The $P_3$ block.

The analysis of $P_3$

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

is nearly identical to that of $P_1$. The invariance

$$R_3P_3R_3=P_3$$

and structure

$$P_3= \left( \begin{array}{l\vert l} q_3 & \hat q_a \\ \hline \hat q_a^t & \tilde Q_3 \end{array}\right)$$

where $\hat q_a$ is an odd row vector, or $\hat q_a J = -\hat q_a$ and

$$JQ_3J=Q_3,$$

obtain the result that the capacitance matrix has the same eigenvalues as $P_3$. The eigenvectors are

$$\hat V=\left(\begin{array}{r} \Theta \\ -\Theta \\ \Theta \\ -\Theta \end{array}\right).$$

We have that

$$\Theta=\left(\begin{array}{l} \theta_0 \\ \hat \theta \end{array}\right)$$

If $\theta_0\neq 0$ then it is true that $\hat\theta$ is odd.

$$\hat \theta=-J\hat\theta.$$

If $\theta_0 = 0$ then it is true that $\hat\theta$ is even.

$$\hat \theta=J\hat\theta.$$

In this case the even $\hat\theta$ are even eigenvectors of $\tilde Q_3$. The odd eigenvectors can be used to find the (odd) eigenvectors of $P_3$ by the same type of construction used for $P_1$

By construction it is possible to show by reduction over the symmetry that if $m-1=2k$ there are $k$ even and $k$ odd eigenvectors of $\tilde Q_3$. If $m-1=2k-1$ there are $k$ even and $k-1$ odd eigenvectors.

Therefore, it is possible to account for all the eigenvectors of $P_3$. The even eigenvectors form what we have called space 4. The odd eigenvectors form space 1.