Timeline for Special considerations when using the Woodbury matrix identity numerically
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2023 at 9:30 | comment | added | Luca Citi | Note that the matrix in the comment above this one is called "bordered" or "arrowhead" matrix and there may be solvers that exploit this structure that you could look into. | |
| Nov 10, 2011 at 8:22 | comment | added | Federico Poloni | If you really can't avoid the inverse, then sparse LU is your only option. You can still try to use the extended matrix trick: the $(1,1)$ block of $\begin{bmatrix}A & U\\\\ U^* & B\end{bmatrix}^{-1}$ should be the inverse you're looking for. | |
| Nov 9, 2011 at 12:56 | comment | added | Kiyo | I do need the inverse. It's a statistical code map making code, but you can think of it as adding up a bunch of Fisher matrices. | |
| Nov 8, 2011 at 22:30 | comment | added | Federico Poloni |
Do you really need an inverse, or do you only need to solve linear system with it (i.e., compute $A^{-1}b$ or $c^T A^{-1}$)? Chances are you only need the second; in this case, your choice for a general linear system is between sparse LU and iterative solvers (like GMRES), depending on the size and sparseness of the matrix. Both can exploit the structure of the augmented matrix I am suggesting. In any case, don't use inv() or compute an inverse explicitly unless you really have to.
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| Nov 8, 2011 at 15:09 | comment | added | Kiyo | Can this be done fast the way SMW lets you? Or do I have to do the whole rank(A) + rank(B) inversion? I guess I'm a little confused about what you are proposing. | |
| Nov 8, 2011 at 13:54 | comment | added | Federico Poloni | No. When you do SMW, you constrain the inversion to be done "starting from a certain block", which is not necessarily the best thing to do. Think for instance to the case in which $B$ is ill-conditioned. When you invert the big matrix, say, with LU, suitable pivoting is done so that you never have trouble with ill-conditioned subblocks; on the other hand, if you go with SMW, you start by inverting $B$, which is troublesome. | |
| Nov 8, 2011 at 13:45 | comment | added | Kiyo | Since Woodbury is derived from block wise matrix inversion, shouldn't your alternative be equivalent? | |
| Nov 8, 2011 at 8:18 | history | answered | Federico Poloni | CC BY-SA 3.0 |