etc.?

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Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Oct 2, 2015 at 19:42	comment	added	Dmitry Kerner		You might be interested in the normal/canonical forms of matrices. In particular: "normal forms that depend continuously on the matrix entries". (unlike the Jordan form, which is not continuous under small deformations.) e.g. V.I. Arnol'd, "On matrices depending on parameters"(1971) and further citations
Sep 8, 2010 at 11:59	comment	added	darij grinberg		Am I wrong in saying that the original question can be reworded as "what linear algebra still holds over arbitrary commutative rings instead of fields"? Then, the best place to start is Chapter 1 of Atiyah/Macdonald. But if you are looking for a big collection of linear algebra over rings, and you can read some French, this is for you: hlombardi.free.fr/publis/A---PTFCours.pdf
Sep 8, 2010 at 4:02	answer	added	mathreader		timeline score: 2
Sep 7, 2010 at 2:56	vote	accept	Theo Johnson-Freyd
Sep 7, 2010 at 2:55	comment	added	Theo Johnson-Freyd		@Victor Protsak: no, I have no applications in mind. My research is elsewhere, and the question is idle curiosity, asking after trying to make sense of the linked closed question. But your second comment essentially answers my motivating example, by leaving only the necessary discontinuity and taking out the spurious part that arises from using Gaussian elimination.
Sep 7, 2010 at 2:52	comment	added	Theo Johnson-Freyd		@Pete: yes, it's probably better not to put "algebraic" in the title. I meant it in the sense of "algebraic function", but the question is more --- VP says "convoluted" --- than that. @alex: yes, although if there were a notion in which the solution could be understood as continuous, that'd be awesome. I recognize that much of problem solving does not give out answers that are completely polynomial/rational/etc. in the parameters, but sometimes an algorithm that doesn't obviously give out such answers in fact does (e.g. inverting a matrix by Gaussian elimination).
Sep 7, 2010 at 2:48	history	edited	Theo Johnson-Freyd	CC BY-SA 2.5	changed the title
Sep 6, 2010 at 17:15	comment	added	Pete L. Clark		May I suggest that you rephrase the title of the question a bit? The suggestion that there is some part of linear algebra which is "not algebraic" seems unnecessarily perplexing.
Sep 6, 2010 at 9:14	comment	added	J. M. isn't a mathematician		Well, for one thing, the provably stable algorithms of numerical linear algebra all rely on the judicious use of orthogonal (unitary in the complex case) matrices. Yes, Gaussian elimination is a fluke.
Sep 6, 2010 at 8:39	comment	added	Victor Protsak		Concerning you motivation: the key step is to complement a subspace, and for a real vector space it's not necessary to use row reduction, you can compute the orthogonal complement instead. Indeterminacy arises because the dimension of the subspace, i.e. the rank of A, is only semialgebraic.
Sep 6, 2010 at 8:25	comment	added	Victor Protsak		The usual requirements to an algorithm are (a) speed; (b) numerical stability. Are you interested in those + (c) algebraic (rational, etc) dependence on the parameters, or is it a purely theoretical question about which problems in LA admit an answer expressible by an algebraic formula?
Sep 6, 2010 at 8:25	answer	added	Charles Matthews		timeline score: 12
Sep 6, 2010 at 8:20	comment	added	alex		I don't follow the statement that there are algorithms for computing inverses which run "algebraically" as opposed to "discontinuosly." Consider the problem of solving the equation $ax=1$. The solution is: if $a \neq 0$, output $x=1/a$, else output "no solution." This depends "discontinuously" on $a$, so I gather you would dislike it, but it is the correct answer, and any algorithm for solving systems of equations or computing inverses has to check if $a \neq 0$ somehow.
Sep 6, 2010 at 7:44	history	asked	Theo Johnson-Freyd	CC BY-SA 2.5