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Let H be a Hilbert space and $A$ and $B$ be bounded operators in $H$. How can I prove that $AB - BA = I$ is not possible ? Probably this is as easy as in the matrices case, but I couldn't prove it. Can you help me please ?

Thank you :)

thetruth
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    More generally, if you're interested, Kleinecke-Shirokov: if $C=AB-BA$ and if $C$ commutes with $A$, then $C$ is quasinilpotent. That's problem 232 in Halmos' Hilbert Space Problem Book. – Julien May 10 '13 at 20:40

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A more general question has been asked and answered on here before. The bounded operators on a Hilbert space form a Banach Algebra. A beautiful argument of H. Wielandt shows that the equation $AB-BA = I$ is not possible in any normed algebra. I won't repeat it here, but it follows by induction that if $AB -BA = I,$ we have $AB^{n}-B^{n}A = nB^{n-1}$ for all positive integers $n.$ This implies that $2\|A\| \|B \| \geq n$ for every $n,$ a contradiction.

Geoff Robinson
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