If a Lie subalgebra $L \leq gl_n(\Bbb C)$ is nilpotent, does it follow that any matrix $x \in L$ is nilpotent (i.e. there is $n>0$ such that $x^n = 0$) ?
I know that the adjoint $ad_x$ of $x$ is nilpotent, and if $x$ is nilpotent then $ad_x$ is nilpotent, but I don't know about my case.
Take a Cartan subalgebra of $\mathfrak{sl}_n(\mathbb{C})\subseteq \mathfrak{gl}_n(\mathbb{C})$ consisting of diagonal matrices of trace zero. This is an abelian subalgebra, hence nilpotent. The diagonal matrices, however, need not be nilpotent; for example $d=diag(1,-1,0,\ldots ,0)$.
Consider any block decomposition, and consider the set of matrices that are diagonal in this block decomposition, and, in each block, upper triangular with constant diagonal. This is a nilpotent Lie sublgebra of $\mathfrak{gl}_n$.
Moreover, a subalgebra of $\mathfrak{gl}_n(C)$ ($C$ any algebraically closed field of characteristic zero) is nilpotent iff it is conjugate (by some matrix in $\mathrm{GL}_n(C)$) into one of the previous subalgebras.
