# Solving an Eigenvalue Problem for Large n Matrix

• Pyrokenesis
In summary, the eigenvalues of a symmetric matrix are the same as the eigenvalues of its transpose, det(A-&lambda;I)=det(A^T-&lambda;I). Additionally, for non-symmetric matrices, the eigenvectors are not necessarily the same.f

#### Pyrokenesis

I am having trouble with the following question. (Just hoping to get some guidance, recommended texts etc.):

"Consider an eigenvalue problem Ax = &lambda;x, where A is a real symmetric n*n matrix, the transpose of the matrix coincides with the matrix, (A)^T = A. Find all the eigenvalues and all the eigenvectors. Assume that n is a large number."

Any help would be fantastic!

Originally posted by Pyrokenesis

"Consider an eigenvalue problem Ax = &lambda;x, where A is a real symmetric n*n matrix, the transpose of the matrix coincides with the matrix, (A)^T = A. Find all the eigenvalues and all the eigenvectors. Assume that n is a large number."

Don't know how much help I can be, but since I am studying the same material at the moment, I will help with what I can.

The eigenvalues of A are equal to the eigenvalues of A^T because det(A-&lambda;I)=det(A^T-&lambda;I). The diagonal/trace stays the same here.

(I am guessing on the next part, as our book does not cover this)
Normally, for non-symmetric matrices the eigenvectors are not the same. However, in your case, since A^T=A then (and I am guessing here) I would tend to believe that Laplace expansions would end up yielding equal eigenvectors as well.

Originally posted by samoth
Don't know how much help I can be, but since I am studying the same material at the moment, I will help with what I can.

The eigenvalues of A are equal to the eigenvalues of A^T because det(A-&lambda;I)=det(A^T-&lambda;I). The diagonal/trace stays the same here.

(I am guessing on the next part, as our book does not cover this)
Normally, for non-symmetric matrices the eigenvectors are not the same. However, in your case, since A^T=A then (and I am guessing here) I would tend to believe that Laplace expansions would end up yielding equal eigenvectors as well.

If A^T = A then A is symmetric, no? Does this help?

Yes, A is symmetric when A^T=A.

First of all, I was wrong about the eigenvectors of A and A^T being the same. They are not. However, I cannot help as to why, as our text offers only two sentences in this matter. Further, I do not believe I am at a level of knowledge upon which speculation would prove fruitful. Hmm.. let's see. I can give you some links that will hopefully be of some help.

We are using a book by Gilbert Strang from MIT. He has quite a bit of information on his/the books website, as well as fully recorded lectures. Here is his site.

http://web.mit.edu/18.06/www/

I am sorry I can be of little help with this, but I hope this can help you shed some light on the problem.

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Thanx bro,

I know, its a tough subject, cheers for the link. Now that all other coursework is out of the way I will crack on with this and post my findings when I find something.

Good luck with your course as well,

cheers,

Dexter

Originally posted by Pyrokenesis

I will crack on with this and post my findings when I find something.

Please do, as I am quite curious now.
Good luck as well with your course!