Diagonalisation of and

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Diagonalisation of ${A^\star}^\prime$ and

Eigenvalue decomposition of ${A^\star}^\prime$ is possible for $r\neq x$ and $r\neq y$ . Write ${A^\star}^\prime$ in short notation:

$\begin{displaymath} {A^\star}^\prime=\left( \begin{array}{ccc} -r & 0 & 0\\ r^\prime & -x & 0\\ r^\prime & 0 & -y \end{array} \right) \end{displaymath}$

(101)

The eigenvalues are

with eigenvectors ( $r\neq x, x\neq y$ )

$\begin{displaymath} a_r=\left(\begin{array}{c} 1 \\ \frac{r^\prime}{x-r} \\ \fr... ...t), a_y=\left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right) \end{displaymath}$

(102)

The degenerate case

has eigenvectors

, $a_{xy}^1=a_x$ , $a_{xy}^2=a_y$ . Then ${A^\star}^\prime = Q D Q^{-1}$ with $D=\rm {diag}(-r, -x, -y)$ and

$\begin{displaymath} Q=\left( \begin{array}{ccc} 1 & 0 & 0\\ \frac{r^\prime}... ...1 & 0\\ -\frac{r^\prime}{y-r} & 0 & 1 \end{array} \right) \end{displaymath}$

(103)

The inverse of ${A^\star}^\prime$ will be needed:

$\begin{displaymath} ({A^\star}^\prime)^-=\left( \begin{array}{ccc} -1/r & 0 &... ...x & 0\\ r^\prime/ry & 0 & -1/y \end{array} \right)\quad . \end{displaymath}$

(104)

Similarly $A=\tilde Q\,D\,\tilde Q^-$ with

and

$\begin{displaymath} \tilde Q=\left( \begin{array}{ccc} 1 & -{r\over x-r} & -{... ...r} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)\quad . \end{displaymath}$

(105)

Markus Mayer 2009-06-22