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This question already has an answer here:

I'm really embarrassed to ask but what is the nilradical of the Lie algebra $\mathfrak{gl}_n(\mathbb{C})$, i.e. the set of ad-nilpotent elements of $\mathfrak{gl}_n(\mathbb{C}) = \mathrm{Mat}_n(\mathbb{C})$$? This must be standard knowledge but I couldn't find a reference.

Clearly, all nilpotent matrices and all diagonal matrices are in the nilradical. What else?

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marked as duplicate by Dietrich Burde, Jonas Meyer, Claude Leibovici, Chris Godsil, Juniven Apr 20 '17 at 12:56

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migrated from mathoverflow.net Nov 19 '14 at 17:37

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  • $\begingroup$ it is zero. This is really a consequence of definitions, and these questions are better suited to math stackexchange $\endgroup$ – Venkataramana Nov 19 '14 at 17:16
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    $\begingroup$ @Venkataramana Um, actually it's the scalar matrices. $\endgroup$ – Ben Webster Nov 19 '14 at 17:37
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You're using the wrong definition of nilradical: it's the largest nilpotent ideal in the Lie algebra. For $\mathfrak{gl}_n$, the only proper ideals are the trace-free matrices $\mathfrak{sl}_n$ and the scalar matrices $\mathbb{C}\cdot I$. The former is not nilpotent (it's simple), and the latter is. So the scalar matrices are the nilradical.

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The Lie algebra $\mathfrak{g}=\mathfrak{gl}_n(\mathbb{C})$ is reductive, i.e., we have $\mathbb{g}=[\mathfrak{g},\mathfrak{g}]\oplus Z(\mathfrak{g})$, where $[\mathfrak{g},\mathfrak{g}]$ is semisimple (in fact it is $\mathfrak{sl}_n(\mathbb{C})$ in this case), and the center is abelian and equals the solvable radical and the nilradical. So the nilradical here equals $nil(\mathfrak{g})=Z(\mathfrak{g})=\mathbb{C}id$.

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