Bob2

Your hint is just basic rule for vector multiplication, just like with real numbers, that (6*3)*7=6*(3*7) holds. Something similar also holds for matrices and vectors. I think I now managed to prove that they have the same eigenvectors: Was this what you meant? Now the rest of the question seems still quite difficult, because I do not really see the link between what I just proved and what I have to prove (see above in previous post). Namely what do I know about labda2, apart from it's definition. In other words: when I am in a proof how that this specific eigenvector I have in the proof is really labda2 and not something else. I hope you know what I mean. Otherwise I will ry to explain it again.

Thanks Timo. It better be , not for your statement to work. Agree, thanks I'm not sure what you need the proof for. I'd try a more specific definition of what convergence towards eigenvector means. Or in other words, saying that is fine, but I'd motivate more why you look at in the sense of why the convergence of this term towards the 2nd eigenvector means that converges to (a multiple of) . Lay uses in his book Linear Algebra and Its applications a similar proof for the proof/explanation of the power method. Moreover the question is not explictily proof, but it says explain, so a semi formal proof will suffice, I think. Strictly speaking, it is possible that is orthogonal to the eigenvector of the 2nd largest eigenvalue. This is fortunately not the case in the situation. What if there is more than one eigenvector to the 2nd largest eigenvalue? There are always more eigenvectors, all multiples are also eigenvectors. But you probably mean that the geometric multiplicity is not one. I don't know how this works in relation to the power method, because in all examples I have seen there is just one approximation for an eigenvector. So I probably don't have to worry about this. There is another question, that I have to solve: I don't know how the to show that 2 matrices have the same eigenvectors. If you would have to show that 2 matrices have the same eigenvalues, then you would have to proof that they have the same characteristic equation. But how about eigenvectors? Thanks for your help.

I thought it was zero, because due to the fact that v1 is orthogonal to x0, there is no component in the v1 direction when you write x0 as a linear combination of v1,...,v12. The full proof would then be: Could you check whether the rest is correct? Thanks a lot!

[math] \alpha_0 [/math] is 0 right?

Timo, If I understand you correct you mean this: But how does this help me to solve the question? Where can I use the orthogonality of x0 to v? Thanks for your reply.

Hi everyone, I have to solve the following problem, but I can't find the solution: A is an 12x12 symmetric matrix. Let v be an eigenvector for lambda1. Explain why the power method (in almost all cases) converges to the eigenvalue lambda2, (the eigenvalue of A with the seconde largest absolute eigenvalue), if one starts with a vector x0, which is orthogonal to v. They give the following hint: note that A is symmetric, and write x0 as a linear combination of eigenvectors Thanks a lot for your replies. It would really help me if someone knows how to solve this question. Kind regards, Bob