How do I find the eigenvectors for matrices V & T with known eigenvalues?

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Homework Help Overview

The discussion revolves around finding eigenvectors for two matrices, V and T, given their eigenvalues. Participants are exploring the methodology for determining eigenvectors and clarifying the implications of their calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of finding eigenvectors, questioning the correctness of the original poster's methodology and the role of certain variables in their calculations. There is also a focus on the representation of eigenvectors and the implications of their multiplicative constants.

Discussion Status

The conversation is active, with participants providing feedback on each other's understanding and methods. Some guidance has been offered regarding the nature of eigenvectors and their normalization, though no consensus has been reached on specific steps or methods.

Contextual Notes

Participants are navigating potential confusion regarding the setup of the problem, particularly with the matrices involved and the interpretation of eigenvalues. There is also mention of normalization and the representation of eigenvectors, indicating a focus on the properties of these mathematical entities.

suckstobeyou
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Hi, I've got these two matrices (V & T) and omega square, which is what I have found to be the eigenvalues. Could anyone tell me if this is the way to find the eigenvectors for these matrices and if they are correct?
Thanks...
http://img305.imageshack.us/img305/6937/ok4zd.jpg
 
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I'm a little confused with your methodology...I believe it's the T matrix that is throwing me off. Why is there an 'm' in the vector? Are you multiplying an identity matrix by 'm' just to get rid of the 1/m in the omega value?

If my understanding is correct it seems what you have done is fine except that I believe that your subtraction is backwards. I havn't done it myself to see if it makes a differce. But when finding eigenvectors I believe that you must have Vij - (wn)^2*Tij. Like I said, I havn't run through this on pencil and paper so I don't know how much of an effect this will make.

Otherwise, if what I think you are doing is correct, everything looks right to me.
 
yes I think that's why I multiplied m. One other question is x2

[1 ]
[-1]

or

[-1]
[1 ]

how can you know which one it should be? or maybe those two are the same?

thanks again...
 
Eigenvectors are determined up to a multiplicative constant, so [1 -1] is the same eigenvector as [-1 1] as far as the eigenvalue goes. Now, if they form a degenerate subspace (i.e. they both have eigenvalue 2) things get slightly more interesting, but not really.
 
abszero said:
Eigenvectors are determined up to a multiplicative constant, so [1 -1] is the same eigenvector as [-1 1] as far as the eigenvalue goes. Now, if they form a degenerate subspace (i.e. they both have eigenvalue 2) things get slightly more interesting, but not really.

exactly. Even more cool is the fact that they don't even have to be [-1,1]/[1,-1]! They could be anything you want as long as a1=-a2 or -a1=a2! It could be [4,523,231,-4,523,231] or [-87,87], the possibilities are endless! Math is cool!
 
Eigenvectors are determined up to a multiplicative constant, so [1 -1] is the same eigenvector as [-1 1] as far as the eigenvalue goes. Now, if they form a degenerate subspace (i.e. they both have eigenvalue 2) things get slightly more interesting, but not really.
We can say this more precisely: either of these eigenvectors constitute a basis for the eigenspace associated with this eigenvalue.

Or, if you prefer, [1 -1] and [-1 1] both span the same subspace.
 
I see, so how do you normalize these eigenvectors?
 
Is this near close to the normalized form?
http://img124.imageshack.us/img124/4827/scan39nn.jpg

Should I multiply the scalars into the vectors or leave them as is?

Thanks again...
 
Last edited by a moderator:
Your normalized vectors look good to me. I'd multiply the scalars into the vectors (though it doesn't really matter).
 

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