# Prove that no two complex matrices have unit commutator

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. –  Arturo Magidin May 19 '12 at 5:27
Oh...${}{}{}{}{}$ So if $\operatorname{char} k = p$, then $k^{np}$ may have such $X$ and $Y$, right? –  Alexei Averchenko May 19 '12 at 5:30
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. –  Arturo Magidin May 19 '12 at 5:45
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. –  JLA May 19 '12 at 6:12
This can be done in characteristic $p$ with suitable $p \times p$ matrices. See the answers at math.stackexchange.com/questions/99175/… –  KCd May 19 '12 at 6:36