Talk:Inverse of Product

For an elementary Linear Algebra student like me, it's not at all obvious this page's results apply to matrices as well. I'd like to either add a separate page for matrices and link to this page as a proof, or add it as a corollary here, or just comment that it applies to matrices somewhere on this page. Thoughts? --GFauxPas 08:25, 23 February 2012 (EST)
 * I think the fundamentally unclear (not intuitively correct) thing here is the associativity of matrices (which is indeed a bit fussy, if you really insist on picking a basis for your linear operator (that is, if you just calculate rather than imagine it as simply a linear function; that's probably your elementary-studentness)). So, if you develop the perspective of seeing matrix multiplication as composition of linear mappings, it will dawn. I would take the separate page route. --Lord_Farin 09:27, 23 February 2012 (EST)
 * Oh, okay. We're actually right now learning about different ways to look at matrix multiplication, but I don't have the full intuition yet. Oh, and also, I didn't know what a semigroup is. And of course I can click on all the links, but now you're expecting the reader learn the meanings of monoid, semigroup, magma, closure, and algebraic structure. Sure, we have links for all those, but the only concept of those that we covered in class was closure. --GFauxPas 09:46, 23 February 2012 (EST)
 * Note that from Ring of Square Matrices over Ring is Ring it follows that this result also applies to the inverse of a matrix product - but it assumes that abstract algebra has already been covered.
 * However, that butters no proverbial parsnips with an elementary Linear Algebra student for whom abstract algebra is three semesters away, so an independent proof of this result purely in the context of matrix manipulations would be an excellent idea.
 * As LF says, on a separate page - on which the instant three-line Abstract Algebraic approach can be appended as a Proof 2 which will link to this one. Actually, I was somewhat surprised that I hadn't already put that proof up. --prime mover 17:29, 23 February 2012 (EST)

This page merits an analogous page for composition of mappings, not burdened with the fact that there is the category of sets behaving like an algebraic structure. --Lord_Farin 18:09, 31 March 2012 (EDT)


 * Not sure quite what you mean ... do you suggest an Also see to Inverse of Composite Bijection? --prime mover 18:15, 31 March 2012 (EDT)


 * Nvm, the Google had me by not interpreting 'composition' also as 'composite' for a search term; therefore, I couldn't find the pages, while I was quite sure they would exist. --Lord_Farin 18:18, 31 March 2012 (EDT)
 * Just noted that Inverse of Composite Bijection does also apply to general pre-images, but apparently isn't phrased in that context. Also, maybe the disambiguation Inverse of Composite may be merited. --Lord_Farin 18:19, 31 March 2012 (EDT)


 * Possibly. Maybe this might also be covered by Inverse of Composite Relation. The reason we haven't got Preimage of Subset under Composite Mapping is because it's not generally a mapping, and in the context into which I was working, there was no immediate need for the concept. Inverse of Composite is on the plan.


 * Insertion: Preimage of Subset under Composite Mapping is now indeed in place -- used to be "Inverse of Composite Mapping", now deleted.


 * As for the word "composition", again, the question needs to be "composition of what" and by the time you've written "inverse of composition of mappings" you realise you've used two more syllables, one more word and n more letters than you needed to ... which is why (despite the fact that the initial definition was "Composition of Mappings") the language evolved in the direction it did. Perhaps there's a case for a rename to / redirect from "Definition:Composite Mapping" etc. and similar redirects from /renames to "Inverse of Composition of (whatever)" to "Inverse of Composite (whatever)" to enhance findability.
 * Good call. --prime mover 01:08, 1 April 2012 (EDT)


 * Posted Preimage of Subset under Composite Mapping, that should close this issue. --Lord_Farin 07:43, 5 April 2012 (EDT)