$1-ab$ 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

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    $\begingroup$ There is a theorem to the effect that if you can prove a non-commutative identity by power series, it is true in an abstract ring. But I can never remember the reference. (My intuition is that this is analogous to proving that a polynomial identity holds over C via an analytical or topological argument to show that it must hold identically.) $\endgroup$ Jul 12, 2010 at 18:28
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    $\begingroup$ I believe that the paper to which Qiaochu refers is D. Krob, in 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, in Formal Power Series and Algebraic Combinatorics (New Brunswick, NJ, 1994), DIMACS Series Discrete Math. Theoret. Comput. Sci. 24, American Mathematical Society, 1996, pp. 159-169. $\endgroup$ Jul 12, 2010 at 19:18
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    $\begingroup$ @Qiaochu: I'm not quite sure what hypotheses are being put on your ring, so apologies if this comment misses the point: but it is well known that there are certain identities which cannot hold in Banach algebras but do hold in arbitrary complex algebras. The standard example I was given is $qp-pq=I$. $\endgroup$
    – Yemon Choi
    Jul 12, 2010 at 20:31
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    $\begingroup$ The theorem of Krob I mention above is roughly the following. Any rational identity that holds in every ring (with identity) is "trivial", i.e., an algebraic consequence of $(1-a)^{-1}=1+a+a^2+\cdots$, and conversely. A nice example is $$ (1+x)(1-yx)^{-1}(1+y) = (1+y)(1-xy)^{-1}(1+x). $$ It is clearly true if we can expand $(1-yx)^{-1}=1+yx+(yx)^2+\cdots$ and similarly for $(1-xy)^{-1}$. Therefore it holds in every ring. (Of course it is assumed that the inverses exist.) It is rather tricky to prove this identity from scratch. $\endgroup$ Jul 12, 2010 at 21:31
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    $\begingroup$ Bill: Neither the title nor the first few lines of the question visible from the "questions" tab give any clue as to its content (in particular, "old chestnut" implied it's an unsolved elementary puzzle). It's sheer luck that I happened to read it at all. May I suggest using more descriptive titles in the future? $\endgroup$ Jul 14, 2010 at 4:43

2 Answers 2


The best way that I know of interpreting this identity is by generalizing it:


Note that this is both more general than the original formulation (set $\lambda=1$) and equivalent to it (rescale). Now the geometric series argument makes perfect sense in the ring $R((\lambda^{-1}))$ of formal Laurent power series, where $R$ is the original ring or even the "universal ring" $\mathbb{Z}\langle a,b\rangle:$

$$ (\lambda-ba)^{-1}=\lambda^{-1}+\sum_{n\geq 1}\lambda^{-n-1}(ba)^n=\lambda^{-1}(1+\sum_{n\geq 0}\lambda^{-n-1}b(ab)^n a)=\lambda^{-1}(1+b(\lambda-ab)^{-1}a).\ \square$$

A variant of $(*)$ holds for rectangular matrices of transpose sizes over any unital ring: if $A$ is a $k\times n$ matrix and $B$ is a $n\times k$ matrix then

$$(\lambda I_n-BA)^{-1}=\lambda^{-1}(I_n+B(\lambda I_k-AB)^{-1}A).\qquad\qquad(**)$$

To see that, let $a = \begin{bmatrix}0 & 0 \\ A & 0\end{bmatrix}$ and $b= \begin{bmatrix}0 & B \\ 0 & 0\end{bmatrix}$ be $(n+k)\times (n+k)$ block matrices and apply $(*).\ \square$

Here are three remarkable corollaries of $(**)$ for matrices over a field:

  • $\det(\lambda I_n-BA) = \lambda^{n-k}\det(\lambda I_k-AB)\qquad\qquad\qquad$ (characteristic polynomials match)
  • $AB$ and $BA$ have the same spectrum away from $0$
  • $\lambda^k q_k(AB)\ |\ q_k(BA)\qquad\qquad\qquad\qquad\qquad\qquad\qquad $ (compatibility of the invariant factors)

I used a noncommutative version of $(**)$ for matrices over universal enveloping algberas of Lie algebras $(\mathfrak{g},\mathfrak{g'})$ forming a reductive dual pair in order to investigate the behavior of primitve ideals under algebraic Howe duality and to compute the quantum elementary divisors of completely prime primitive ideals of $U(\mathfrak{gl}_n)$ (a.k.a. quantizations of the conjugacy classes of matrices).


