How to show that there do not exist bounded operators $A$ and $B$ such that $AB-BA=I$

Asked Dec 04 '13 at 03:56

Active Dec 04 '13 at 03:56

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Suppose $X$ is a Banach space, $A$ and $B$ are bounded operators on $X$. Then how to show that it's impossible to have $$AB-BA=I$$ If $A$ and $B$ are matrix, then we just take trace of both side to obtain contradiction,since $trAB=trBA$. Furthermore, this can only hold for bounded operators since if allow $A$ and $B$ to be unbounded, we have $[\frac{d}{dx},x]=I$

Thanks in advance.

asked Dec 04 '13 at 03:56

Tomas

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There is a proof at http://math.stackexchange.com/questions/54397/the-identity-cannot-be-a-commutator-in-a-banach-algebra – Unnecessary Apostrophe's Dec 04 '13 at 04:03
thanks very much, I didn't notice that before. – Tomas Dec 04 '13 at 04:07

How to show that there do not exist bounded operators $A$ and $B$ such that $AB-BA=I$

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