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I'm reading a scientific article. It says that:
Consider the following matrix:
$$[T]=C\left( \begin{array}{ccc} 1 & 0 & \rho \\ 0 & \eta & 0 \\ \rho ^* & 0 & \zeta \\ \end{array} \right)$$
The eigenvalues of $[T]$ will be:
$$\lambda_1=\frac{1}{2} C \left(\zeta +1+\sqrt{\Delta}\right)$$ $$\lambda_2=\frac{1}{2} C \left(\zeta +1-\sqrt{\Delta}\right)$$ $$\lambda_3=C \eta$$
And the corresponding eigenvectors will be:
$$k_1=\begin{pmatrix}\frac{\sqrt{\Delta}-\zeta +1}{2 \rho ^*}\\0\\1\end{pmatrix}$$ $$k_2=\begin{pmatrix}-\frac{\sqrt{\Delta}+\zeta -1}{2 \rho ^*}\\0\\1\end{pmatrix}$$ $$k_3=\begin{pmatrix}0\\1\\0\end{pmatrix}$$

Usind the shorthand notation $\Delta = 4 \left| \rho \right| ^2+(\zeta -1)^2$ then $$\frac{k_{11}}{k_{21}}=-\frac{(\zeta-1-\sqrt{\Delta})^2}{4|\rho|^2}$$
I don't even understand what is $\frac{k_{11}}{k_{21}}$ to start and compute it to see whether it is calculated correctly or not?

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    $\begingroup$ I'm guessing $k_{11}$ refers to the first entry of $k_1$, and $k_{21}$ refers to the first entry of $k_2$, but I'm not sure. $\endgroup$ Commented Jan 16, 2016 at 23:16
  • $\begingroup$ Wait, what's $\xi$? It only appears in the last equation. $\endgroup$ Commented Jan 16, 2016 at 23:24
  • $\begingroup$ Oh , that is zeta. K made a mistake. Please edit it. I'm on mobile now and I can't edit $\endgroup$ Commented Jan 16, 2016 at 23:26
  • $\begingroup$ @AkivaWeinberger $\endgroup$ Commented Jan 16, 2016 at 23:27
  • $\begingroup$ I'm on mobile, too. I don't know why that makes it harder for you to edit things. I'm using the app; maybe you're using a browser? In any case, I edited it. $\endgroup$ Commented Jan 16, 2016 at 23:39

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$k_{11}$ appears to be the first entry in the eigenvector $k_1$, and $k_{21}$ appears to be the first entry in the eigenvector $k_2$. The quotient $k_{11}/k_{21}$ and its expression in the paper should be what you obtain after substituting these two elements and simplifying that quotient algebraically.

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  • $\begingroup$ Well I've tried so far to simplify the fraction and get that answer but I couldn't. Could you show me how do you get that last simplified expression for k11/k21 ? $\endgroup$ Commented Jan 16, 2016 at 23:24
  • $\begingroup$ @sepideh Multiply the numerator and denominator by $\sqrt{\Delta}-(\zeta - 1)$ and expand $\Delta$. $\endgroup$
    – amd
    Commented Jan 17, 2016 at 0:48

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