 Solution

The Killing form is ${\kappa }_L(x,y)=\mathrm{tr}({\mathrm{ad}}_{L}x{\mathrm{ad}}_{L}y)$. Symmetry follows from $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Since $\mathrm{ad}[z,x]=[\mathrm{ad}z,\mathrm{ad}x]$, cyclicity of trace gives the invariance identityIf $x\in I^{\perp }$ and $y\in I$, then $[z,y]\in I$, so the right side is zero. Hence $[z,x]\in I^{\perp }$ for every $z$, proving that $\boxed{I^{\perp }\text{ is an ideal}}$.

For $x\in I$, use a basis of $I$ followed by a complementary basis of $L$. Because $[x,L]\subseteq I$, the matrix of ${\mathrm{ad}}_{L}x$ has blocks $\left(\begin{array}{cc}{\mathrm{ad}}_{I}x& B_x\\ 0& 0\end{array}\right)$. Multiplying two such matrices showsThe restriction statement has both arguments in $I$; it does not identify the two forms for arbitrary elements of $L$ outside $I$.

Now use the ordered basis in the question for a central ideal $J$. The images of the $u_j$ in the quotient have adjoint matrix $A$, so the $u$-coordinate block of their images is $A$; arbitrary $J$-components form $B$. Each $v_j$ is central, so its image under every adjoint map is zero. Thus the matrix is exactly $\left(\begin{array}{cc}A& 0\\ B& 0\end{array}\right)$. Products have top-left block $A_x{A}_y$ and zero bottom-right block. Taking traces givesFinally, a nonzero finite-dimensional nilpotent Lie algebra has nonzero centre by (1)(b). Induct on its dimension: the Killing form descends to the quotient by its centre, which is again nilpotent of smaller dimension, and is therefore zero by induction. The zero algebra starts the induction. Apply this to the nilpotent ideal $I$ and then use the restriction identity to obtain $\boxed{{\kappa }_L(x,y)=0\text{ for }x,y\in I}$. This avoids the false general claim that the product of two nilpotent linear maps must be nilpotent.