Solution
= Solution
Let $\rho:\mathfrak g\to\mathfrak{gl}(V)$ be a finite-dimensional <Lie algebra representation>. The form denoted $B_V$ is
$$
\boxed{B_V(X,Y)=\operatorname{tr}_V\bigl(\rho(X)\rho(Y)\bigr).}
$$
It is the <Trace form of a Lie algebra representation>; the <Killing form> $B$ without a subscript is specifically the case $V=\mathfrak g$, $\rho=\operatorname{ad}$. The distinction matters: the form of the trivial representation cannot detect whether the algebra is semisimple.
Linearity of $\rho$ and <trace> proves bilinearity, and cyclicity of the <trace> gives symmetry. With $A=\rho(X)$, $D=\rho(Y)$ and $C=\rho(Z)$, the representation identity gives
$$
\begin{aligned}
B_V([X,Y],Z)&=\operatorname{tr}((AD-DA)C)\\
&=\operatorname{tr}(ADC-ACD)=\operatorname{tr}(A(DC-CD))\\
&=B_V(X,[Y,Z]).
\end{aligned}
$$
Thus it is an <invariant bilinear form on a Lie algebra>, and in particular the <Adjoint representation> preserves the <Killing form>.
The <Cartan solvability criterion> has two useful formulations. For a finite-dimensional complex <Lie algebra> $\mathfrak g$,
$$
\boxed{\mathfrak g\text{ is solvable}\quad\Longleftrightarrow\quad B(\mathfrak g,[\mathfrak g,\mathfrak g])=0.}
$$
Its matrix version says that a <Lie subalgebra> $L\subseteq\mathfrak{gl}(V)$ is solvable precisely when $\operatorname{tr}(xy)=0$ for all $x\in[L,L]$ and $y\in L$. We prove the matrix version first, with ordinary <trace> in $V$.
If $L$ is solvable, the <Lie theorem> makes all its matrices upper triangular. Their <commutators> are strictly upper triangular, so multiplying such a matrix by an upper triangular one still has zero diagonal and hence zero <trace>. This proves the easy direction.
Conversely assume the trace-orthogonality condition and fix $x\in[L,L]$. We show that all <eigenvalues> of $x$ vanish. Use its <Additive Jordan decomposition> $x=x_s+x_n$ with $[x_s,x_n]=0$. On the <generalized eigenspace> $V_\lambda$ of $x$, define an auxiliary endomorphism $y$ to be $\overline\lambda I$. This $y$ need not be in $L$; we only need control of its <commutator> with $L$.
On $\operatorname{Hom}(V_\mu,V_\lambda)$, $\operatorname{ad}x_s$ acts by $\lambda-\mu$ and $\operatorname{ad}y$ by $\overline\lambda-\overline\mu$. Choose a <polynomial> $q$ with $q(0)=0$ and $q(\lambda-\mu)=\overline{\lambda-\mu}$ at the finitely many distinct differences. <Polynomial interpolation> supplies it because equal differences have equal conjugates. Therefore $\operatorname{ad}y=q(\operatorname{ad}x_s)$.
The <adjoint compatibility of additive Jordan decomposition> identifies $\operatorname{ad}x_s$ as the semisimple part of $\operatorname{ad}x$. Elementary <Jordan–Chevalley decomposition> gives $\operatorname{ad}x_s=p(\operatorname{ad}x)$ for a <polynomial> $p$ with $p(0)=0$. Hence $\operatorname{ad}y$ is a <polynomial> in $\operatorname{ad}x$ with zero constant term. Since $\operatorname{ad}x$ maps $L$ into $[L,L]$ and preserves that <Lie algebra ideal>, we obtain
$$
[y,L]\subseteq[L,L].
$$
No assumption that $x_s$, $x_n$ or $y$ belongs to $L$ was made.
Write $x=\sum_i[a_i,b_i]$ with $a_i,b_i\in L$. Cyclicity and the assumed orthogonality now give
$$
\operatorname{tr}(xy)=\sum_i\operatorname{tr}([a_i,b_i]y)=\sum_i\operatorname{tr}(a_i[b_i,y])=0.
$$
On the other hand the nilpotent parts have zero <trace> on each <generalized eigenspace>, so
$$
\operatorname{tr}(xy)=\sum_\lambda (\dim V_\lambda)\lambda\overline\lambda=\sum_\lambda (\dim V_\lambda)|\lambda|^2.
$$
Thus all $\lambda$ are zero and $x$ is a <nilpotent endomorphism>. This is the <Conjugate-spectrum proof of Cartan solvability>.
The permitted <Engel theorem>, in the form needed here, states: a finite-dimensional <Lie subalgebra> of <endomorphisms> in which every element is nilpotent has a nonzero vector annihilated by all its elements, and iteration on quotients makes every element simultaneously strictly upper triangular. Apply it to $[L,L]$. The strictly upper triangular algebra is nilpotent, so $[L,L]$ is a <Nilpotent Lie algebra> and therefore solvable. Since $L/[L,L]$ is abelian, the <derived series of a Lie algebra> of $L$ terminates too. This proves the matrix criterion.
Apply it to $L=\operatorname{ad}\mathfrak g$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\operatorname{ad}\mathfrak g$ is solvable. The <kernel> of $\operatorname{ad}$ is the <center of a Lie algebra>, which is abelian, so $\mathfrak g$ is solvable as well: once the derived series maps to zero it is central, and its next term vanishes. Conversely a solvable $\mathfrak g$ has solvable adjoint image, proving the abstract <Cartan solvability criterion> in both directions.
A finite-dimensional <Lie algebra> is a <semisimple Lie algebra> when it has no nonzero solvable <Lie algebra ideal>, equivalently its <solvable radical> is zero. Let $R=\operatorname{rad}B=\{x:B(x,\mathfrak g)=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\operatorname{ad}x$ induces zero on $\mathfrak g/R$, and the block <trace> gives $B_R(x,y)=B_{\mathfrak g}(x,y)=0$, where $B_R$ is the adjoint <Killing form> of $R$ itself. The <Cartan solvability criterion> therefore makes $R$ solvable. If $\mathfrak g$ is semisimple, $R=0$.
Conversely suppose $B$ is a <nondegenerate bilinear form>. If a nonzero solvable ideal existed, its last nonzero derived term $A$ would be an abelian ideal of $\mathfrak g$. For $a\in A$ and $x\in\mathfrak g$, the operator $\operatorname{ad}a\,\operatorname{ad}x$ has image in $A$ and is zero on $A$, so its square and its <trace> are zero. Thus $B(A,\mathfrak g)=0$, contradicting nondegeneracy. This is <Abelian ideals lie in the radical of the Killing form>. We conclude the <Cartan criterion for semisimplicity>:
$$
\boxed{\mathfrak g\text{ is semisimple}\quad\Longleftrightarrow\quad B\text{ is nondegenerate}.}
$$