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})$$? This must be standard knowledge but I couldn't find a reference.
Clearly, all nilpotent matrices and all diagonal matrices are in the nilradical. What else?
On
The Lie algebra $\mathfrak{g}=\mathfrak{gl}_n(\mathbb{C})$ is reductive, i.e., we have $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]\oplus Z(\mathfrak{g})$, where $[\mathfrak{g},\mathfrak{g}]$ is semisimple (in fact it is $\mathfrak{sl}_n(\mathbb{C})$ in this case), and the center is abelian and equals the solvable radical and the nilradical. So the nilradical here equals $$\operatorname{nil}(\mathfrak{g})=Z(\mathfrak{g})=\mathbb{C}\cdot \operatorname{id}.$$
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.