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        general >> wanted >> Linear Algebra ---- eigenvalues/eigenvectors
        
(Message started by: MonicaMath on Mar 3^rd, 2009, 3:00pm)

Title: Linear Algebra ---- eigenvalues/eigenvectors
Post by MonicaMath on Mar 3^rd, 2009, 3:00pm

hi,,

could anyone help me !!

What is the relation between the eigenvalues of the matrix A and thos for the matrix B, where

A= [ b a+c 0 0
a b a+c 0
0 a b a+c
0 0 a b ]

and
B =[ d e 0 0
e d e 0
0 e d e
0 0 e d ]

with d ,e are related to a, b , and c.

These are both tridiagonal matrices.
So Im trying to find the eigenvalues and eigenvectors of a nonsymmetric tridiagonal matrix A by finding the eigenvalues and eigenvectors of a symmetric tridiagonal matrix B, with some relation between d,e and b,a,and c .

so, what u suggest ??

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by Eigenray on Mar 4^th, 2009, 1:08am

Please don't post the same problem to multiple forums.

The idea is probably that for any a,b,c, you can find d,e such that B is similar to A.

If this is the case, you need tr A = tr B and det A = det B. This gives you two equations to relate a,b,c with d,e. Under these conditions, it turns out A and B actually are similar (assuming e http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0). In fact, show that there is a diagonal matrix D such that AD=DB.

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by MonicaMath on Mar 4^th, 2009, 12:54pm

HI,

Thank you for your replay,

I solved the problem, and both A and B have the same eigenvalues, but the eigevectore differ !

my solution is as follows,

d will be: d =b - 2a,
and: e=sqrt(a^2 + ac ),

A and B have the same trace and same determinant, but there still some wrong ??

( take as an example a=2, b=3, c=-7
so d=3, and e=sqrt(-5) )

so what is my mistake ??

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by towr on Mar 4^th, 2009, 1:47pm

on 03/04/09 at 12:54:29, MonicaMath wrote:

d will be: d =b - 2a?

Correct me if I'm wrong, but isn't the trace the sum of the elements on the diagonal? And therefore since we have 4 times b on the diagonal in A, and 4 times d on the diagonal in B, we must have d=b.

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by Eigenray on Mar 4^th, 2009, 1:56pm

If A and B are similar there will be an isomorphism that takes the eigenvectors of one to the eigenvectors of the other. That is, suppose B = P^-1AP. Then v is an eigenvector of B if and only if Pv is an eigenvector of A (with the same eigenvalue).

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by MonicaMath on Mar 4^th, 2009, 2:06pm

Sorry for my misstype ..

d= - c - 2a , (not b - 2a)

and e is still the same.

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by MonicaMath on Mar 4^th, 2009, 2:09pm

Thanks Mr. Eigenray ,,,
you are right .... but how I can fine a formula for that matrix P ?

if you can help me then my problem will be solved ...

thanks

Title: Re: Linear Algebra ---- eigenvalues/eigenvectors
Post by Eigenray on Mar 4^th, 2009, 2:15pm

The relation should be d = b, e² = a(a+c). As a hint, it turns out you can take P to be a diagonal matrix. Since any scalar multiple of P will also work, you can assume the upper-left entry is a 1. Then there are only three variables you should be able to solve for to get AP = PB.