Talk:Transformation of Unit Matrix into Inverse

How sure are we that:
 * $\mathbf{A}=\mathbf{\breve{E}_1 \breve{E}_2 \cdots \breve{E}_{t-1} \breve{E}_t}$

implies
 * $\breve{\mathbf{A} }= \mathbf{\breve{\breve{E} }_1 \breve{\breve{E} }_2 \cdots \breve{\breve{E} }_{t-1} \breve{\breve{E} }_t}$

?

That is can we say that $\left({\mathbf A \mathbf B}\right)^{-1} = \mathbf A^{-1} \mathbf B^{-1}$? Isn't it $\left({\mathbf A \mathbf B}\right)^{-1} = \mathbf B^{-1} \mathbf A^{-1}$?

In which case your line above would be:


 * $\breve{\mathbf{A} }= \mathbf{\breve{\breve{E} }_t \breve{\breve{E} }_{t-1} \cdots \breve{\breve{E} }_2 \breve{\breve{E} }_1}$

am I right? - pm


 * Not only are you right, but the proof wouldn't even work otherwise. Thanks. --GFauxPas 06:14, 26 February 2012 (EST)


 * Re: "transformation" being ambiguous, what should be the title of this page? --GFauxPas 06:22, 26 February 2012 (EST)


 * "Construction of Inverse Matrix by Elementary Row Operations" would work for me. I've never been happy with viewing binary operations as "transforming one thing by applying another" - I find it can obscure what's really going on, where you are "combining" two things to get a third. (There is some small discussion on this in the Binary Operation page.) In this context I'm not sure it's helpful, but if you like it then no worries.--prime mover 08:04, 26 February 2012 (EST)
 * I have no particular attachment to the word transform, I'm just first getting accustomed to matrix jargon and am trying to sound fancy --GFauxPas 08:09, 26 February 2012 (EST)

I think it would be sensible to put this in the Algorithm namespace (it does exist, not?) because it is effectively an algorithm for computing the inverse. Maybe with transclusions of subpages... One could then call it 'Matrix Inversion by Row Operations' or whatever. --Lord_Farin 06:26, 26 February 2012 (EST)


 * I was thinking that the algorithm would come later, and that I should link to this page to prove the algorithm's effectiveness. The algorithm does indeed exist and that's where I was going, but I was afraid the page would be too long. See my sandbox. --GFauxPas 06:28, 26 February 2012 (EST)


 * Not sure if this proof has any use beyond proving the algorithm effective... if you agree, then it could harmlessly be moved to a subpage of the page of the algorithm. This suggestion partially comes from the fact that I can't really think how to name it.


 * If you insist, you could put it in a broader context to 'Row Operations Commute with Matrix Right-Multiplication' (which actually seems rather trivial and hands a simpler proof (multiplying by $A^{-1}$ on both sides...)). --Lord_Farin 06:57, 26 February 2012 (EST)


 * I wasn't planning on using this for anything other than showing the effectiveness of the algorithm, but, er, I've yet to figure out how to transclufigrate pages, so I don't know how to make subpages. Multiplying both sides by $\mathbf{A^{-1}}$ wouldn't work for the algorithm, though,... right? --GFauxPas 07:38, 26 February 2012 (EST)


 * As you assume $A$ to be invertible, it would actually work to show effectiveness. Then, the algorithm can be applied to actually find the explicit form of $A^{-1}$. Or that's how I see it. --Lord_Farin 07:52, 26 February 2012 (EST)


 * Well, part of the algorithm will be that "if the reduced row echelon form isn't $\mathbf{I}$, then the matrix isn't invertible. Why? Because since the reduced row echelon form is unique, . Actually, it may be easy to turn this proof into an iff statement, as we have Elementary Matrix is Invertible. --GFauxPas 08:07, 26 February 2012 (EST)


 * I see this as: there needs to be a page on how to convert a matrix to its inverse, which would contain an algorithm (or several algorithms, transcluded) and, as appropriate, this page should be a proof of the fact that it is an algorithm.


 * I'm not sure we do have an algorithm namespace - I'm not sure we need one. I'd lump them in with proofs, myself. --prime mover 08:04, 26 February 2012 (EST)

I have adjusted the proof to show that $\mathbf{A} \sim \mathbf {I} \implies \exists \mathbf{A}^{-1}$, so it's more interesting now. --GFauxPas 08:41, 26 February 2012 (EST)