I am doing research on the structure in the Schur complements and find an interesting phenomenon:

Suppose that A is from 5--pt laplacian. If I use nested dissection ordering and multifrontal method to compute the LU factorization and then check the last schur complement block, it has low-rank for the off-diagonal blocks.

But, when I use the same method to factorize $A - \lambda I$, where $\lambda$ is some positive value near the eigenvalues of A, then the last schur complement does not have the low-rank property.

I do not know whether the indefinite will change the structure in the schur complement or not. Can anyone provide some reference for this topic?


1 Answer 1


Welcome to the wonderful world of Helmholtz equations. Replace $\lambda \ge 0$ with $\omega^2$ and you are describing a factorization of the Helmholtz equation. You may be interested in this paper, which covers this exact issue. There is also a nice review paper which explains why Helmholtz equations are hard.

  • $\begingroup$ In Ying's paper, he showed that for 2D problem, the schur complement should have the low rank property. He only claims that for 3D problem,the low-rank property is not significant. My problem is a 2D problem, but it dose not have low-rank. $\endgroup$ Commented Apr 23, 2012 at 18:47
  • $\begingroup$ @Willowbrook: I think that you should give it a more careful read. The low-rank property is only argued to hold for 1d subproblems of the 2D problem, and only in the case where an absorbing boundary condition is used. If you introduce one into your formulation, I think that your off-diagonal ranks will significantly decrease, though they should still grow significantly with the problem size. $\endgroup$ Commented Apr 23, 2012 at 18:50

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