= Solution
Let $\widehat T_a=\operatorname{ad}_{T_a}$ on the <Lie algebra> itself. The <Jacobi identity> says, for every $X$,
$$
[T_a,[T_b,X]]-[T_b,[T_a,X]]=[[T_a,T_b],X].
$$
Therefore $[\widehat T_a,\widehat T_b]=c^c_{ab}\widehat T_c$, proving that the displayed structure-constant matrices form the <Adjoint representation of a Lie algebra>.
The bilinear form $\kappa_{ab}=\operatorname{tr}(\widehat T_a\widehat T_b)$ is the <Killing form>. It is symmetric by cyclicity of <trace>. Its invariance follows by expanding the two commutators:
$$
\begin{aligned}
\kappa([T_c,T_a],T_b)+\kappa(T_a,[T_c,T_b])
&=\operatorname{tr}([\widehat T_c,\widehat T_a]\widehat T_b+\widehat T_a[\widehat T_c,\widehat T_b])\\
&=\operatorname{tr}([\widehat T_c,\widehat T_a\widehat T_b])=0.
\end{aligned}
$$
In components this is precisely
$$
\boxed{\kappa_{db}c^d_{ca}+\kappa_{ad}c^d_{cb}=0.}
$$
Equivalently, the <adjoint action> preserves the bilinear form, so it is an <invariant tensor>.
The invariant abelian subalgebra is an abelian <ideal of a Lie algebra> $\mathfrak a$. Assume it is nonzero, as required for the conclusion. For $X\in\mathfrak a$ and $Y\in\mathfrak g$, the map $\operatorname{ad}_X\operatorname{ad}_Y$ sends $\mathfrak g$ into $\mathfrak a$, because it is an ideal. It vanishes on $\mathfrak a$, because $\operatorname{ad}_Y$ preserves the ideal and $\operatorname{ad}_X$ kills it. Relative to $\mathfrak a$ and a vector-space complement, this map has zero diagonal blocks, so its <trace> is zero. Thus
$$
\kappa(X,Y)=0\qquad(X\in\mathfrak a,\ Y\in\mathfrak g).
$$
This proves that <abelian ideals lie in the radical of the Killing form>. Choose a nonzero $X=x^aT_a$ in the ideal. Then $\kappa_{ba}x^a=0$ for every $b$, so \b[the coordinate vector $x$ is a nonzero null eigenvector of the Killing matrix]. The nonzero-ideal qualification cannot be dropped: the zero ideal is abelian even in a semisimple algebra with nondegenerate <Killing form>.
Now suppose $\kappa$ is invertible and let $C_\kappa=\kappa^{bc}t_bt_c$ in any <Lie algebra representation>. In matrix notation invariance is $A_a^T\kappa+\kappa A_a=0$, where $(A_a)^d{}_b=c^d_{ab}$. Multiplying by the inverse form gives $A_a\kappa^{-1}+\kappa^{-1}A_a^T=0$. Hence
$$
\begin{aligned}
[t_a,C_\kappa]
&=\kappa^{bc}(c^d_{ab}t_dt_c+c^d_{ac}t_bt_d)\\
&=(c^b_{ad}\kappa^{dc}+\kappa^{bd}c^c_{ad})t_bt_c=0.
\end{aligned}
$$
No commutativity of the representation matrices has been assumed. This proves the centrality of the <quadratic Casimir operator> formed with the inverse <Killing form>. <Schur lemma> makes it a scalar on every finite-dimensional complex <irreducible representation>.
For <SU(2)>, make the normalization explicit. In a Hermitian angular-momentum basis with $[J_a,J_b]=i\epsilon_{abc}J_c$, the structure-constant matrices obey
$$
\kappa_{ab}=\sum_{c,d}(i\epsilon_{adc})(i\epsilon_{bcd})=2\delta_{ab}.
$$
Thus $C_\kappa=\tfrac12\sum_aJ_a^2$. In the compact anti-Hermitian basis $T_a=-iJ_a$, the brackets have real structure constants $\epsilon_{abc}$ and the <Killing form of the SU(2) Lie algebra> is instead $-2\delta_{ab}$; the inverse contraction gives the same $C_\kappa$. This sign change is a basis convention, not a different spectrum.
In the complex highest-weight basis $(H,E^+,E^-)$ used earlier, the same <Killing form> and its inverse are
$$
\kappa=\begin{pmatrix}8&0&0\\0&0&4\\0&4&0\end{pmatrix},\qquad \kappa^{-1}=\begin{pmatrix}1/8&0&0\\0&0&1/4\\0&1/4&0\end{pmatrix}.
$$
Consequently
$$
C_\kappa=\frac{H^2}{8}+\frac{E^+E^-+E^-E^+}{4}.
$$
On a highest vector of weight $m=2j$, the ladder calculation gives $E^+E^-v=mv$ and $E^-E^+v=0$. Its <Casimir eigenvalue> is therefore
$$
\boxed{\frac{m^2}{8}+\frac m4=\frac{m(m+2)}8=\frac12j(j+1),\qquad j=0,\tfrac12,1,\ldots.}
$$
Centrality gives the same value throughout the irreducible multiplet. This is the <Killing-normalized SU(2) quadratic Casimir>; the familiar angular-momentum operator $\sum_aJ_a^2$ has eigenvalue $j(j+1)$ and is twice this inverse-Killing-form contraction.
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