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Let $a,b,c,x$ be elements of a unital non-commutative ring. Assume $c$ is an inverse of $1-ab$: $$ c(1-ab) = 1$$

How can I find an inverse for $1-ba$? What I tried:

Denote the unknown by $x$. Then $x (1-ba) = 1 = x - xba$. I tried to replace $1$ with $c(1-ab)$ but it didn't help because I cannot solve for $x$. I also tried to subtract $x (1-ba) $ from $ c(1-ab)$ but couldn't solve for $x$ either. Any suggestions?

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3 Answers 3

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Hint $ $ The inverse can be slickly derived via geometric power series as below (straightforward algebra then proves that it is indeed an inverse).

$$\begin{eqnarray} \rm (1-ba)^{-1} &=&\rm 1+ \color{#0a0}b\color{#c00}a + \color{#0a0}b\color{c00}{ab}\color{#c00}a + \color{#0a0}b\color{0a0}{abab}\color{#c00}a +\,\cdots\\ &=&\rm 1+ \color{#0a0}b\:\! (1\:\! +\, \color{c00}{ab}\ \ +\ \ \color{0a0}{abab}\,\ +\,\cdots\,)\:\!\color{#c00}a\\ &=&\rm 1+ \color{#0a0}b\:\! (1\,-\,ab)^{-1}\color{#c00}a\end{eqnarray}\qquad\qquad$$

Halmos asks why this works in a famous expository article (excerpted below). When I was a student I became interested in this. It turns out that one can give good (and rigorous) explanations. It can be proved that all such rational identities are essentially consequences of geometric power series expansions. For references see this Mathoverflow question. See also Paul Cohn, A remark on the quasi-inverse of a product, Illinois J. Math, 2003. Cohn wrote this paper in reply to my question regarding his viewpoint on this topic.

Geometric series. In a not necessarily commutative ring with unit (e.g., in the set of all $3 \times 3$ square matrices with real entries), if $\,1 - ab\,$ is invertible, then $\,1 - ba\,$ is invertible. However plausible this may seem, few people can see their way to a proof immediately; the most revealing approach belongs to a different and distant subject.

Every student knows that $\,1 - x^2 = (1 + x) (1 - x),\,$ and some even know that $\,1 - x^3 =(1+x +x^2) (1 - x).\,$ The generalization $\,1 - x^{n+1} = (1 + x + \cdots + x^n) (1 - x)\,$ is not far away. Divide by $\,1 - x\,$ and let $\,n\,$ tend to infinity; if $\,|x| < 1,\,$ then $\,x^{n+1}$ tends to $\,0,\,$ and the conclusion is that $\frac{1}{1 - x} = 1 + x + x^2 + \cdots.\,$ This simple classical argument begins with easy algebra, but the meat of the matter is analysis: numbers, absolute values, inequalities, and convergence are needed not only for the proof but even for the final equation to make sense.

In the general ring theory question there are no numbers, no absolute values, no inequalities, and no limits - those concepts are totally inappropriate and cannot be brought to bear. Nevertheless an impressive-sounding classical phrase, "the principle of permanence of functional form", comes to the rescue and yields an analytically inspired proof in pure algebra. The idea is to pretend that $\,\frac{1}{1 - ba}\,$ can be expanded in a geometric series (which is utter nonsense), so that $\,(1 - ba)^{-1} = 1 + ba + baba + bababa + \cdots\,$ It follows (it doesn't really, but it's fun to keep pretending) that $\,(1 - ba)^{-1} = 1 + b (1 + ab + abab + ababab + \cdots) a.\,$ and, after one more application of the geometric series pretense, this yields $\,(1 -ba)^{-1} = 1 + b (1 - ab)^{-1} a.\,$

Now stop the pretense and verify that, despite its unlawful derivation, the formula works. If, that is, $\, c = (1 - ab)^{-1},\,$ so that $\,(1 - ab)c = c(1 - ab) = 1,\,$ then $\,1 + bca\,$ is the inverse of $\,1 - ba.\,$ Once the statement is put this way, its proof becomes a matter of (perfectly legal) mechanical computation.

