ conductors enclosed by linear dielectrics, then we can write 'because of linearity'(when there is no free charge in the dielectrics):
$$\tag{1} Q_i = \sum_j k_{ij}V_j$$
From this they can introduce the coefficients of capacitance. An other source that states the same (in a slightly different way) can be found here at p.114.
It feels like a trivial question, but I don't really see how this follows 'because of lineairty'. I hope someone can clarify this for me. I made already some reasoning that will get me there perhaps, I mention it here for if that is the case, but it can be a dead track too. (Please ignore this attempt if it is not the right way to get to $(1)$)
I think they mean by a linear dielectric that:
$$\vec{D} = \epsilon\vec{E}$$
So there follows for the potential (in one dielectric) that:
$$\tag{2}\nabla\cdot(\epsilon\nabla V)  = 0$$ 
So I figured that we can solve for $V$ by superposition:
$$\tag{3} V = \tilde{V}_1+\tilde{V}_2+...+\tilde{V}_n$$
Where $\tilde{V}_i$ satisfies $(2)$ in every dielectric and $\tilde{V}_i = 0$ at conductors $j \neq i$ and $ \tilde{V}_i = V_i $ at conductor $i$.
Now $Q_i$ is given by:
$$\tag{4} Q_i = \int_i{\vec{D}\cdot\vec{da}} = \int_i{-\epsilon\frac{\partial{V}}{\partial{n}}dA}$$
Where $\epsilon$ can change if we go from one dielectric to another and with the symbol $\int_i$  I mean a surface integral about conductor $i$.
Now it feels that with $(3)$ and $(4)$ I can derive $(1)$ but I have no clue how.
 A: I think that "because of linearity" should be read as "because of the superposition principle" (which does rely on the linear response of the dielectrics). You do not need to go into the detail of the field distribution:


*

*As in your example, set all potentials to 0, except for $V_i$. Denote this situation by $(i)$. The corresponding charges on each conductor $j$, $Q_j^{(i)}$ are proportional to $V_i$ (by superposition: $2V_i$ yields $2 Q_j^{(i)}$) and we can define $k_{ji} = Q_j^{(i)}/V_i$.

*Let us now apply simultaneously $V_i$ and $V_m$, with $m \neq i$. By the same superposition principle, $Q_j^{\text{tot}} = Q_j^{(i)} + Q_j^{(m)}$. Applying all potentials and using the definition of $k_{ji}$ above yields the desired relation.

A: I first show that we can write
$$\Phi_j=\sum_{k=1}^NP_{jk}Q_k\tag{1}$$
for a given configuration of conductors; then it it will be straightforward to deduce 
$$ Q_i = \sum_j k_{ij}V_j$$.
To prove $(1)$ we draw on two concepts: principle of superposition, and the uniqueness of the solutions of electrostatics problems.
Consider $N$ isolated conductors. First, assume that there exists a charge $Q_1$ on the first conductor, and others have zero net charge. From the linearity of equations of electrostatics we can deduce that the if we increase the charge density everywhere by a factor of $a$, the electric field and electric potential will change by the same factor everywhere. 
Uniqueness theorems tell us that if an electric field $\vec E$ results in a charge of $Q$ on a conductor, a field $\alpha \vec E$ will give $\alpha Q$; i.e., in our system, the total charge $Q$ on each conductor must increase by the same factor. This means we can write the potential everywhere as $\Phi(\vec r)=P(\vec{r}) Q_1$.
Putting a charge $Q_i$ on the $i$th conductor and invoking superposition,  we'll arrive at equation $(1)$. Solving this linear system for $Q_i$s will give the desired result:
$$ Q_i = \sum_j k_{ij}V_j$$
