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For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.

A linear map $\alpha$ is a nilpotent endomorphism if ${\alpha }^q=0$ for some positive integer $q$, and a semisimple endomorphism if it is diagonalizable over $\mathbb{C}$. Decompose $V$ into its generalized eigenspaces $V_{\lambda }=\ker (\alpha -\lambda I{)}^d$, where $d=\dim V$. Define ${\alpha }_s$ on $V_{\lambda }$ to be $\lambda I$, and set ${\alpha }_n=\alpha -{\alpha }_s$. Then ${\alpha }_s$ is diagonalizable, ${\alpha }_n$ is nilpotent, and both preserve these vector subspaces and commute. ThusFor uniqueness, suppose $\alpha =S+N$ with $S$ semisimple, $N$ nilpotent and $SN=NS$. Both commute with $\alpha$, so preserve each $V_{\lambda }$. On an eigenspace of $S$ of eigenvalue $\mu$ inside $V_{\lambda }$, the map $\alpha$ is $\mu I+N$ and has only the eigenvalue $\mu$. Since the same subspace lies in $V_{\lambda }$, $\mu =\lambda$. Hence $S=\lambda I$ on $V_{\lambda }$, proving $S={\alpha }_s$ and $N={\alpha }_n$. This is the additive Jordan–Chevalley decomposition.

We need a polynomial consequence of this decomposition. Hermite interpolation supplies a polynomial $p$ with $p(\alpha )={\alpha }_s$, by prescribing $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^d)$ for each eigenvalue. For any endomorphism $T$, its semisimple part can likewise be expressed as a polynomial in $T$ with zero constant term: if zero is an eigenvalue, its interpolation condition already forces this; if not, add the independent condition $p(0)=0$.

On $\mathrm{End}(V)$, the maps $\mathrm{ad}{\alpha }_s$ and $\mathrm{ad}{\alpha }_n$ commute. The first is diagonalizable, with eigenvalues $\lambda -\mu$ on $\mathrm{Hom}(V_{\mu },V_{\lambda })$. The second is nilpotent, sincewhich vanishes for $k\ge 2q-1$ when $N^q=0$. Uniqueness therefore proves adjoint compatibility of additive Jordan decomposition: $\mathrm{ad}{\alpha }_s=(\mathrm{ad}\alpha {)}_s$.

Now assume $\alpha \in M$. The condition $[\alpha ,W]\subseteq U\subseteq W$ implies that both $W$ and $U$ are invariant under $\mathrm{ad}\alpha$, and every polynomial in $\mathrm{ad}\alpha$ with zero constant term maps $W$ into $U$. Define $\beta$ to be multiplication by $\overline{\lambda }$ on $V_{\lambda }$. It commutes with $\alpha$. On $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}\beta$ acts by $\overline{\lambda -\mu }$. Polynomial interpolation on the finite set of differences gives $\mathrm{ad}\beta =q(\mathrm{ad}{\alpha }_s)$ with $q(0)=0$. The preceding paragraph then expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Consequently $[\beta ,W]\subseteq U$, so $\beta \in M$.

The assumed trace orthogonality nilpotence lemma now follows directly. On $V_{\lambda }$, $\alpha =\lambda I+{\alpha }_n$ and $\beta =\overline{\lambda }I$, while the matrix trace of the nilpotent restriction of ${\alpha }_n$ is zero. HenceEvery summand is nonnegative, so every eigenvalue of $\alpha$ is zero. Its Jordan–Chevalley decomposition therefore has ${\alpha }_s=0$, and $\alpha$ is nilpotent. Notice that neither $U$ nor $W$ was required to be a Lie subalgebra.

Let $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ be a finite-dimensional Lie algebra representation. The form denoted $B_V$ isIt is the Trace form of a Lie algebra representation; the Killing form $B$ without a subscript is specifically the case $V=\mathfrak{g}$, $\rho =\mathrm{ad}$. 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 givesThus 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 $\mathfrak{g}$,Its matrix version says that a Lie subalgebra $L\subseteq \mathfrak{gl}(V)$ is solvable precisely when $\mathrm{tr}(xy)=0$ 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 $\mathrm{Hom}(V_{\mu },V_{\lambda })$, $\mathrm{ad}{x}_s$ acts by $\lambda -\mu$ and $\mathrm{ad}y$ 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 $\mathrm{ad}y=q(\mathrm{ad}{x}_s)$.

