Semisimplicity of Lie algebras and non-degeneracy of associated bilinear forms of representations Let $ \mathfrak{g} $ be a finite-dimensional Lie algebra over a field of characteristic $ 0 $. For a finite-dimensional representation $ \rho $, we define a bilinear form $ B_\rho $ on $ \mathfrak{g} $ by
$$
B_{\rho}(x, y) = \operatorname{Tr}(\rho(x) \circ \rho(y))
\qquad (x, y \in \mathfrak{g}).
$$
$ B_\rho $ is called the bilinear form associated to $ \rho $. Note that the Killing form is the bilinear form associated to the adjoint representation.
If $ \mathfrak{g} $ is semisimple, $ B_\rho $ is non-degenerate for every finite-dimensional faithful representation $ \rho $ of $ \mathfrak{g} $ (Bourbaki, Lie Groups and Lie Algebras, I.6.1, Proposition 1).
Question. Does the converse of the above statement hold? That is, if $ B_\rho $ is non-degenerate for every finite-dimensional faithful representation $ \rho $ of $ \mathfrak{g} $, can we conclude that $ \mathfrak{g} $ is semisimple?
(If in addition we assume that $ \mathfrak{g} $ has a trivial center, the adjoint representation $ \mathfrak{g} $ is faithful, and hence $ \mathfrak{g} $ is semisimple by the assumption and Cartan’s criterion.)
 A: $\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}$The first preliminary remark is that $\mathfrak{z}(\g)\cap [\g,\g]$ is in the kernel of $B_\rho$ for every rep $\rho$. Indeed, for this we can pass to an algebraic closure, block-trigonalize the representation with irreducible diagonal blocks: on each (irreducible) diagonal block, this central intersection acts by scalars of trace zero, hence by zero.
So, combined with Ado's theorem, the assumption ensures that $\mathfrak{z}(\g)\cap [\g,\g]=0$.
Se can write $\g$ as $\mathfrak{z}(\g)\times\h$. By Ado's theorem choose a faithful rep for $\h$, and take the direct sum with a faithful rep for the central factor so that it acts by strict upper triangular matrices. The direct sum of those reps is a faithful rep of the direct product, with degenerate associated bilinear form unless $\mathfrak{z}(\g)=0$.
Hence, $\g$ has trivial center, and this case is already mentioned by the OP, using the adjoint representation.
(Note that the result trivially implies Ado's theorem, so we can't expect to bypass it.)
