Find $(1-ba)^{-1}$ when $c=(1-ab)^{-1} $ in ring $R$.

Question

For $R$ is a ring has identity element. $a,b\in R$ and $c=(1-ab)^{-1}$ . Find $(1-ba)^{-1}$.

In my question : $R$ is a ring not real field . I haven't ideal to find the inverse element of $(1-ba)^{-1}$ From supposition. — Hung nguyen, Apr 11 '13 at 14:21

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

Hint $\ $ The answer is easily discoverable by examining geometric formal power series expansions, as below. It is simple algebra to prove that the derived formula is correct. $$\rm\begin{eqnarray} (1\!−\!ba)^{−1} &=&\rm 1+ba+b\color{#C00}{ab}a+b\color{#0A0}{abab}a+\ \cdots \\ &=&\rm 1+b(1\!+\,\color{#C00}{ab}\ \ \,+\ \ \color{#0A0}{abab}\ \ +\ \cdots)\,a\\ &=&\rm \ \ldots\end{eqnarray}$$

Remark $\ $ This is an old chestnut which Halmos made famous by asking for an explanation why the formal manipulations work. Some interesting explanations are known. See here for more.

score 2 · Answer 2 · edited Dec 23 '22 at 22:45

2

$x$ is the inverse of $1 - ab$. $(1 - ba)bxa = b(1-ab)xa = ba$. Therefore, $(1 - ba)(1 + bxa) = 1 - ba + ba = 1$. It is easy to verify that $(1 + bxa)(1 - ba) = 1$. Hence, $1 - ba$ is invertible.

edited Dec 23 '22 at 22:45

Jonas Hardt

535

answered Jan 31 '17 at 07:18

tony

769

Find $(1-ba)^{-1}$ when $c=(1-ab)^{-1} $ in ring $R$.

2 Answers2