0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Akiva Weinberger Jan 16 '16 at 23:16
  • $\begingroup$ Wait, what's $\xi$? It only appears in the last equation. $\endgroup$ – Akiva Weinberger Jan 16 '16 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 $\endgroup$ – Sepideh Abadpour Jan 16 '16 at 23:26
  • $\begingroup$ @AkivaWeinberger $\endgroup$ – Sepideh Abadpour Jan 16 '16 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. $\endgroup$ – Akiva Weinberger Jan 16 '16 at 23:39
1
$\begingroup$

$k_{11}$ 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.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Sepideh Abadpour Jan 16 '16 at 23:24
  • $\begingroup$ @sepideh Multiply the numerator and denominator by $\sqrt{\Delta}-(\zeta - 1)$ and expand $\Delta$. $\endgroup$ – amd Jan 17 '16 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.