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

The eigenvalues are $(-r, -x, -y)$ with eigenvectors ( $r\neq x, x\neq y$)
\begin{displaymath}
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)
\end{displaymath} (55)

Introduce
\begin{displaymath}
q_{rr}=\frac{x-r}{r}, \quad q_{xx}=\frac{y-x}{x}, \quad q_{yr}=\frac{x}{y-r}\quad .
\end{displaymath} (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
\begin{displaymath}
Q=\left(
\begin{array}{ccc}
q_{rr} & 0 & 0 \\
1 & q_{xx...
...q_{xx}^--q_{yr}) & -q_{xx}^- & 1
\end{array}
\right)\quad .
\end{displaymath} (57)

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

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

Then using the result for ${A^\star}^\prime$ it is
\begin{displaymath}
\tilde Q=\left(
\begin{array}{ccc}
q_{rr}^- & -q_{rr}^- q...
...
0 & q_{xx} & 1 \\
0 & 0 & 1
\end{array}
\right)\quad ,
\end{displaymath} (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:

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


next up previous
Next: Special results for EFM(2) Up: EFM(2) Previous: Model specification
Markus Mayer 2009-06-22