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

[suckstobeyou]

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

 

[-Christastic-]

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.

 

[suckstobeyou]

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...

 

[abszero]

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.

 

[-Christastic-]

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!

 

[Hurkyl]

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.

 

[suckstobeyou]

I see, so how do you normalize these eigenvectors?

 

[suckstobeyou]

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...

 

[Doc Al]

Your normalized vectors look good to me. I'd multiply the scalars into the vectors (though it doesn't really matter).