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How to solve an eigenvalue problem with a constant?

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Old   April 6, 2013, 04:24
Default How to solve an eigenvalue problem with a constant?
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The eigenvalue problem I would like to solve is Lq=\lambda q +a, where a has the same dimension with q.

How to solve this kind of problem? Thanks.

Shu
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Old   April 6, 2013, 04:48
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Filippo Maria Denaro
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Originally Posted by shubiaohewan View Post
The eigenvalue problem I would like to solve is Lq=\lambda q +a, where a has the same dimension with q.

How to solve this kind of problem? Thanks.

Shu
Maybe I do not understand well your question ...but if you have an eigenvalue \lambda and q is an eigenvector, then it is required that the vector a is parallel to q, right?
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Old   April 6, 2013, 10:46
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Maybe I do not understand well your question ...but if you have an eigenvalue \lambda and q is an eigenvector, then it is required that the vector a is parallel to q, right?
Dear FMDenaro

Thank you. I start to understand what you mean by a should be parallel to q. BUt in general, a could make any angle w.r.t q. Since q is unknown for the moment, we don't know what's the angle a makes with q...

Shu

Last edited by shubiaohewan; April 6, 2013 at 11:47.
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Old   April 7, 2013, 01:00
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Originally Posted by shubiaohewan View Post
The eigenvalue problem I would like to solve is Lq=\lambda q +a, where a has the same dimension with q.

How to solve this kind of problem? Thanks.

Shu
Isn't this equivalent to solving (L-\lambda I)q= a? This is just a linear system that can be solved a number of ways.
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Old   April 8, 2013, 11:25
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Originally Posted by TheRenegade View Post
Isn't this equivalent to solving (L-\lambda I)q= a? This is just a linear system that can be solved a number of ways.
But this linear system is still a little special. There are n+1 unknowns (n = size q and "+ 1" is \lambda) while you will only find n unknowns (e.g. q) by solving the linear system. In other words, there is one q for every \lambda you can think of.

Anyway, there are n \lambda for which you will not only find one but infinitely many solutions. These are the Eigenvalues and these are the same as in (L-\lambda I)q= 0 because Eigenvalue only means "the value of \lambda that makes L linearly dependent". You will find a linearly dependent matrix by setting its determinant to 0, which is the "characteristic equation" you might have heard of. In that case, an Eigenvector is one of these infinitely many solutions (e.g. if the Eigenvector for a 2x2 matrix is [1;2] it could also be [2;4] and so on).

Back to your question: You could find some kind of Eigenvector and Eigenvalue, which is the same as in the original Eigenvalueproblem and additionally you will find infinitely many other solutions for infinitely many other \lambda. In the real Eigenvalueproblem, these infinitely many solutions are all 0, which is why you search for the only non-zero solution. But in your case most of them will be non-zero solutions and only you are able to know what this might mean.

My hones advice: You should check if this equation is really what you want/need to solve!

Good Luck!
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