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There is a problem given in a representation theory textbook:

Prove that for any finite-dimensional complex vector space $V$ there are no $X, Y \in \operatorname{End}V$ such that $[X, Y] = \mathrm{id}$.

I tried looking at $\mathbb{C}[X, Y]$ and the ideal of $\operatorname{End}V$ generated by $XY - YX$, but so far to no avail. I could use a hint.

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    **Hint.** Consider the trace.2012-05-19
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    Oh...${}{}{}{}{}$ So if $\operatorname{char} k = p$, then $k^{np}$ may have such $X$ and $Y$, right?2012-05-19
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    Well, if the characteristic is $p$, then the trace won't settle it. I don't know if it is possible or not. I know it *can* happen in infinite dimensions.2012-05-19
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    There are two important properties of the trace that can be used to prove this: The trace is linear, and two operators commute in the argument of the trace.2012-05-19
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    This can be done in characteristic $p$ with suitable $p \times p$ matrices. See the answers at http://math.stackexchange.com/questions/99175/solutions-to-the-matrix-equation-mathbfab-ba-i-over-general-fields2012-05-19
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    @KCd, awesome! I went through the case $(\mathbb{Z}/2)^2$ by hand, and it's great to know how deep this problem really is!2012-05-19

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