A partial recursion for the construction of eigenstates.

In the analysis of the capacitance matrix eigenspaces we found the real symmetric matrices $\tilde Q_1$, $\tilde Q_2$ and $H_2$. $\tilde Q_1$ and $Q_2$ have an additional invariance.

$$\tilde Q_1=J\tilde Q_1J$$

$$\tilde Q_2=J\tilde Q_2J$$

It is possible to develop a direct method for the calculation of eigenvectors and eigenvalues that applies to general real symmetric systems and will preserve the invariance of real symmetric systems $\tilde Q_1$ and $\tilde Q_2$. We consider the case of real symmetric $Q$ with symmetry

$$Q=JQJ$$

The general case is immediate.

We partition $Q$ as follows.

$$Q= \left( \begin{array}{l\vert l\vert l} q_1 & \hat q & q_k... ... & J\hat q^t \\ \hline q_k & \hat q J & q_1 \end{array}\right)$$

where $Q_a$ is real symmetric and

$$JQ_aJ=Q_a.$$

To find an even eigenvector and eigenvalue of $Q$

$$Q\hat V_e=\Omega \hat V_e$$

we set

$$\hat V_e=\left(\begin{array}{r} u_0 \\ \hat u_e \\ u_0 \end{array}\right)$$

where $\hat u_e$ is even. $Q_a$ has a full set of even and odd eigenvectors.

$$Q_a\hat u_j^e=\sigma_j^e\hat u_j^e$$

$$Q_a\hat u_j^o=\sigma_j^o\hat u_j^o$$

Expand $\hat u_e$ in the basis $\{\hat u_j^e\}$.

$$\hat u_e=\sum f_j \hat u_j^e$$

Expand $\hat q^t$ in the full set of eigenvectors.

$$\hat q = \sum b^e_j \hat u_j^e + \sum b_j^o \hat u_j^o$$

Find there is a solution iff:

$$f_j^e=\frac{2u_0b_j^e}{\Omega - \sigma_j^e}$$

for $j=1,2,\ldots$, and

$$\Omega=q_1+q_k + 2\sum\frac{b_j^eb_j^e}{\Omega-\sigma_j^e}.$$

The result for the odd eigenvector is:

$$f_j^o=\frac{2u_0b_j^o}{\Omega - \sigma_j^o}$$

for $j=1,2,\ldots$, and

$$\Omega=q_1-q_k + 2\sum\frac{b_j^ob_j^o}{\Omega-\sigma_j^o}.$$

Therefore we have a separate recursion for the even and odd eigenvalues and eigenvectors.

The $k$ $\{\sigma_j^e\}$ are the real eigenvalues associated with the $k$ $\{u_j^e\}$. There are $k+1$ real solutions of the equation for $\Omega$. Of these $k-1$ interlace the $\sigma_j$. One is greater than and one is less than the set of $\{\sigma_j^e\}$.

We note that the function $G(x)$

$$G(x)=x-a -\sum \frac{c_j^2}{x-\sigma_j}$$

has a positive derivative on the real axis. Assume that $\{\sigma_j < \sigma_{j+1}\}.$ In the interval $(\sigma_j,\sigma_{j+1})$ $G(x)$ goes to $-\infty$ when $x\rightarrow \sigma_j$ and goes to $\infty$ when $x\rightarrow \sigma_{j+1}$. In the interval $(-\infty,\sigma_0)$ $G(x)$ goes to $-\infty$ when $x\rightarrow -\infty$ and goes to $\infty$ when $x\rightarrow \sigma_{0}$. In the interval $(\sigma_k,\infty)$$G(x)$ goes to $\infty$ when $x\rightarrow \infty$ and goes to $-\infty$ when $x\rightarrow \sigma_{k}$. The convergence of Newton's method to find the zeros of this function would be rapid.