 Solution

Let $L_A$ and $R_A$ be left and right multiplication by $A$ on the vector space $M_n(\mathbb{C})$. Then $\mathrm{ad}A=L_A-R_A$. On matrix units, or from $M_n(\mathbb{C})={\mathbb{C}}^n\otimes ({\mathbb{C}}^n{)}^{\ast }$, one obtainsFor example, the coefficient of $E_{ij}$ in $A{E}_{ij}B$ is $A_{ii}{B}_{jj}$; summing over $i,j$ proves the mixed-trace identity. Expanding the product of the two adjoint operators therefore gives the Killing form of the general linear Lie algebra:For $A,B$ in the special linear Lie algebra, both individual traces vanish. Furthermore $M_n(\mathbb{C})={\mathfrak{sl}}_n\oplus \mathbb{C}I$, and every adjoint operator kills the scalar summand, so its product has the same trace on $M_n$ and on ${\mathfrak{sl}}_n$. Consequently the Killing form of the special linear Lie algebra isNo uniqueness theorem for invariant bilinear forms on a Lie algebra is needed: this calculation establishes the constant directly. For the degenerate case $n=1$, the Lie algebra is zero, both forms vanish, and proportionality does not determine a unique constant.