The eigenvalue problem

The eigenvalue problem for the capacitance matrix

$$C_n\hat V=\lambda\hat V$$

now reduces by

$$\hat V = (F_n\otimes I_m)\hat \Theta$$

to the block diagonal

$${\rm diag}(P_1,P_2,\ldots,P_n)\hat \Theta = \lambda\Theta.$$

Therefore, the eigenvalues and eigenvectors of $C_n$ are determined by the systems

$$P_j\theta_j=\lambda_j\theta_j.$$

for $j=1,2,\dots,n$. Since the $P_j$are symmetric or hermitian the $\lambda_j$ are real.

With $z_j=w^{(j-1)}$,

1. For real $z_j$, $\lambda_j$ is a simple eigenvalue of $C_n$.
2. For complex $z_j$, $P^*_j=P_j=\overline{P}_{n+2-j}$ implies that $\lambda_j$ is a double eigenvalue of $C_n$.

For the equilateral triangle with $w=e^{\imath 2\pi/3}$

$$P_1=A_1+A_2+A_2^t$$

$$P_2=A_1+\overline{w}A_2+wA_2^t$$

$$P_3=A_1+wA_2+\overline{w}A_2^t$$

where $A_1=A_1^t$ and $P_3=\overline{P_2}$.

For the square with $w=e^{\imath 2\pi/4}$

$$P_1=A_1+A_2+A_2^t+A_3$$

$$P_2=A_1+\imath(A_2-A_2^t) -A_3$$

$$P_3=A_1-(A_2+A_2^t)+A_3$$

$$P_4=A_1-\imath(A_2-A_2^t) -A_3$$

where $A_1=A_1^t$, $A_3=A_3^t$ and $P_4=\overline{P_2}$.