How Do You Solve Modified Eigenvalue Problems Like Lq=\lambda q + a?

[jollage]

I know that eigenvalue problem like $Lq=\lambda q$ could be easily solved by eig command in Matlab.

But how to solve a problem like $Lq=\lambda q + a$, where $a$ has the same dimension with the eigenfunction $q$?

Thanks a lot in advance.

Jo

 

[jollage]

OK, by reading the article On Inhomogeneous Eigenvalue Problems by Mattheij and Soderlind, this kind of the eigenvalue problem could be solved.

Then I have a new problem, which is $Lq=\lambda q + b\lambda^2$, where b has the same dimension with q. Note that now $\lambda$ is squared and unknown.

How to solve this problem? Thank you.

Jo

 

[Mark44]

Doesn't the article you cited give any insight? It seems odd to me that the authors would trade one problem (solving for eigenvalues in Lq = λq + a) for what seems to be a harder problem, without providing some direction.

 

[jollage]

Doesn't the article you cited give any insight? It seems odd to me that the authors would trade one problem (solving for eigenvalues in Lq = λq + a) for what seems to be a harder problem, without providing some direction.

Dear Mark,

Thank you. My apologies. I only read half of that paper, till where I have found the way to attack the problem Lq = λq + a directly. The remaining part of the paper introduces a method called power iteration.

I thought this kind of problem could be easily and directly solved with some tricks that I still do not know. (like the quadratic eigenvalue problem to be solved with a new eigenfunction, which is elegant and smart). I have a feeling that problem like these two I posted here could always be solved using some iteration methods by guessing and adjusting then guessing... I want to know the elegant and smart way, the so-called tricks.

The second problem is more complex since now λ is squared in the third term while the first problem has only a constant additional term.

Maybe I should finish that paper. Otherwise if you have some experiences and tricks, please share. Thank you.

Jo