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

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

$${A^\star}^\prime=\left( \begin{array}{ccc} -r & 0 & 0\\ r^\prime & -x & 0\\ r^\prime & 0 & -y \end{array} \right)$$ (101)

The eigenvalues are $(-r, -x, -y)$ with eigenvectors ( $r\neq x, x\neq y$)

$$a_r=\left(\begin{array}{c} 1 \\ \frac{r^\prime}{x-r} \\ \fr... ...t), a_y=\left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)$$ (102)

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

$$Q=\left( \begin{array}{ccc} 1 & 0 & 0\\ \frac{r^\prime}... ...1 & 0\\ -\frac{r^\prime}{y-r} & 0 & 1 \end{array} \right)$$ (103)

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

$$({A^\star}^\prime)^-=\left( \begin{array}{ccc} -1/r & 0 &... ...x & 0\\ r^\prime/ry & 0 & -1/y \end{array} \right)\quad .$$ (104)

Similarly $A=\tilde Q\,D\,\tilde Q^-$ with $D = diag(A)$ and

$$\tilde Q=\left( \begin{array}{ccc} 1 & -{r\over x-r} & -{... ...r} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right)\quad .$$ (105)