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Eigenstructure

Starting with the defining equation for an eigenvector $ \underline{e}$ and its corresponding eigenvalue $ \lambda$,

$\displaystyle \mathbf{A}\underline{e}_i = {\lambda_i}\underline{e}_i,\quad i=1,2
$

we get, using Eq.$ \,$(J.10),

$\displaystyle \left[\begin{array}{cc} gc & c-1 \\ [2pt] gc+g & c \end{array}\ri...
...n{array}{c} {\lambda_i} \\ [2pt] {\lambda_i}\eta_i \end{array}\right]. \protect$ (J.14)

We normalized the first element of $ \underline{e}_i$ to 1 since $ g\underline{e}_i$ is an eigenvector whenever $ \underline{e}_i$ is. (If there is a missing solution because its first element happens to be zero, we can repeat the analysis normalizing the second element to 1 instead.)

Equation (J.14) gives us two equations in two unknowns:

$\displaystyle gc+\eta_i(c-1)$ $\displaystyle =$ $\displaystyle {\lambda_i}
\protect$ (J.15)
$\displaystyle g(1+c) +c\eta_i$ $\displaystyle =$ $\displaystyle {\lambda_i}\eta_i$ (J.16)

Substituting the first into the second to eliminate $ {\lambda_i}$, we get

\begin{eqnarray*}
g+gc+c\eta_i &=& [gc+\eta_i(c-1)]\eta_i = gc\eta_i + \eta_i^2 ...
...{g\left(\frac{1+c}{1-c}\right)
- \frac{c^2(1-g)^2}{4(1-c)^2}}.
\end{eqnarray*}

As $ g$ approaches $ 1$ (no damping), we obtain

$\displaystyle \eta_i = \pm j\sqrt{\frac{1+c}{1-c}} \qquad \hbox{(when $g=1$)}.
$

Thus, we have found both eigenvectors:

\begin{eqnarray*}
\underline{e}_1&=&\left[\begin{array}{c} 1 \\ [2pt] \eta \end{...
...t{g\left(\frac{1+c}{1-c}\right)
- \frac{c^2(1-g)^2}{4(1-c)^2}}
\end{eqnarray*}

They are linearly independent provided