= Solution
The <Killing form> is $\kappa_L(x,y)=\operatorname{tr}(\operatorname{ad}_Lx\,\operatorname{ad}_Ly)$. Symmetry follows from $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Since $\operatorname{ad}[z,x]=[\operatorname{ad}z,\operatorname{ad}x]$, cyclicity of <trace> gives the invariance identity
$$
\kappa_L([z,x],y)=-\kappa_L(x,[z,y]).
$$
If $x\in I^\perp$ 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 $\operatorname{ad}_Lx$ has blocks $\begin{pmatrix}\operatorname{ad}_Ix&B_x\\0&0\end{pmatrix}$. Multiplying two such <matrices> shows
$$
\boxed{\kappa_L(x,y)=\kappa_I(x,y)\quad(x,y\in I).}
$$
The 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 of a Lie algebra>[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 $\begin{pmatrix}A&0\\B&0\end{pmatrix}$. Products have top-left block $A_xA_y$ and zero bottom-right block. Taking traces gives
$$
\boxed{\kappa_L(x,y)=\kappa_{L/J}(\bar x,\bar y).}
$$
Finally, a nonzero finite-dimensional <nilpotent Lie algebra> has nonzero <center of a Lie algebra>[centre] by (1)(b). Induct on its <dimension>: the <Killing form> descends to the quotient by its <center of a Lie algebra>[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 of a Lie algebra>[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.
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