Magic

by math_explorer, Jan 25, 2013, 1:17 PM

If $1 - ab$ is invertible in a ring then $1 - ba$ is invertible with inverse $b(1-ab)^{-1}a + 1$ (okay, this can be motivated but then it wouldn't be magic anymore)

abstract algebra

one-liner

algebra

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hmm its motivated by matrices right
EDIT : also noncommutative rings are hard darn

This post has been edited 2 times. Last edited by dinoboy, Jan 25, 2013, 3:45 PM

by dinoboy, Jan 25, 2013, 3:37 PM

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Prof's motivation:

$\frac{1}{1-ab} \approx 1 + ab + abab + \cdots = x$

$\frac{1}{1-ba} \approx 1 + ba + baba + \cdots = bxa + 1$

I think you can make it rigorous with a homomorphism from the ring of formal power series or something.

Whatever, it's still magic.

by math_explorer, Jan 26, 2013, 12:43 AM

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darn that's pretty tricky
the motivation from matrices is that $A^{-1}(I - AB)A = I - BA$ when $A$ is invertible and then go bash
I guess it doesn't actually use matrices, but similarities in matrices helps motivate that. Conjugates in a group do as well I guess.

This post has been edited 2 times. Last edited by dinoboy, Jan 26, 2013, 6:22 AM

by dinoboy, Jan 26, 2013, 6:16 AM

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Quote:

I think you can make it rigorous with a homomorphism from the ring of formal power series or something.

Hmm, I'm curious how exactly does this work?

by math154, Dec 2, 2013, 6:41 AM

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"I think you can make it rigorous" means "I don't actually know how to make it rigorous but it kind of looks like it".

Looking back (especially adding the fact that it's unobvious enough for you to ask me this), I think I was wrong...

by math_explorer, Dec 2, 2013, 9:42 AM

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I just realized I had actually bookmarked a MathOverflow thread about this. I don't understand everything that's going on there, but maybe you will.
http://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem

by math_explorer, Jun 13, 2014, 1:49 AM