Killing form is symmetric

Statement
Suppose $$L$$ is a finite-dimensional fact about::Lie algebra over a field $$F$$ and $$\kappa$$ is the fact about::Killing form on $$L$$. Then, $$\kappa$$ is symmetric, i.e.:

$$\! \kappa(x,y) = \kappa(y,x) \ \forall \ x,y \in L$$

Facts used

 * 1) uses::Trace of product of two linear transformations is independent of their order: For any two linear transformations $$A,B$$ from a vector space to itself, the trace of $$AB$$ is the same as the trace of $$BA$$.

Related facts

 * Trace of product of two linear transformations is independent of their order
 * Associativity-like relation between Killing form and Lie bracket

Proof
By definition, we have:

$$\kappa(x,y) = \operatorname{Tr}(\operatorname{ad}x \circ \operatorname{ad}y)$$

Similarly, we have:

$$\kappa(y,x) = \operatorname{Tr}(\operatorname{ad}y \circ \operatorname{ad}x)$$

Setting $$A = \operatorname{ad}x$$ and $$B = \operatorname{ad}y$$, we get, from fact (1), that $$\kappa(x,y) = \kappa(y,x)$$.