The adjoint compatibility of additive Jordan decomposition identifies $\mathrm{ad}{x}_s$ as the semisimple part of $\mathrm{ad}x$. Elementary Jordan–Chevalley decomposition gives $\mathrm{ad}{x}_s=p(\mathrm{ad}x)$ for a polynomial $p$ with $p(0)=0$. Hence $\mathrm{ad}y$ is a polynomial in $\mathrm{ad}x$ with zero constant term. Since $\mathrm{ad}x$ maps $L$ into $[L,L]$ and preserves that Lie algebra ideal, we obtainNo 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 giveOn the other hand the nilpotent parts have zero trace on each generalized eigenspace, soThus 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 $L=\mathrm{ad}\mathfrak{g}$. The condition on $B$ is exactly the matrix condition on $L$. Thus $\mathrm{ad}\mathfrak{g}$ is solvable. The kernel of $\mathrm{ad}$ 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 $R=\mathrm{rad}B=\{x:B(x,\mathfrak{g})=0\}$. Invariance makes $R$ an ideal. For $x,y\in R$, $\mathrm{ad}x$ 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 $\mathrm{ad}a\mathrm{ad}x$ 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:

Work over $\mathbb{C}$. Decompose $V$ as the direct sum of the generalized eigenspaces $V_{\lambda }$ of $X$. Define $X_s$ to be $\lambda I$ on $V_{\lambda }$ and set $X_n=X-X_s$. Then $X_s$ is a diagonalisable endomorphism, $X_n$ is a nilpotent endomorphism, and they commute. For uniqueness, any commuting decomposition $X=S+N$ has $S$ and $N$ commuting with $X$, hence preserving each $V_{\lambda }$. Decompose $V_{\lambda }$ further into eigenspaces of $S$. On a nonzero such space with eigenvalue $\sigma$, the operator $X=\sigma I+N$ has only the eigenvalue $\sigma$, so $\sigma =\lambda$. Since $S$ is diagonalizable, $S=\lambda I$ throughout $V_{\lambda }$. Thus $S=X_s$ and $N=X_n$. This is the Additive Jordan decomposition.

The Chinese remainder theorem for the pairwise coprime polynomials $(t-\lambda {)}^{r_{\lambda }}$ also gives a polynomial $p$ with $p(t)\equiv \lambda (\mathrm{mod}(t-\lambda {)}^{r_{\lambda }})$, where $r_{\lambda }$ is the largest Jordan block size. Consequently $X_s=p(X)$ and $X_n=X-p(X)$. This polynomial description shows that both parts preserve every $X$-invariant subspace.

On $\mathrm{Hom}(V_{\lambda },V_{\mu })$ the operator $\mathrm{ad}{X}_s$ is the scalar $\mu -\lambda$, so it is diagonalizable. The nilpotence of commutation by a nilpotent endomorphism makes $\mathrm{ad}{X}_n$ nilpotent. They commute, since $[\mathrm{ad}{X}_s,\mathrm{ad}{X}_n]=\mathrm{ad}[X_s,X_n]=0$. Uniqueness of the Additive Jordan decomposition therefore gives the adjoint compatibility of additive Jordan decomposition:Now let $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ be a complex semisimple Lie algebra and $X\in \mathfrak{g}$. Since the semisimple part of $\mathrm{ad}X$ is a polynomial in $\mathrm{ad}X$, it preserves $\mathfrak{g}$. Thus $[X_s,\mathfrak{g}]\subseteq \mathfrak{g}$. By the Weyl complete reducibility theorem, the Adjoint representation of $\mathfrak{g}$ on $\mathrm{End}(V)$ has a decomposition $\mathrm{End}(V)=\mathfrak{g}\oplus M$ into invariant subspaces. Write $X_s=y+z$ with $y\in \mathfrak{g}$ and $z\in M$. For $a\in \mathfrak{g}$, the vector $[z,a]=[X_s,a]-[y,a]$ lies in $\mathfrak{g}$, and invariance of $M$ puts it in $M$ as well. Hence $[z,a]=0$.

Decompose $V={⨁}_j{U}_j$ into Irreducible Lie algebra representations. Each $U_j$ is preserved by $X_s=p(X)$ and by $y$, hence by $z$. The Schur lemma makes $z{|}_{U_j}=c_{j}I$. A semisimple Lie algebra is a perfect Lie algebra, so every representing element of $\mathfrak{g}$ has zero trace on every $U_j$. Also $\mathrm{tr}(X_s{|}_{U_j})=\mathrm{tr}(X{|}_{U_j})$, because $X_n$ is nilpotent there. It follows that $c_j\dim U_j=\mathrm{tr}(z{|}_{U_j})=0$. In characteristic zero, $c_j=0$, so $z=0$. Therefore $X_s\in \mathfrak{g}$ and $X_n=X-X_s\in \mathfrak{g}$. This proves that semisimple matrix Lie algebras are closed under additive Jordan decomposition, including representations with repeated isomorphic irreducible summands.

Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.