How would you solve this tantalizing Halmos problem?

Question

$1-ab$ 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

Answer 1

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

$$(\lambda-ba)^{-1}=\lambda^{-1}+\lambda^{-1}b(\lambda-ab)^{-1}a.\qquad\qquad\qquad(*)$$

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).

---

Addendum

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 $y.\ \square$

Answer 2

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.