Linear algebra/numerical analysis: Power method question(s)

[fluidistic]

Homework Statement

Ok I understand how to find an approximate value for the largest eigenvalue of a given matrix A. I use the power method (or the normalized one) to find an eigenvector associated to the approximate largest (in the sense that its modulus is the largest) eigenvalue.
Then I use Rayleigh's quotient as mentioned in http://college.cengage.com/mathematics/larson/elementary_linear/4e/shared/downloads/c10s3.pdf.
However looking in wikipedia about Rayleigh's quotient, it seems that the matrix A must be Hermitian (see http://en.wikipedia.org/wiki/Rayleigh_quotient).

In my assignment I'm been given a matrix A which is NOT Hermitian! So I wonder 2 things I'd like an anwer:
1)Can I still use Rayleigh's quotient, why or why not?
2)I'm more than sure there's another way to get the greatest eigenvalue from the approximate eigenvector, that does not involve Rayleigh's quotient but does involve an infinity norm. I've searched on the web about this and found really nothing. If you know it, please let me know what is this alternative.
Thanks a lot in advance.

 

[lanedance]

the example in your notes is not a symmetric matrix - Ex3

if you think about it geometrically, it should still work - each time you multiple by the matrix, the component in the direction of the dominant eigenvector gets scaled the most

the thing about hermitian matricies, are that they're guaranteed to have an orthonormal basis of eigenvectors, so the method above should converge pretty quickly

 

[fluidistic]

the example in your notes is not a symmetric matrix - Ex3

if you think about it geometrically, it should still work - each time you multiple by the matrix, the component in the direction of the dominant eigenvector gets scaled the most

the thing about hermitian matricies, are that they're guaranteed to have an orthonormal basis of eigenvectors, so the method above should converge pretty quickly

Thanks for your reply. I think I understand what you mean.
If someone could help me with part 2) I'd be glad too.