Timeline for A special eigenvalue problem
Current License: CC BY-SA 3.0
17 events
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| Aug 22, 2017 at 14:22 | comment | added | Surb | @yarchik It makes sense to worry because the critical point of $x^TAx/x^TBx$ are solutions of the equation $(A+A^T)x=\lambda (B+B^T)x$. | |
| Aug 22, 2017 at 14:19 | comment | added | yarchik | @IgorKhavkine The Rayleigh quotient should probably be $\lambda=x^TAx/(x^TBx)$ as in symmetric $A$ and $B$ quotient minimization. However, I worry that there could be some issues related to the fact that matrices are not symmetric. | |
| Aug 22, 2017 at 13:52 | comment | added | Igor Khavkine | @yarchik, presumably you would proceed analogously to how a conjugate gradient solver is used in an eigenvalue iteration for the symmetric case, except that at each step your eigenvalue estimate would be $\lambda \sim y^T A x / (y^T B x)$. But I'm far from an expert on this. | |
| Aug 22, 2017 at 12:09 | comment | added | Surb | Moreover, note that to compute an eigenpair of $AB^{-1}$ you do not need to explicitly compute the inverse of $B$. Just use a power method where you solve a linear system at each iteration. | |
| Aug 22, 2017 at 11:59 | comment | added | Surb | Omitting the requirement that $x^T$ must also be left eigenvector, do you know some iterative method for solving $Ax=\lambda Bx$ for general $A,B$? | |
| Aug 22, 2017 at 11:45 | comment | added | yarchik | @IgorKhavkine This could be a good idea. Assuming a good starting guess is known, what would be the procedure for iterative refinement ? | |
| Aug 22, 2017 at 10:24 | comment | added | Igor Khavkine | Unless you have some independent way to make good starting guesses to feed into an iterative eigen-solver, you might need to check the full eigenvalue spectrum before identifying those $\lambda$ for which the corresponding right- and left-eigenvector pair is of the form $x$ and $x^T$. | |
| Aug 22, 2017 at 10:20 | comment | added | Igor Khavkine | What you could reasonably do is to selectively solve for triples $(\lambda,x,y^T)$ where $x$ and $y^T$ are respectively right- and left-eigenvectors with the same eigenvalue $\lambda$. The biconjugate gradient method (or its stabilized version) iteratiely solves simultaneous linear systems of the form $Ax=b$, $y^TA=c^T$. So it might be a good drop-in replacement for the conjugate gradient method in such an iterative simultaneous eigen-solver. But, as others have mentioned, you have no guarantee that $y=x$. | |
| S Aug 22, 2017 at 10:19 | history | suggested | Rodrigo de Azevedo |
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| Aug 22, 2017 at 10:14 | review | Suggested edits | |||
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| Aug 22, 2017 at 9:05 | comment | added | Federico Poloni | @DirkLiebhold I think that the main difficulty here is knowing which eigenpairs to look for. | |
| Aug 22, 2017 at 8:42 | comment | added | Dirk | I googled a bit and it seems that there are some results on the eigenvalues of a product (in your case: $AB^{-1}$), especially when the matrices are symmetric and positive definite. You might be able to find an algorithm that allows for the computation of these eigenvalues without actually computing the product or the inverse if you look in this direction. Unfortunately, I'm not familiar enough with these topics to help you more. | |
| Aug 22, 2017 at 8:36 | comment | added | yarchik | @DirkLiebhold Yes, it is possible. There are, however, practical problems with this approach: i) for large matrices this might be inefficient; ii) since B is not symmetric in general, one needs to do SVD on B, which is an expensive operation. Also, for the transformed $A$, a general algorithm is needed, while only specific solutions are required. | |
| Aug 22, 2017 at 8:28 | comment | added | Dirk | If $B$ is invertible (e.g. when it is positive definite), can't you just put it on the other side to turn everything into one (or two, if you also want $x^T$) classical eigenvalue problems? | |
| Aug 22, 2017 at 8:24 | comment | added | yarchik | @FedericoPoloni I do know that there are nontrivial solutions. The reason for that is, however, the physical background of my problem. At the moment I cannot say what are the respective mathematical theoretical properties. On the other hand, one can always construct matrices $A$ and $B$ that have the desired solutions. | |
| Aug 22, 2017 at 8:16 | comment | added | Federico Poloni | Do you have any theoretical property on $A$ and $B$ that guarantees that there are nontrivial solution? Because I am afraid that the general case is that there is none. | |
| Aug 22, 2017 at 7:50 | history | asked | yarchik | CC BY-SA 3.0 |