= Solution
Let $R_a=\operatorname{ad}(T_a)$, where $\operatorname{ad}(X)Y=[X,Y]$. In the given basis, the <Adjoint representation of a Lie algebra> is
$$
(R_a)^c{}_b=f^c{}_{ab}.
$$
The <Jacobi identity> gives $[\operatorname{ad}X,\operatorname{ad}Y]=\operatorname{ad}[X,Y]$, so $[R_a,R_b]=f^c{}_{ab}R_c$, as required for a <Lie algebra representation>. The <Killing form> is the symmetric <Trace form of a Lie algebra representation>
$$
\boxed{\kappa_{ab}=\operatorname{tr}(R_aR_b)=f^c{}_{ad}f^d{}_{bc}.}
$$
Cyclicity of the <trace> proves its invariance directly:
$$
\begin{aligned}
\kappa([T_a,T_b],T_c)+\kappa(T_b,[T_a,T_c])
&=\operatorname{tr}\bigl([R_a,R_b]R_c+R_b[R_a,R_c]\bigr)\\
&=\operatorname{tr}(R_aR_bR_c-R_bR_cR_a)=0.
\end{aligned}
$$
In components this is precisely $\boxed{\kappa_{dc}f^d{}_{ab}+\kappa_{bd}f^d{}_{ac}=0}$, so the <Killing form> is an <invariant bilinear form on a Lie algebra>.
Choose the infinitesimal <gauge transformation> convention $\delta\phi=\omega\phi$, with $\omega(x)=\omega^a(x)t_a$, and write $A_\mu=A_\mu^at_a$. A <gauge covariant derivative> must satisfy $D'_\mu=UD_\mu U^{-1}$ for $U=1+\omega+O(\omega^2)$; hence
$$
A'_\mu=UA_\mu U^{-1}-(\partial_\mu U)U^{-1},\qquad
\boxed{\delta A_\mu=-\partial_\mu\omega+[\omega,A_\mu].}
$$
Equivalently $\delta A_\mu^a=-\partial_\mu\omega^a+f^a{}_{bc}\omega^bA_\mu^c$. Expanding $\delta(D_\mu\phi)$ makes the cancellation transparent:
$$
\begin{aligned}
\delta(D_\mu\phi)
&=(\partial_\mu\omega)\phi+\omega\partial_\mu\phi
+(-\partial_\mu\omega+[\omega,A_\mu])\phi+A_\mu\omega\phi\\
&=\omega(\partial_\mu\phi+A_\mu\phi).
\end{aligned}
$$
Thus $\boxed{\delta(D_\mu\phi)=\omega^at_aD_\mu\phi}$: it transforms in the same representation as $\phi$. In <adjoint covariant derivative> notation, $\delta A_\mu=-D_\mu^{\mathrm{ad}}\omega$.
The commutator of two <gauge covariant derivatives> defines the <gauge field strength>:
$$
[D_\mu,D_\nu]\phi=F_{\mu\nu}^at_a\phi,\qquad
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu].
$$
Thus
$$
\boxed{F^a_{\mu\nu}=\partial_\mu A^a_\nu-\partial_\nu A^a_\mu+f^a{}_{bc}A^b_\mu A^c_\nu.}
$$
Conjugating the derivative commutator gives $F'_{\mu\nu}=UF_{\mu\nu}U^{-1}$, so the <gauge field strength> transforms homogeneously in the <Adjoint representation>:
$$
\boxed{\delta F_{\mu\nu}=[\omega,F_{\mu\nu}],\qquad
\delta F^a_{\mu\nu}=f^a{}_{bc}\omega^bF^c_{\mu\nu}.}
$$
Even if the matrices $t_a$ in a particular matter representation are not faithful, the displayed component formula defines the Lie-algebra-valued curvature independently of that representation.
Only $g_{(ab)}$ contributes to the <Yang-Mills theory> Lagrangian, since $F^a_{\mu\nu}F^{b\mu\nu}$ is symmetric in $a,b$. Thus take $g_{ab}$ to be a constant real <symmetric bilinear form>. Its infinitesimal variation is
$$
\delta\mathcal L=-\frac14\omega^a\bigl(g_{dc}f^d{}_{ab}+g_{bd}f^d{}_{ac}\bigr)
F^b_{\mu\nu}F^{c\mu\nu}.
$$
Arbitrary local field strengths and gauge parameters give the necessary and sufficient <invariant gauge kinetic form> condition
$$
\boxed{g_{dc}f^d{}_{ab}+g_{bd}f^d{}_{ac}=0,\qquad g_{ab}=g_{ba}.}
$$
An antisymmetric part is unconstrained but contributes nothing. A nondegenerate kinetic term additionally needs a <nondegenerate bilinear form>; positive energy additionally needs a positive internal metric in the signature $(+---)$ convention. Neither extra condition follows solely from <gauge invariance>. For a compact simple real <Lie algebra>, $-\kappa$ is positive definite and is the standard choice up to a positive factor.
For the intended complex-simple setting, the <Killing form> is nondegenerate. Consequently write any <invariant bilinear form on a Lie algebra> uniquely as $g(X,Y)=\kappa(SX,Y)$. Invariance of both forms implies
$$
\kappa\bigl((S\operatorname{ad}Z-\operatorname{ad}Z\,S)X,Y\bigr)=0
$$
for every $X,Y,Z$, so $S$ commutes with the <Adjoint representation>. A subspace invariant under all adjoint maps is an ideal, hence simplicity makes the adjoint representation irreducible. By <Schur lemma>, $S=cI$, giving the <uniqueness of an invariant bilinear form on a simple Lie algebra>:
$$
\boxed{g_{ab}=c\kappa_{ab}.}
$$
The same conclusion holds for compact simple real algebras by complexification, with $c$ real; nondegeneracy excludes $c=0$, and positive energy selects $c<0$ in the stated convention.
There is a genuine <real-simple exception to uniqueness of the Killing form> if “simple” is read as an arbitrary real simple algebra. Regard $\mathfrak{sl}_2(\mathbb C)$ as real. It is real simple: its complexification consists of two complex simple factors exchanged by conjugation, so a conjugation-stable ideal is zero or the whole algebra. Both $\operatorname{Re}\kappa_{\mathbb C}$ and $\operatorname{Im}\kappa_{\mathbb C}$ are real symmetric invariant forms, while $\kappa_{\mathbb R}=2\operatorname{Re}\kappa_{\mathbb C}$. With $H=\operatorname{diag}(1,-1)$, the <Killing form of the special linear Lie algebra> gives $\kappa_{\mathbb C}(H,H)=8$ and $\kappa_{\mathbb C}(H,iH)=8i$. Therefore the imaginary part cannot be a real multiple of $\kappa_{\mathbb R}$. The uniqueness conclusion requires the customary complex-simple or compact-real interpretation.
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