I want to show that the Killing $k$ form of $u(\mathbb{C},3)$ is such that $k(x,y)=0$ $\forall x,y \in u(\mathbb{C},3)$.
I have used the basis {$e_{12}, e_{13}, e_{23}$}, and found that the adjoint matrices are given by $$ad(e_{12})=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, ad(e_{13})=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},ad(e_{23})=\begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
I then found that the matrix of the killing form is just the $0$ matrix, but I was wondering is this enough to say that $k(x,y)=0 \hskip 0.5em \forall x,y \in u(\mathbb{C},3)$?