= Solution
The <Killing form> is the bilinear form
$$
\boxed{B(X,Y)=\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y),\qquad\operatorname{ad}X(Z)=[X,Z].}
$$
The <Adjoint representation> is linear, giving bilinearity, and the cyclic property of <trace> gives symmetry. To prove invariance, abbreviate the adjoint matrices of $X,Y,Z$ by $A_X,A_Y,A_Z$. The <Jacobi identity> gives $A_{[X,Y]}=[A_X,A_Y]$, and cyclic trace then gives
$$
\begin{aligned}
B([X,Y],Z)&=\operatorname{tr}\bigl((A_XA_Y-A_YA_X)A_Z\bigr)\\
&=\operatorname{tr}\bigl(A_X(A_YA_Z-A_ZA_Y)\bigr)=B(X,[Y,Z]).
\end{aligned}
$$
Thus $\boxed{B([X,Y],Z)=B(X,[Y,Z])}$, equivalently $B([X,Y],Z)+B(Y,[X,Z])=0$. This is infinitesimal invariance of the <Killing form>, with no choice of basis affecting the trace.
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