Killing form on ideal equals restriction of Killing form

Statement
Suppose $$F$$ is a field, $$L$$ is a finite-dimensional Lie algebra over $$F$$ and $$\kappa$$ is the fact about::Killing form on $$L$$. Suppose $$I$$ is an ideal of $$L$$. Then, the restriction of $$\kappa$$ to $$I$$ equals the Killing form on $$I$$, viewed as a Lie algebra.

Related facts

 * Cartan's first criterion
 * Cartan's second criterion
 * Abelian ideal is degenerate for Killing form
 * Killing form on subalgebra not equals restriction of Killing form

Proof
Given: A finite-dimensional Lie algebra $$L$$ over a field $$F$$. $$I$$ is an ideal of $$L$$ and $$\kappa$$ is the Killing form on $$L$$.

To prove: The restriction of $$\kappa$$ to $$I$$ equals the Killing form on $$I$$.

Proof: Choose a basis for $$I$$ and extend it to a basis for $$L$$. With this basis, for any $$x \in I$$, the matrix for $$\operatorname{ad}(x)$$ is of the form:

$$\begin{pmatrix} \operatorname{ad}(x)_I & * \\ 0 & 0 \end{pmatrix}$$.

Thus, the top left part is the same as the matrix for $$\operatorname{ad}(x)$$ on $$I$$.

Thus, the matrix for $$\operatorname{ad}(x) \circ \operatorname{ad}(y)$$ is:

$$\begin{pmatrix} \operatorname{ad}(x)_I \circ \operatorname{ad}(y)_I & * \\ 0 & 0\end{pmatrix}$$

The trace of this matrix equals the trace of $$\operatorname{ad}(x)_I \circ \operatorname{ad}(y)_I$$, so $$\kappa(x,y)$$ equals the Killing form on $$x,y$$ in $$I$$.