Let $R$ be an associative ring with unit (but not necessarily commutative). Let $x$, $y$ be elements in $R$ such that $1−xy$ is invertible. Show that $1 − yx$ is also invertible element in $R$.

What I've tried to do is let $t$ denote the inverse of 1 − $xy$ and try to construct the inverse of 1 − $yx$ as an expression of $x,y,$ and $t$. I pretended that $x$, $y$ are (k × k) matrices with small entries such that $(1−x)$$^-$$^1$ = 1 + $x$ + $x^2$ + ...

Thanks to Alex in the comments, I see that if $R$ were commutative, then 1−$xy$=1−$yx$ the problem would be trivial. Any advice on how to proceed would be greatly appreciated, thanks in advance!

  • $\begingroup$ The non-commutativity of $R$ is not supposed to be used. It is mentioned only to draw attention on the fact that you may not assume commutativity (which often simplifies proofs). In particular, if $R$ were commutative, then $1-xy = 1 - yx$ and the problem would be trivial. $\endgroup$
    – Alex M.
    Mar 21, 2017 at 19:43
  • $\begingroup$ Ah that's a good point, I was thinking it was just a trivial proof. I now see the error in my ways. $\endgroup$ Mar 21, 2017 at 19:49

1 Answer 1


The question has many nice answers given at Mathoverflow; here is another one: if $\rm\:(1\!-\!ab)^{-1}\:$ exists then $\rm\:(1\!-\!ba)^{-1}\:$ exists, and has the value

$$\rm\begin{eqnarray} (1\!−\!ba)^{−1} &=&\rm 1+ba+b\color{#C00}{ab}a+b\color{#0A0}{abab}a+\ \cdots \\ &=&\rm 1+b(1\!+\,\color{#C00}{ab}\ \ \,+\ \ \color{#0A0}{abab}\ \ +\ \cdots)\,a\\ &=&\rm 1+b(1\!−\!ab)^{-1} a\end{eqnarray}$$

Richard Stanley gave the identity $$(1+a)(1-ba)^{-1}(1+a)=(1+b)(1-ab)^{-1}(1+a),$$ which also can be proved easily.

  • $\begingroup$ Thanks so much, Dietrich! Very helpful $\endgroup$ Mar 21, 2017 at 20:05
  • $\begingroup$ I don't understand it: what meaning do you give to the infinite sum $1 + x + x^2 + x^3 + \dots$, absent a notion of convergence (and not working in a ring of formal series)? $\endgroup$
    – Alex M.
    Mar 21, 2017 at 20:06
  • 1
    $\begingroup$ @AlexM. This is exactly what the MO question in the first part explains. You should read it, it is fun. $\endgroup$ Mar 21, 2017 at 20:07
  • $\begingroup$ @AlexM: One can still use this argument even if you don't give meaning to the infinite sum -- as soon as you have a reason to think that some particular ring element $x$ is an inverse of $(1-ba)$, one can make a rigorous proof simply by checking whether $x(1-ba)=1=(1-ba)x$. $\endgroup$
    – user14972
    Mar 21, 2017 at 20:14

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