Timeline for Solving $(G^TA^{-1}G)x = b$ without inverting $A$
Current License: CC BY-SA 3.0
10 events
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| May 20, 2012 at 8:15 | comment | added | Arnold Neumaier | This depends a lot on which iterative solver you are using, and whether both rows and columns your matrix is well-scaled. As this now gets farr off the original question, let us further discuss it by email. (My address can be found on my home page.) | |
| May 19, 2012 at 8:30 | comment | added | Costis | Hi Arnold. Thanks, this indeed does work! I tested it with some very small test examples, and it works great. However, my iterative solver is having huge issues inverting the augmented matrix. While it takes only about 80 iterations (a few seconds) to solve a system of the form $Ax=b$ with the original A matrix, the system with the augmented matrix (which is 2n+m x 2n+m or 2n-m x 2n-m using @wolfgang-bangerth 's approach) takes over tens of thousands of iterations (several hours) to solve for one RHS. Are there any strategies for accelerating the convergence? perhaps a preconditioner? | |
| May 17, 2012 at 12:30 | comment | added | Arnold Neumaier | Introduce $y:=A^{-1}Gx$ and $z:=A^{-H}By$, and proceed in analogy to the case worked out. (POssibly you also need to factor $B$ into full rank matrices and introduce an additional intermediate vector.) | |
| May 16, 2012 at 21:58 | comment | added | Costis | Hi guys! Thanks for all the replies; this place is great! An extension to the original question: Assume now that I have $(G^TA^{-H}BA^{-1}G)x=b$, where G and A have the same meaning as in the original question but B is a rank deficient nxn matrix (same size as A) and the whole $G^TA^{-H}BA^{-1}G$ is full rank. How would you go about solving the new system, since now the inverse of B does not exist so you cannot have $AB^{-1}A^H$. I don't think it would work simply with the pseudoinverse of B either. | |
| May 16, 2012 at 19:45 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
remark on exploiting the 0-1 structure.
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| May 16, 2012 at 10:02 | comment | added | Arnold Neumaier | The condition number of the two systems is generally quite unrelated; it depends very much on what $G$ is. - I added to my answer information on how to exploit complex symmetry. | |
| May 16, 2012 at 10:01 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added detials after knowing complex symmetry
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| May 16, 2012 at 9:25 | vote | accept | Costis | ||
| May 16, 2012 at 9:25 | comment | added | Costis | Thank you! A is complex symmetric. Is there reason to expect the condition of the augmented matrix to be worse than that of the original matrix $A$? If m is small, the augmented matrix is only marginally larger in size than A, so I would suspect that solving this augmented system iteratively should not be much tougher than solving a system with A? | |
| May 16, 2012 at 8:59 | history | answered | Arnold Neumaier | CC BY-SA 3.0 |