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Jan 26, 2021 at 3:45 answer added Reid.Atcheson timeline score: 1
Sep 23, 2020 at 18:11 history edited Yaroslav Bulatov
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Sep 23, 2020 at 18:06 answer added Abdullah Ali Sivas timeline score: 4
Sep 23, 2020 at 16:23 history edited Yaroslav Bulatov CC BY-SA 4.0
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Sep 23, 2020 at 6:00 history tweeted twitter.com/StackSciComp/status/1308647298193395712
Sep 22, 2020 at 21:27 comment added Wolfgang Bangerth As a point in terminology, you're really looking at families of matrices with varying sizes. So a family of matrices all of which have have eigenvalues equal to just the elements of the same, small set (in other words, with a relatively small number of eigenvalues but growing multiplicities) can be solved in $O(N)$. That's because for all of the typical iterative methods, the number of iterations necessary is bounded by the number of distinct eigenvalues.
Sep 22, 2020 at 20:49 comment added Yaroslav Bulatov by "fast" I mean that linear solver runtime is linear in the number of rows, while matrix inverse runtime is quadratic
Sep 22, 2020 at 20:39 comment added Yaroslav Bulatov Either one -- since you can find inverse by solving k linear systems, having a fast linear solver will also give a fast inversion routine, and vica versa
Sep 22, 2020 at 20:11 comment added Federico Poloni Are you using "invert" as in "solve a linear system with $A$" or as "compute the entries of the inverse of $A$"?
Sep 22, 2020 at 19:23 comment added Yaroslav Bulatov Came across an interesting overview in Ch.1 rasmuskyng.com/rjkyng-dissertation.pdf, another easy case seems to be "symmetric M-matrix", case when $DMD$ is diagonally dominant for some diagonal $D$
Sep 22, 2020 at 18:50 history edited Yaroslav Bulatov CC BY-SA 4.0
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Sep 22, 2020 at 18:24 history edited Yaroslav Bulatov CC BY-SA 4.0
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Sep 22, 2020 at 18:18 history asked Yaroslav Bulatov CC BY-SA 4.0