eigs using 'smallestabs' vs scalar
4 次查看(过去 30 天)
显示 更早的评论
Hello,
I have noticed that for some cases of using the eigs command to solve a generalized eigenvalue problem, the smallest non-zero eigenvalue and its corresponding eigenfunction obtained when using 'smallestabs' are complex. However (for the same problem), when targeting the smallest non-zero eigenvalue using a real scalar, the resulting eigenvalue and eigenvector are real. Is there a reason for the inconsistency between eigs(A,B,k,'smallestabs') and eigs(A,B,k,scalar) when targeting the same eignevalue?
Thanks
2 个评论
Christine Tobler
2022-2-16
There really shouldn't be any difference between those two calls. Would you be able to put some input matrices where this happens?
采纳的回答
Christine Tobler
2022-2-18
Hi Jack,
I had initially misunderstood that you were getting different results when passing in 'smallestabs' vs. passing in the numeric scalar 0, which would have been a bug in eigs. Now I understand the question is about differences in the eigenvalues computed between having 'smallestabs' (equivalent to 0) passed in as opposed to 1.248 (close to the targeted eigenvalue).
I couldn't reproduce seeing 1.248 +/- 0.0003i myself, but this is likely due to some machine-dependent round-off. The choice of sigma has an impact on round-off error because the first step that EIGS does is to compute the LU factorization of A - sigma*B, and linear system solves with this matrix are then used in the algorithm to determine the eigenvalues.
Because of this, a small change can have an impact on round-off errors, and this can be relatively large when the shift is close to an eigenvalue (which 0 is, since one eigenvalue is computed as about 1e-12). So for a nearly singular matrix like A, any choice of sigma might be an improvement, even -2 for example, since it means the matrix A - sigma*B that's used isn't close to singular anymore.
There isn't really a systematic reason about returning real vs. complex numbers here. In the special case of a symmetric input A and symmetric positive defined B, the eigenvalues are always real. But for a real nonsymmetric A like here, the eigenvalues are either real or complex conjugate pairs. Because there are two eigenvalues close to 1.2483, it's possible that they get split off into a complex conjugate pair, which is still a short distance from the two real values.
更多回答(0 个)
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!