Timeline for How to solve a system of linear equations without storing the matrix?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2012 at 14:25 | comment | added | Brian Borchers | If the matrix isn't sparse, and the cost of getting individual matrix entries is large compared to the cost of accessing an element of a matrix stored in conventional dense matrix form, then iterative methods are going to be horribly slow in practice. | |
| Dec 16, 2012 at 14:22 | comment | added | Brian Borchers | Let me clarify what I meant here- "being able to get an arbitrary element M(i,j) at little cost" isn't very useful. If you don't know where the nonzero elements are in the matrix, then you have to check every single one to find the nonzeros. If you do happen to know where the nonzero elements are, and you can compute them quickly, then you could use this as a way to do matrix vector multiplications in an iterative method. | |
| Dec 7, 2012 at 15:54 | comment | added | Ruslan | As already said above, being able to get matrix elements is enough to compute matrix vector products. The whole matrix consists mostly of non-zero elements, so it's not useful to search for any zero ones. As for structure, the matrix is an effective Hamiltonian, so what I can say for sure is only that it's Hermitian. As for origin of the system of equations, I'm trying to use inverse iteration with shift to find specific eigenvectors of this matrix, so Y is current approximation of eigenvector, and X is next step approximation. | |
| Dec 7, 2012 at 14:48 | comment | added | Emil Jeřábek | Being able to get elements of the matrix is useful to compute matrix vector products, for instance. Am I missing something? | |
| Dec 7, 2012 at 14:21 | history | answered | Brian Borchers | CC BY-SA 3.0 |