Let $G$ be a complex algebraic group embedded into $GL_{n}(\mathbb{C})$$\operatorname{GL}_{n}(\mathbb{C})$. A criterion for $G$ to be reductive is the following. Let $\mathfrak{g}$ be the Lie algebra of $G$, and let $B$ be the trace form of $\mathfrak{gl}_{n}(\mathbb{C})$ restricted to $\mathfrak{g}$: $B(x,y) = Tr(xy)$$B(x,y) = \operatorname{Tr}(xy)$. Then $G$ is reductive if and only if $B$ is non-degenerate.
This criterion was used in an answer to one of my previous questions. I think I also know how to prove it. It follows because the Lie algebra of the unipotent radical, $\mathfrak{r}$, is the kernel of $B$. The reason for this is roughly as follows: using a Levi decomposition, the Lie algebra $\mathfrak{g}$ decomposes (as a vector space) into $\mathfrak{g} = \mathfrak{r} \oplus \mathfrak{z} \oplus \mathfrak{s}$, where $\mathfrak{z}$ is abelian (the Lie algebra of a torus) and $\mathfrak{s}$ is semisimple. Since $\ker(B)$ is solvable, it must be contained in $\mathfrak{r} \oplus \mathfrak{z}$, which is the radical. But since $\mathfrak{z}$ is the Lie algebra of a torus, $B$ restricts to $\mathfrak{z}$ to be positive definite. On the other hand, we can choose a basis so that $\mathfrak{r} \oplus \mathfrak{z}$ lies in the upper triangular matrices, and this implies that $\mathfrak{r} \subseteq \ker B$.
I haven't been able to find this result anywhere in the literature. I am wondering if anyone knows where I can find it?
