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I encountered the following question as a question about linear operators. If $AB$ and $BA$ are invertible for some $A, B \in \frak{L}$$(E)$ where $E$ is a Banach space, then show that $A$ and $B$ are also invertible.

I worked this out by showing that $A$ is bijective and its inverse is linear and bounded. But that required making explicit the action of $A$ and $B$ as operators on an underlying space. So my question is whether there is an 'algebraic' way of showing the statement is true, i.e. whether if $ab$ and $ba$ are invertible elements in a Banach algebra imply $a$ and $b$ invertible.

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Let $C$ be the inverse of $AB$ and $D$ be the inverse of $BA$.

Then: $I=C(AB)=(CA)B$ and $I=(BA)D=B(AD)$. With $X=CA$ and $Y=AD$ we therefore have

$XB=I=BY$.

This shows that $B$ is invertible. A similar proof shows that $A$ is invertible.

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  • $\begingroup$ Geez, it's really simple, now I feel really dumb $\endgroup$ – E.Lim Dec 2 '16 at 10:43

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