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I was reading this lecture notes from MIT OCW on capacitance .
It says

$V_i =\sum_j P_{ij}Q_{j} $ where the constants $P_{ij}$ are determined by the geometry of the conductors.
This matrix can be inverted, so we can write $Q_i =\sum_j C_{ij}V_{j} $
where as a matrix, $$ \textbf{C} = \textbf{ P}^{-1} $$ , or equivalently $C_{ij}P_{jk} = > δ_{ik}$ .


How can we say by certainity that the matrix $\textbf{C}$ will be invertible(I don't doubt the validity, just wondering how to prove)? Is there any theorem or proof for it?

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Considering the internal electrostatic energy in a system of conductors, one can prove that those two matrices are positive-definite and, hence, invertible.

A detailed derivation of the properties of the elements of the capacitance matrix can be found in

R. M. Fano, L. J. Chu, and R. B. Adler, Electromagnetic fields, energy, and forces, MIT Press, 1968.

Here, I'll just sketch the idea. The electrostatic energy $U$ of a system of $n$ charged conductors is given by

$$U = \sum_{i=1}^n Q_iV_i = \sum_{j=1}^n\sum_{j=1}^n P_{ij}Q_iQ_j = \boldsymbol{Q}^\mathrm{T}\boldsymbol{P}\boldsymbol{Q},$$

where $\boldsymbol{Q} = (Q_1,\ldots,Q_n)^\mathrm{T}$ and $\mathrm{T}$ denotes transposition.

Since the electrostatic energy is always positive, $U\ge 0$, and since it can be shown that $P_{ij} = P_{ji}$ (reciprocity), the above equation implies that $\boldsymbol{P}$ is a symmetric positive semi-definite matrix. Excluding the degenerate case where all the elements are zero, $\boldsymbol{P}$ is positive definite.

A positive definite matrix has positive eigenvalues and it is thus invertible. In fact, if $\lambda$ is an eigenvalue of $\boldsymbol{P}$ with eigenvector $\boldsymbol{Q}$ we have $U = \boldsymbol{Q}^\mathrm{T}\boldsymbol{P}\boldsymbol{Q} = \lambda \boldsymbol{Q}^\mathrm{T}\boldsymbol{Q} = \lambda ||\boldsymbol{Q}||^2>0$, from which $\lambda > 0$.

More on the invertibility of positive definite matrices can be found in a linear algebra book, e.g.,

S. Roman, Advanced linear algebra, Springer, 1992.

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