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This question already has an answer here:

if $a,b$ are elements of a unital algebra $A,$ then $1-ab$ is invertible if and only if $1-ba$ is invertible.

because if $1-ab\ $ has inverse $x$ , then $1-ba\ $ has inverse $1+bxa$. but how ??

$$(1-ab)x=x(1-ab)=1$$

then $$(1-ba)(1+bxa)=1+bxa-ba-babxa$$

but how the expression on right hand side equal to $1$.

any hint ??

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marked as duplicate by José Carlos Santos, Saad, Arnaud Mortier, Ethan Bolker, Xander Henderson Mar 31 '18 at 0:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ I'm almost sure this theme is already present on math.SE (LMGTFY? Haven't got the time.). $\endgroup$ – Hanno Mar 29 '18 at 10:08
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From $(1-ab)x=1$ we get $abx=x-1$. This and $(1-ba)(1+bxa)=1+bxa-ba-b(abx)a$ give

$(1-ba)(1+bxa)=1$.

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