suppose solving sequential generalized eigenvalue problems

$$A_i x= \lambda Bx, i=1,2,3,\ldots $$

In general setting, we always need to perform LU for matrix B (preconditioned) before to apply the rest iterative algorithm. Is there a numerical library(I have programming experiences with PETSc+SLEPc) or a toolkit that can allow me to separate those two parts, thus to perform LU only once?

By default, LU factorization of $B$ is by direct solver, whose costs may be somewhat comparable, I suppose.

Update: thanks to Arnold, but I want to modify my problem a little, where $B$ has a null vector s.t. $B\mathbf{1}=\mathbf{0},\quad \mathrm{rank}(B) = n-1$ where $A_i,B$ are both $n\times n$ sparse symmetric matrix

  • $\begingroup$ Are $A$ and $B$ dense or sparse? $\endgroup$
    – Dan
    May 23 '12 at 22:30
  • $\begingroup$ Are you sure you mean preconditioned? I see no point to preconditioning B if you are using a direct solver to obtain an LU decomposition for it. $\endgroup$
    – Costis
    May 23 '12 at 23:07
  • 1
    $\begingroup$ Which eigenvalues are you looking for and with which method? If you are looking for small eigenvalues using shift-and-invert, then won't you need a preconditioner for $K_i = A_i - \alpha B$? If that is the case, then no, there is not a reliable way to reuse a factorization of $K_i$ when solving with $K_j$ (though if they are very small or if $K_j$ is well approximated by $K_i + (\text{low rank})$, then there are opportunities. $\endgroup$
    – Jed Brown
    May 24 '12 at 3:30
  • $\begingroup$ @Costis precondition here is for iterative eigenvalue solver. $\endgroup$
    – bobye
    May 24 '12 at 13:24
  • $\begingroup$ @JedBrown yeah, I have already invert the problem, and expected to find large eigenvalues. Arnold give a solution when B is nonsingular, but unfortunately B has a null space, $B \mathbf{1} = \mathbf{0}$. I hope to have some way to work around shift transform. $\endgroup$
    – bobye
    May 24 '12 at 13:27

Factor $PB=LU$ yourself and write a routine for evaluating $L^{-1}PAU^{-1}x$ given $x$ (using two backsolves). Then you can solve the problem $(L^{-1}PAU^{-1}-\lambda I)z=0$ with a standard iterative solver for the ordinary eigenvalue problem.

If $B$ is singular, compute a left null space basis consisting of the rows of $M$, and a right null space basis consisting of the columns of $N$, so that $MB=0$ and $BN=0$. Then you can replace the eigenvalue problem by the modified problem with the matrices $A'=\pmatrix{A & sAN\\MA & sC}$ and $B'=\pmatrix{B & AN\\MA & C}$, with $C$ and $s$ arbitrary. If $x$ solves the original eigenvalue problem then $x'=\pmatrix{x\\0}$ is an eigenvector of the new problem with the same eigenvalue. Now the kernel of the matrix $\pmatrix{B\\MA}$ is trivial since otherwise the eigenvalue problem is ill-posed. This implies that $B'$ is a nonsingular matrix. Thus one can apply the preceding to the modified problem.

The new eigenvalue problem also has the eigenvalue $s$, attained for all vectors of the form $x'=\pmatrix{0\\z}$. Therefore one should choose $s$ such that it lies somewhere in the middle of the expected spectrum.

  • $\begingroup$ Very useful, but I found in my situation, $B$ has null space given by $\mathbf{1}$(in fact, stiff matrix derived from FEM), if I apply shift rule $B-\alpha A_i$, I would compute LU each time. Is there a way to work around? $\endgroup$
    – bobye
    May 24 '12 at 13:21
  • $\begingroup$ You can probably change the near zero entry in the $U$ factor so something of order $\sqrt{\epsilon}$ times the norm of $U$ without seriously degrading performance. $\endgroup$ May 24 '12 at 14:47
  • $\begingroup$ @bobye: I updated my answer to account for the singular case without any approximation. $\endgroup$ May 25 '12 at 12:35
  • $\begingroup$ many thanks, I think it is a very reasonable approach, that does work for me. I will try to implement. $\endgroup$
    – bobye
    May 26 '12 at 2:02

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