Trace forms of faithful representations of semisimple Lie algebras are nondegenerate

Given: A finite-dimensional Lie algebra $\mathfrak{h}$ over a characteristic-zero field $k$ with $\mathrm{rad}(\mathfrak{h})=0$, and a faithful finite-dimensional representation $\rho :\mathfrak{h}\to \mathfrak{g}\mathfrak{l}(V)$. Write $B_{\rho }(x,y)=\mathrm{tr}(\rho (x)\rho (y))$ for its trace form and $\mathfrak{r}=\{x\in \mathfrak{h}:B_{\rho }(x,y)=0\text{ for every }y\in \mathfrak{h}\}$ for its radical.

Symmetry and invariance. $B_{\rho }$ is bilinear and symmetric, and $B_{\rho }([z,x],y)+B_{\rho }(x,[z,y])=0$ for all $x,y,z\in \mathfrak{h}$ (Trace forms are symmetric and invariant).

Radicals of invariant forms are ideals. If $\mathfrak{i}$ is an ideal of a Lie algebra carrying a symmetric invariant bilinear form $B$, then ${\mathfrak{i}}^{\perp }$ is an ideal; in particular $\mathfrak{r}={\mathfrak{h}}^{\perp }$ is an ideal of $\mathfrak{h}$ (Orthogonal complements under invariant forms are ideals, Lie subalgebras, ideals, and center).

Cartan's solvability criterion. Let $\mathfrak{l}\subseteq \mathfrak{g}\mathfrak{l}(V)$ be a finite-dimensional linear Lie algebra over a characteristic-zero field. If $\mathrm{tr}(xy)=0$ for all $x\in [\mathfrak{l},\mathfrak{l}]$ and $y\in \mathfrak{l}$, then $\mathfrak{l}$ is solvable (Cartan's solvability criterion).

The solvable radical. The solvable radical $\mathrm{rad}(\mathfrak{h})$ is the largest solvable ideal of $\mathfrak{h}$: it is solvable and contains every solvable ideal (Solvable radical), and semisimplicity means $\mathrm{rad}(\mathfrak{h})=0$ (Simple, semisimple, and reductive Lie algebras).

Faithfulness transfers solvability. If $\rho$ is injective and bracket preserving then $\rho (D^j\mathfrak{r})=D^j\rho (\mathfrak{r})$ for every $j$, so $\mathfrak{r}$ is solvable as soon as the linear Lie algebra $\rho (\mathfrak{r})$ is solvable; ideals and quotients of a semisimple algebra are again semisimple (Ideals and quotients of semisimple Lie algebras).