On the Invertibility of $1-ab$ in a Unital Ring

Theorem. Let $R$ be a ring with identity, and let $a,b \in R$. Then
$$
1-ab \text{ is invertible } \iff 1-ba \text{ is invertible}.
$$

Idea. If $x^n = 0$, then
$$
1 - x^n = 1,
$$
and hence
$$
(1-x)(1+x+x^2+\cdots+x^{n-1}) = 1.
$$
Thus the inverse of $1-x$ is
$$
1+x+x^2+\cdots+x^{n-1}.
$$

Motivated by this formal identity, one may heuristically think of the inverse of $1-ab$ as
$$
c = 1 + ab + (ab)^2 + \cdots,
$$
and then imagine the inverse of $1-ba$ as
$$
1 + ba + (ba)^2 + \cdots = 1 + bca,
$$
which suggests the correct formula and can then be verified directly.

Proof: Suppose that
$$
(1-ab)^{-1} = c.
$$
Then
$$
(1-ab)c = c(1-ab) = 1.
$$

Now observe that
$$
(1-ba)bca = bca - babca = b(1-ab)ca = b \cdot 1 \cdot a = ba.
$$

Therefore,
$$
(1-ba)(1+bca) = 1 - ba + (1-ba)bca = 1 - ba + ba = 1.
$$

Similarly, one checks that
$$
(1+bca)(1-ba) = 1.
$$

Hence $1-ba$ is invertible, and
$$
(1-ba)^{-1} = 1 + b(1-ab)^{-1}a.
$$

The cover image of this article was taken in Irkutsk Oblast, Russia.