$1-ab$ 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**

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 $y.\ \square$

**Attribution***Source : Link , Question Author : Bill Dubuque , Answer Author : Victor Protsak*