Levi decomposition of $\mathfrak{gl}_n(\mathbb{K})$

Question

Im trying to find the Levi decomposition of $\mathfrak{gl}_n(\mathbb{K})$ where $\mathbb{K}$ has characteristic zero. By Levi's theorem $\mathfrak{gl}_n(\mathbb{K})=Rad(\mathfrak{gl}_n(\mathbb{K} )+S$ where $Rad$ is the solvable radical and S is semisimple.

I know that $Rad{$\mathfrak{gl}_n(\mathbb{K})$ is the set of scalar matrix, but how I can find this set $S$?

Answer 1

For a reductive Lie algebra $L$ we have the Levi decomposition $L=[L,L]\oplus Z(L)$. Here the commutator subalgebra is semisimple, i.e., a Levi subalgebra, and the center $Z(L)$ is abelian, i.e., the solvable radical of the Levi decomposition. For $L=\mathfrak{gl}_n(K)$ we have $$ S=[L,L]=\mathfrak{sl}_n(K), \quad Z(L)=K\cdot I_n. $$