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Topic: Linear Algebra  eigenvalues/eigenvectors (Read 3290 times) 

MonicaMath
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Linear Algebra  eigenvalues/eigenvectors
« on: Mar 3^{rd}, 2009, 3:00pm » 
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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 ??


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Eigenray
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #1 on: Mar 4^{th}, 2009, 1:08am » 
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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 0). In fact, show that there is a diagonal matrix D such that AD=DB.

« Last Edit: Mar 4^{th}, 2009, 1:11am by Eigenray » 
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MonicaMath
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #2 on: Mar 4^{th}, 2009, 12:54pm » 
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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 ??


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towr
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #3 on: Mar 4^{th}, 2009, 1:47pm » 
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on Mar 4^{th}, 2009, 12:54pm, MonicaMath wrote: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.


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Eigenray
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #4 on: Mar 4^{th}, 2009, 1:56pm » 
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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^{1}AP. Then v is an eigenvector of B if and only if Pv is an eigenvector of A (with the same eigenvalue).


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MonicaMath
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #5 on: Mar 4^{th}, 2009, 2:06pm » 
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Sorry for my misstype .. d=  c  2a , (not b  2a) and e is still the same.


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MonicaMath
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #6 on: Mar 4^{th}, 2009, 2:09pm » 
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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


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Eigenray
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Re: Linear Algebra  eigenvalues/eigenvectors
« Reply #7 on: Mar 4^{th}, 2009, 2:15pm » 
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The relation should be d = b, e^{2} = 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 upperleft entry is a 1. Then there are only three variables you should be able to solve for to get AP = PB.


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