Killing form on subalgebra not equals restriction of Killing form

Statement
The following is possible: $$L$$ is a finite-dimensional Lie algebra over a field $$F$$, and $$\kappa$$ is the Killing form on $$L$$. $$A$$ is a subalgebra of $$L$$, and the restriction of $$\kappa$$ to $$A$$ is not equal to the Killing form on $$A$$ as a Lie algebra.

Related facts

 * Killing form on ideal equals restriction of Killing form: If the subalgebra is an ideal, then the Killing form on it does equal the restriction of the Killing form on the whole algebra.
 * Nilpotent ideal is degenerate for Killing form: In particular, the Killing form, restricted to a nilpotent ideal, is zero. Hence, for any subalgebra contained in a nilpotent ideal (even if that subalgebra is not itself an ideal) both its own Killing form and the Killing form obtained by restriction from the whole algebra are trivial.

Proof
Consider a two-dimensional Lie algebra $$L$$ over $$F$$ with basis $$x,y$$ and $$[x,y] = y$$. Let $$A$$ be the subalgebra of $$L$$ generated by $$x$$. Then:


 * $$\! \kappa_A(x,x) = 0$$, since $$\operatorname{ad}(x)_A = 0$$.
 * $$\! \kappa_L(x,x) = 1$$, since, in basis $$\{ x,y \}$$:

$$\operatorname{ad}(x)_L = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$.