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

Diagonalisation of ${A^\star}^\prime$ is possible for $\kappa^\star_r\neq \kappa^\star_x, \kappa^\star_x\neq \kappa^\star_y$. For ease of notation, write ${A^\star}^\prime$ as follows:

$${A^\star}^\prime=:\left( \begin{array}{ccc} -r & 0 & 0\\ r & -x & 0\\ 0 & x & -y \end{array} \right)$$ (54)

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

$$a_r=\left(\begin{array}{c} \frac{x-r}{r} \\ 1 \\ \frac{x}{y... ...t), a_y=\left(\begin{array}{c} 0 \\ 0 \\ 1\end{array}\right)$$ (55)

Introduce

$$q_{rr}=\frac{x-r}{r}, \quad q_{xx}=\frac{y-x}{x}, \quad q_{yr}=\frac{x}{y-r}\quad .$$ (56)

For $r\neq x, x\neq y$ diagonalisation is possible, ${A^\star}^\prime = Q D Q^{-1}$, with $D=\rm {diag}(-r, -x, -y)$ and

$$Q=\left( \begin{array}{ccc} q_{rr} & 0 & 0 \\ 1 & q_{xx... ...q_{xx}^--q_{yr}) & -q_{xx}^- & 1 \end{array} \right)\quad .$$ (57)

Similarly, the diagonalisation of $A = \tilde Q\,\tilde D\,\tilde Q^-$ is achieved. Write $A$ in short notation,

$$A=:\left( \begin{array}{ccc} -r & r & 0\\ 0 & -x & x\\ 0 & 0 & -y \end{array} \right)\quad .$$ (58)

Then using the result for ${A^\star}^\prime$ it is

$$\tilde Q=\left( \begin{array}{ccc} q_{rr}^- & -q_{rr}^- q... ... 0 & q_{xx} & 1 \\ 0 & 0 & 1 \end{array} \right)\quad ,$$ (59)

where $q_{rr},q_{xx},q_{yr}$ defined as above except with $x,y,z$ the entries of matrix $A$.

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

$$({A^\star}^\prime)^- = -\left( \begin{array}{ccc} r^{-1... ...0\\ y^{-1} & y^{-1} & y^{-1} \end{array} \right) \quad .$$ (60)