For a finite-dimensional vector space $V$ will the space of nilpotent linear operators on $V$ be closed under the commutator bracket?
2026-04-11 20:13:17.1775938397
Do the nilpotent linear operators form a Lie algebra
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A counterexample exists in dimension $2$:
Let $X=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $Y=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)$, then $XY-YX=\left(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right)$.