1
$\begingroup$

$$ \left[ \begin{array}{@{}ccccc@{}} 0.9& 0.1& 0& 0& 0& 0& \\ 0& 0.9& 0.1& 0& 0& 0& \\ 0& 0& 0.9& 0& 0& 0.1& \\ 0& 0& 0& 0.9& 0.1& 0& \\ 0& 0& 0& 0.1& 0.9& 0& \\ 0.1& 0& 0& 0& 0& 0.9& \\ \end{array} \right] $$

This is a Perron (P) Matrix, in documents i read μ_2 is the second largest eigenvalue of P matrix. here what is the μ_2 or second largest eigenvalue.

$\endgroup$
  • $\begingroup$ I am sorry Jyrki I gave a wrong example i changed my matrix example.but i want ask an eigenvalue is a sum of a row or a column? $\endgroup$ – Toto Mar 25 '13 at 15:02
  • $\begingroup$ No worries. Unfortunately withot the circulant structure it is hard to tell what the eigenvalues are :-/ $\endgroup$ – Jyrki Lahtonen Mar 25 '13 at 15:07
5
$\begingroup$

There is an excellent comment and response, but I'll write them out explicitly.

For the new matrix, you have:

$\lambda_1 = 1$

$\lambda_2 = 1$

$\lambda_3 = 0.9 + 0.1 i$

$\lambda_4 = 0.9 - 0.1 i$

$\lambda_5 = 0.8$

$\lambda_6 = 0.8$

What do you mean by largest eigenvalue in this context? Do you mean in absolute value, in magnitude or using some other measure?

$\endgroup$
  • $\begingroup$ I mean absolute value but not largest ,largest 2nd eigenvalue..Could i ask how can you calculate λ1=1..λ6=0.8 ? Thank you so much $\endgroup$ – Toto Mar 25 '13 at 15:44
  • $\begingroup$ The standard way $|A -\lambda I| = 0$ and then solve for the roots of the characteristic polynomial. Regards $\endgroup$ – Amzoti Mar 25 '13 at 15:52
  • $\begingroup$ You are very welcome. Regards $\endgroup$ – Amzoti Mar 25 '13 at 17:06
  • $\begingroup$ Amzoti I wonder do you have knowledge about primitive matrix,if yes is this matrix is primitive? $\endgroup$ – Toto Mar 25 '13 at 17:08
  • $\begingroup$ Read the 1-st paragraph en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem. If still not clear, this can be a new question on how to determine it. Regards $\endgroup$ – Amzoti Mar 25 '13 at 17:19
2
$\begingroup$

This is a circulant matrix; the Wikipedia article gives the eigenvectors and eigenvalues.

$\endgroup$
  • $\begingroup$ You are right it is my fault let me give a better example or edit the matrice.Thank you Joriki $\endgroup$ – Toto Mar 25 '13 at 14:49

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.