$1-ab$ invertible $\implies$ $1-ba$ invertible has a slick power series "proof" as below, where Halmos asks for an explanation of why this tantalizing derivation succeeds. Do you know one?

*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

http://dx.doi.org/10.1007/BF03022973

Topics in invariant theory, (M.-P. Malliavin, ed.), Lecture Notes in Math., vol. 1478, Springer-Verlag, 1991, pp.215-243. A short discussion also appears in Section 8 of C. Reutenauer, A survey of noncommutative rational series, inFormal Power Series and Algebraic Combinatorics (New Brunswick, NJ, 1994), DIMACS Series Discrete Math. Theoret. Comput. Sci.24, American Mathematical Society, 1996, pp. 159-169. $\endgroup$ – Richard Stanley Jul 12 '10 at 19:18