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 $1/(1-ba)$ can be expanded in a geometric series (which is utter nonsense), so that \[(1-ba)^{-1} = 1 + ba + baba + bababa + ⋯\] It follows (it doesn't really, but it's fun to keep pretending) that \[(1-ba)^{-1} = 1 + b(1 + ab + abab + ababab +⋯)a\] 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.

How would you solve this tantalizing Halmos problem?

The best way that I know of interpreting this identity is by generalizing it: $$(λ-ba)^{-1}=λ^{-1}+λ^{-1}b(λ-ab)^{-1}a.\tag{1}$$ Note that this is both more general than the original formulation (set $λ=1$) and equivalent to it (rescale). Now the geometric series argument makes perfect sense in the ring $R((λ^{-1}))$ of formal Laurent power series, where $R$ is the original ring or even the "universal ring" $ℤ\langle a,b\rangle:$ $$ (λ-ba)^{-1}=λ^{-1}+\sum_{n\geq 1}λ^{-n-1}(ba)^n=λ^{-1}(1+\sum_{n\geq 0}λ^{-n-1}b(ab)^n a)=λ^{-1}(1+b(λ-ab)^{-1}a).\ \square$$ A variant of (*) holds for rectangular matrices of transpose sizes over any unital ring: if $A$ is a $k×n$ matrix and $B$ is a $n×k$ matrix then $$(λ I_n-BA)^{-1}=λ^{-1}(I_n+B(λ I_k-AB)^{-1}A).\tag{**}$$ To see that, let $a = \begin{bmatrix}0 & 0 \\ A & 0\end{bmatrix}$ and $b= \begin{bmatrix}0 & B \\ 0 & 0\end{bmatrix}$ be $(n+k)×(n+k)$ block matrices and apply (*).

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

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

Binomial inverse theorem