The identity $(1+x)(1-yx)^{-1}(1+x)=(1+y)(1-xy)^{-1}(1+x)$ mentioned by Richard Stanley in the comments can be easily proven by the same method: after homogenization, it becomes

$$(\lambda+x)(\lambda^2-yx)^{-1}(\lambda+y)= (\lambda+y)(\lambda^2-xy)^{-1}(\lambda+x).$$

The left hand side expands in the ring $\mathbb{Z}\langle x,y\rangle((\lambda^{-1}))$ as

$$1+\sum_{n\geq 1}\lambda^{-2n}(yx)^n+ \sum_{n\geq 0}\lambda^{-2n}(x(yx)^n+y(xy)^n)+ \sum_{n\geq 1}\lambda^{-2n}(xy)^n,$$

which is manifestly symmetric with respect to $x$ and $y.\ \square$

  • $\begingroup$ Is homogenization needed? Certainly it is a reasonable move to make and might lead to other interesting identities when comparing powers of $\lambda$ on both sides. I am asking just about its logical status in understanding where these formulas naturally "live". $\endgroup$
    – T..
    Jul 14, 2010 at 18:10
  • $\begingroup$ Yes, in this approach, introducing a formal variable $\lambda$ is essential. It's been a while since I thought about the general case; in the situations that I was interested in, the expressions were rational in $\lambda$ and their expansions at $\infty$ lived in in $R((\lambda^{-1})).$ $\endgroup$ Jul 15, 2010 at 2:17

Denote by $R$ the ring containing $a,b$ and a solution $X$ of $(1-ab)X = X(1-ab) = 1$.

The subring of polynomial expressions in $a$, $b$ and $x$ (let us keep it flexible initially whether we want $Z$ or $R$ coefficients) is a quotient of the universal example: the polynomial ring on $a,b$ modulo the equations stating that $X$ inverts $1-ab$. If $(1-ba)$ is invertible in the universal ring this fact will descend to the quotient. The universal example embeds into the same universal gadget made using formal power series in place of polynomials, as the set of series whose terms are eventually zero for large enough degree. In the larger universal ring the inverses of $(1-ab)$ and $(1-ba)$ can be identified explicitly as series in $a$ and $b$, the formal calculations are legitimate, and they show that these two inverses satisfy a relation with coefficients in $Z\langle a,b \rangle$. This relation allows one to define an element $Y$ in $Z\langle a,b,X \rangle$ which is inverse to $(1-ba)$.

Possibly there are buried subtleties one needs to overcome in passing between the different rings, subrings and quotients involved. But at a minimum, this argument shows that convergence questions are superfluous and the series calculations rigorizable.

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    $\begingroup$ More precisely, you want to show that, if the identity $(1+bXa)(1-ba)=1$ holds in the completion of $R$ (with respect to $a$ and $b$), then it holds in $R$. So all you need to do is show that $R$ embeds in its completion. $\endgroup$ Jul 12, 2010 at 19:25
  • $\begingroup$ How do I get angle brackets in TeX to show up in MO? As in Z < a , b >. $\endgroup$
    – T..
    Jul 12, 2010 at 19:26
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    $\begingroup$ For the good of mankind, use $\langle a,b \rangle$ and not < a, b>. $\endgroup$
    – Ryan Reich
    Jul 12, 2010 at 20:56
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    $\begingroup$ There's always \langle and \rangle, which have a funny sound to them. $\endgroup$ Jul 13, 2010 at 2:20
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    $\begingroup$ @JoséFigueroa-O'Farrill, the person whose job it is to make sure no one uses > improperly must be the \rangle wrangler. $\endgroup$
    – LSpice
    Feb 13, 2016 at 0:41

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