[proofplan] The proof is a direct unpacking of the definition of the Killing form. First we verify that the assignment $X\mapsto \operatorname{ad}_X$ is linear, using bilinearity of the Lie bracket. Then linearity of composition in each factor and linearity of trace give bilinearity of $B$. Finally, symmetry follows from the cyclic trace identity $\operatorname{tr}(S\circ T)=\operatorname{tr}(T\circ S)$ for endomorphisms of a finite-dimensional [vector space](/page/Vector%20Space). [/proofplan]

[step:Show that the adjoint map is linear in the Lie algebra element]Let \begin{align*} \operatorname{ad}:\mathfrak g\to \operatorname{End}_k(\mathfrak g) \end{align*} denote the map $X\mapsto \operatorname{ad}_X$. For $X_1,X_2,Y\in\mathfrak g$ and $a,b\in k$, bilinearity of the Lie bracket gives \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=[aX_1+bX_2,Y]. \end{align*} Hence \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=a[X_1,Y]+b[X_2,Y]. \end{align*} By the definition of $\operatorname{ad}_{X_1}$ and $\operatorname{ad}_{X_2}$, this is \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=a\operatorname{ad}_{X_1}(Y)+b\operatorname{ad}_{X_2}(Y). \end{align*} Since this holds for every $Y\in\mathfrak g$, we have \begin{align*} \operatorname{ad}_{aX_1+bX_2}=a\operatorname{ad}_{X_1}+b\operatorname{ad}_{X_2} \end{align*} as elements of $\operatorname{End}_k(\mathfrak g)$.[/step]

[guided]We first isolate the only Lie-algebraic input: the bracket is bilinear. Define \begin{align*} \operatorname{ad}:\mathfrak g\to \operatorname{End}_k(\mathfrak g) \end{align*} by sending $X\in\mathfrak g$ to the endomorphism \begin{align*} \operatorname{ad}_X:\mathfrak g\to\mathfrak g, \end{align*} where $\operatorname{ad}_X(Y)=[X,Y]$ for $Y\in\mathfrak g$. To prove that $\operatorname{ad}$ is linear, fix $X_1,X_2\in\mathfrak g$ and scalars $a,b\in k$. Two endomorphisms of $\mathfrak g$ are equal exactly when they have the same value on every vector $Y\in\mathfrak g$, so let $Y\in\mathfrak g$ be arbitrary. By bilinearity of the bracket in its first variable, \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=[aX_1+bX_2,Y]. \end{align*} Therefore \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=a[X_1,Y]+b[X_2,Y]. \end{align*} Using the definition of the adjoint endomorphisms again, this becomes \begin{align*} \operatorname{ad}_{aX_1+bX_2}(Y)=a\operatorname{ad}_{X_1}(Y)+b\operatorname{ad}_{X_2}(Y). \end{align*} Because $Y$ was arbitrary, the endomorphisms themselves are equal: \begin{align*} \operatorname{ad}_{aX_1+bX_2}=a\operatorname{ad}_{X_1}+b\operatorname{ad}_{X_2}. \end{align*} This proves that $X\mapsto\operatorname{ad}_X$ is a $k$-[linear map](/page/Linear%20Map) from $\mathfrak g$ to $\operatorname{End}_k(\mathfrak g)$.[/guided]

[step:Use linearity of trace to prove bilinearity of the Killing form] Fix $X_1,X_2,Y\in\mathfrak g$ and $a,b\in k$. Using the previous step, linearity of composition in the left factor, and linearity of trace on $\operatorname{End}_k(\mathfrak g)$, we obtain \begin{align*} B(aX_1+bX_2,Y)=\operatorname{tr}((a\operatorname{ad}_{X_1}+b\operatorname{ad}_{X_2})\circ\operatorname{ad}_Y). \end{align*} Thus \begin{align*} B(aX_1+bX_2,Y)=a\operatorname{tr}(\operatorname{ad}_{X_1}\circ\operatorname{ad}_Y)+b\operatorname{tr}(\operatorname{ad}_{X_2}\circ\operatorname{ad}_Y). \end{align*} By the definition of $B$, \begin{align*} B(aX_1+bX_2,Y)=aB(X_1,Y)+bB(X_2,Y). \end{align*} Similarly, for $X,Y_1,Y_2\in\mathfrak g$ and $a,b\in k$, linearity of $Y\mapsto\operatorname{ad}_Y$, linearity of composition in the right factor, and linearity of trace give \begin{align*} B(X,aY_1+bY_2)=\operatorname{tr}(\operatorname{ad}_X\circ(a\operatorname{ad}_{Y_1}+b\operatorname{ad}_{Y_2})). \end{align*} Therefore \begin{align*} B(X,aY_1+bY_2)=aB(X,Y_1)+bB(X,Y_2). \end{align*} So $B$ is bilinear. [/step]

[step:Apply cyclicity of trace to prove symmetry] Let $X,Y\in\mathfrak g$. Since $\mathfrak g$ is finite-dimensional over $k$, the endomorphisms $\operatorname{ad}_X$ and $\operatorname{ad}_Y$ have well-defined traces, and the cyclic trace identity applies to the pair \begin{align*} \operatorname{ad}_X,\operatorname{ad}_Y\in\operatorname{End}_k(\mathfrak g). \end{align*} Thus \begin{align*} B(X,Y)=\operatorname{tr}(\operatorname{ad}_X\circ\operatorname{ad}_Y). \end{align*} By cyclicity of trace, \begin{align*} \operatorname{tr}(\operatorname{ad}_X\circ\operatorname{ad}_Y)=\operatorname{tr}(\operatorname{ad}_Y\circ\operatorname{ad}_X). \end{align*} Using the definition of $B$ again, \begin{align*} B(X,Y)=B(Y,X). \end{align*} Since $X$ and $Y$ were arbitrary, $B$ is symmetric. Together with bilinearity, this proves the theorem. [/step]