eigs() runs faster for more eigenvalues of the same matrix

Question

Ilya Tyuryukanov 2017-9-12

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/356200-eigs-runs-faster-for-more-eigenvalues-of-the-same-matrix

编辑： Walter Roberson 2018-8-12

I'm a bit confused: running eigs() for 3..20 largest eingevalues/vectors of a ~3000x3000 sparse real symmetric matrix causes the following run times (in seconds, on Linux):

1.03 1.42 1.02 1.16 1.11 1.81 3.01 3.13 3.73 3.12 2.89 2.03 1.36 1.32 1.27 1.22 1.07 0.97 1.23

That is, solving for 9..14 eingepairs is more difficult, but the same eigenpairs (may be a tiny bit less accurate) can be obtained by computing the first 15 or 20 eigenvalues...

I'm running eigs with the following options:

opts.issym = 1;
opts.isreal = 1;
opts.disp = 0;
opts.maxit = 10000;
opts.tol = 2.2204e-16;
[V, v] = eigs(adj, k, 'la', opts);

The large max number of iterations is to get rid of possible NaN values in the solution.

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Answer 1

Christine Tobler 2017-9-12

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https://ww2.mathworks.cn/matlabcentral/answers/356200-eigs-runs-faster-for-more-eigenvalues-of-the-same-matrix#answer_281257

The reason is that the "search space" used in each iteration is determined based on the number of eigenvalues requested. By increasing k, the size of the subspace is increased, which in this case means that all eigenvalues can be found much faster. There's no way of knowing the optimal subspace size to use, EIGS just uses max(2*k, 20) by default. So increasing opts.p will have the same effect as asking for more eigenvalues.

The reason EIGS is having problems is that the largest eigenvalues of matrix adj are very close together, and the algorithm used in EIGS depends on a gap between eigenvalues.

An alternative would be to just call eig(full(A)) and compute all eigenvalues and eigenvectors. For this specific matrix, it could be faster than EIGS, since the matrix is real symmetric and has a difficult eigenvalue distribution for EIGS. On my machine, this took 2.5 seconds, while most EIGS calls I tried took about 1.5 seconds.