Why does it all this work? What goes on here? Why does it seem that the formula for the sum of an infinite geometric series is true even for an abstract ring in which convergence is meaningless? What general truth does the formula embody? I don't know the answer, but I note that the formula is applicable in other situations where it ought not to be, and I wonder whether it deserves to be called one of the (computational) elements of mathematics. -- P. R. Halmos [1]

[1] Halmos, P.R. $ $ Does mathematics have elements?
Math. Intelligencer 3 (1980/81), no. 4, 147-153

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Assuming completeness w.r.t. $(a)$ or $(b)$ or just as a heuristic:

$x = (1 - ba)^{-1} = 1 + ba + baba + \cdots = 1 + b(1 + ab + abab + \cdots)a = 1 + bca$

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  • $\begingroup$ What is completeness with respect to $(a)$? This is great I wonder why you call it "heuristic" it seems to be a solid proof. $\endgroup$
    – newb
    Commented Feb 13, 2014 at 13:11
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    $\begingroup$ @newb The steps in between are not justified: the expression $1 + ba + baba + \dots$ does not necessarily converge, for example. But you can pretend that it does in order to help you find the solution (heuristic), and in this case it works. And when you've found $1+bca$, you can immediately check that it's the right answer, regardless of whether the expression converged. $\endgroup$
    – Najib Idrissi
    Commented Feb 13, 2014 at 13:14
  • $\begingroup$ For example in $\mathbb{Z}$ say $a = 1, b = 2$, then $c = -1$. In the steps to find $x$, you would write something like $$x = (1 - 2)^{-1} = 1 + 2 + 4 + 8 + 16 + \dots$$ which doesn't converge. $\endgroup$
    – Najib Idrissi
    Commented Feb 13, 2014 at 13:17
  • $\begingroup$ Thanks. And what does completeness with respect to $(a)$ mean? $\endgroup$
    – newb
    Commented Feb 13, 2014 at 13:24
  • $\begingroup$ @newb: the keyword is $I$-adic topology for an ideal $I$, in this case $(a)$ or $(b)$. $I$-adic completeness essentially means that if the $n$-th partial sum is in $I^n$ then there is a single element which modulo $I^n$ is equal to the partial sums. $\endgroup$
    – doetoe
    Commented Feb 13, 2014 at 13:27
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Let's think in term of symbols, since we want $x$ to have a "simple" form.

Suppose that $x(1-ba)=1$. If $x$ could be writen only with symbols $a,b,c$, the result $x(1-ba)$ would also be written in symbols $a,b,c$, so it is natural to search for $x$ of the form $x=1+y$. Also, $y$ must be chosen so that it "cancels" the symbols $ba$.

Notice that the formula $c(1-ab)$ can be rewritten as $c-cab=1$, that is, $cab=c-1$. With that equality, we can get a term with $b$ and rewrite it as a difference of terms without $b$.

Notice that $(1+y)(1-ba)=(1-ba)+y+yba$. In order to rewrite $yba$ without the $b$, we can search for $y$ of the form $y=zca$ for some $z$, so $yba=z(cba)a=z(c-1)a$. Finally, we want that the following holds: $$(1+zca)(1-ba)=1$$ that is, $1=1-ba+zca-z(cab)a=1-ba+zca-z(c-1)a=1-ba+zca-zca+za=1-ba+za$, which is valid for $z=b$.

Therefore, the natural candidate for the inverse of $1-ba$ is $x=1+bca$, and you can easily show that this is, in fact its inverse.

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  • $\begingroup$ Thanks. Why $x=1+y$ but not $x=1-y$? $\endgroup$
    – newb
    Commented Feb 13, 2014 at 13:12
  • $\begingroup$ More importantly, could you elaborate please on why $c-cab=1$ implies that $y=zca$? $\endgroup$
    – newb
    Commented Feb 13, 2014 at 13:25
  • $\begingroup$ @newb If we wanted $x=1-y$, the sigs would change, but the reasoning would be the same. I edited it for your second question. Is it better now? $\endgroup$
    – Luiz Cordeiro
    Commented Feb 13, 2014 at 13:43
  • $\begingroup$ I still don't really understand most of the steps. I already have trouble believing step one where you say that $x$ can't be a product of $a,b,c$. Is it really true? $\endgroup$
    – newb
    Commented Feb 14, 2014 at 7:56
  • $\begingroup$ In my solution, first I tried guessing what would be the inverse of $1-ba$ just by analizing the possibilities. To simplify things, I assumed $x$ to be a simple formula involving $1,a,b,c$. More exactly, I looked for $x$ of the form $x=\sum_n x^n_1\cdots x^n_{i_n}$, where $x^n_{j}\in\left\{1,a,b,c\right\}$. Then I argued that some term $x^n_1\cdots x^n_{i_n}$ should be $=1$. $\endgroup$
    – Luiz Cordeiro
    Commented Feb 15, 2014 at 0:19

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