Diagonal block of a capacitance matrix is also a capacitance matrix

Question

I am trying to conclude the answer to the following question (which I posed on the mathematical branch of the community):

https://math.stackexchange.com/questions/4739174/linear-system-ax-y-with-partially-known-x-y-and-non-singular-a

in order to do so I need to prove the following:

"Let us assume that we have already established that for an isolated system of conductors $\{c_j\}_{j=1,...,N}$ the following is true: $$ Q = CV $$ where $Q=(Q_j)_{j=1,...,N}$ and $V=(V_j)_{j=1,...,N}$ are the vector representing the potentials and the charges on the conductors and $C$ is known as the capacitance matrix and it is invertible. Hence prove that the diagonal block $E$ of this matrix, consisting of the first $m$ columns and rows, is the capacitance matrix of the system composed by the first $m$ conductors $\{c_j\}_{j=1,...,m}$.

Answer 1

I think that your claim is not correct, unless we have different definitions of capacitance matrices.

The issue is that your sub matrix is a capacitance matrix if you can set the potentials of the $N-m$ conductors to zero (i.e. you ground them). However, since the $N-m$ conductors are still present, they will still influence the influence coefficients of the $m$ conductors. In particular, they will distort the field lines to enforce the Dirichlet boundary conditions. Therefore, you cannot expect the capacitance $m\times m$ submatrix to be the capacitance matrix of the $m$ conductors when there are the $N-m$ conductors are not present.

Back to your original problem, you can use the fact that the capacitance matrix is diagonally dominant, so that the the sub matrix $E$ is still invertible.

Hope this helps.