[proofplan] We prove that every element of the derived algebra $[\mathfrak h,\mathfrak h]$ is nilpotent as an endomorphism of $V$. For a fixed $x \in [\mathfrak h,\mathfrak h]$, we use the Jordan decomposition of $x$ and construct, from an arbitrary rational linear functional on the eigenvalue span, a semisimple endomorphism $b$ whose commutator action on $\mathfrak h$ lands in $[\mathfrak h,\mathfrak h]$. The trace hypothesis then forces a sum of rational squares to vanish, so all eigenvalues of $x$ are zero. [Engel's theorem](/theorems/3798) then makes $[\mathfrak h,\mathfrak h]$ nilpotent, and a Lie algebra with nilpotent derived algebra is solvable. [/proofplan] [step:Fix an element of the derived algebra and decompose its eigenvalue data] Let $x \in [\mathfrak h,\mathfrak h]$. Since $F$ is algebraically closed and $V$ is finite-dimensional, the endomorphism \begin{align*} x: V &\to V \end{align*} has a finite set of distinct eigenvalues $\lambda_1,\dots,\lambda_m \in F$. Let \begin{align*} V_i := \ker\left((x-\lambda_i I)^{\dim_F V}\right) \end{align*} be the generalized eigenspace of $x$ for $\lambda_i$. Then \begin{align*} V = \bigoplus_{i=1}^m V_i. \end{align*} Let $d_i := \dim_F V_i$ for $1 \leq i \leq m$, and let $E \subset F$ be the finite-dimensional $\mathbb Q$-vector subspace spanned by $\lambda_1,\dots,\lambda_m$. If $E=\{0\}$, then all eigenvalues of $x$ are already zero. It remains to rule out the case $E \neq \{0\}$. [/step] [step:Build a semisimple endomorphism from a rational functional on the eigenvalues] Let \begin{align*} f: E &\to \mathbb Q \end{align*} be an arbitrary $\mathbb Q$-[linear map](/page/Linear%20Map). Define the endomorphism \begin{align*} b: V &\to V \end{align*} by requiring \begin{align*} b|_{V_i} = f(\lambda_i) I_{V_i} \end{align*} for each $1 \leq i \leq m$. Thus $b$ is semisimple and preserves each generalized eigenspace $V_i$. We claim that \begin{align*} [b,\mathfrak h] \subseteq [\mathfrak h,\mathfrak h]. \end{align*} [guided] The purpose of $b$ is to encode the rational functional $f$ into an endomorphism of $V$. On the generalized eigenspace for $\lambda_i$, the map $b$ acts by the scalar $f(\lambda_i)$. The key point is not that $b$ lies in $\mathfrak h$; it need not. What we need is stronger in the useful direction: commutators with $b$ should send elements of $\mathfrak h$ into the derived algebra. Let \begin{align*} \operatorname{ad}_x: \mathfrak{gl}(V) &\to \mathfrak{gl}(V), \\ z &\mapsto [x,z] \end{align*} and \begin{align*} \operatorname{ad}_b: \mathfrak{gl}(V) &\to \mathfrak{gl}(V), \\ z &\mapsto [b,z]. \end{align*} Since $x \in \mathfrak h$ and $\mathfrak h$ is a Lie subalgebra, $\operatorname{ad}_x(\mathfrak h) \subseteq \mathfrak h$. The generalized eigenspace decomposition of $V$ induces a decomposition of $\mathfrak{gl}(V)$ into spaces of maps $V_j \to V_i$. On such a block, the semisimple part of $\operatorname{ad}_x$ has eigenvalue $\lambda_i-\lambda_j$, while $\operatorname{ad}_b$ acts by the scalar \begin{align*} f(\lambda_i)-f(\lambda_j)=f(\lambda_i-\lambda_j). \end{align*} Because there are only finitely many differences $\lambda_i-\lambda_j$, Hermite interpolation over $F$ gives a polynomial $p \in F[t]$ such that $p(0)=0$ and such that $p(\operatorname{ad}_x)=\operatorname{ad}_b$ on $\mathfrak{gl}(V)$. The derivative conditions in the interpolation kill the nilpotent parts of the Jordan blocks of $\operatorname{ad}_x$. Since $p(0)=0$, there exists a polynomial $r \in F[t]$ such that \begin{align*} p(t)=t r(t). \end{align*} For every $z \in \mathfrak h$, \begin{align*} [b,z] = p(\operatorname{ad}_x)(z) = \operatorname{ad}_x\bigl(r(\operatorname{ad}_x)(z)\bigr). \end{align*} Because $\operatorname{ad}_x$ preserves $\mathfrak h$, the element $r(\operatorname{ad}_x)(z)$ belongs to $\mathfrak h$. Therefore \begin{align*} [b,z] \in [\mathfrak h,\mathfrak h]. \end{align*} This proves $[b,\mathfrak h]\subseteq[\mathfrak h,\mathfrak h]$. [/guided] Indeed, let \begin{align*} \operatorname{ad}_x: \mathfrak{gl}(V) &\to \mathfrak{gl}(V),& z &\mapsto [x,z], \end{align*} and define $\operatorname{ad}_b$ analogously. On the block $\operatorname{Hom}_F(V_j,V_i)$, the semisimple part of $\operatorname{ad}_x$ has eigenvalue $\lambda_i-\lambda_j$, while $\operatorname{ad}_b$ acts by $f(\lambda_i-\lambda_j)$. By Hermite interpolation, choose $p \in F[t]$ with $p(0)=0$ and with enough vanishing derivative conditions at the finitely many numbers $\lambda_i-\lambda_j$ so that \begin{align*} p(\operatorname{ad}_x)=\operatorname{ad}_b \end{align*} on $\mathfrak{gl}(V)$. Since $p(0)=0$, write $p(t)=t r(t)$ with $r \in F[t]$. For $z \in \mathfrak h$, \begin{align*} [b,z] = p(\operatorname{ad}_x)(z) = [x,r(\operatorname{ad}_x)(z)]. \end{align*} Because $x \in \mathfrak h$ and $\operatorname{ad}_x(\mathfrak h)\subseteq\mathfrak h$, we have $r(\operatorname{ad}_x)(z)\in\mathfrak h$, hence $[b,z]\in[\mathfrak h,\mathfrak h]$. [/step] [step:Use the trace hypothesis to force the weighted eigenvalue sum to vanish] Since $x \in [\mathfrak h,\mathfrak h]$, there exist elements $a_1,\dots,a_N,c_1,\dots,c_N \in \mathfrak h$ such that \begin{align*} x = \sum_{k=1}^N [a_k,c_k]. \end{align*} Using cyclicity of trace, for each $k$ we have \begin{align*} \operatorname{tr}([a_k,c_k]b) &= \operatorname{tr}([b,a_k]c_k). \end{align*} By the previous step, $[b,a_k]\in[\mathfrak h,\mathfrak h]$, and $c_k\in\mathfrak h$. The hypothesis therefore gives \begin{align*} \operatorname{tr}([b,a_k]c_k)=0. \end{align*} Summing over $k$ yields \begin{align*} \operatorname{tr}(xb)=0. \end{align*} On $V_i$, the nilpotent part of $x-\lambda_i I$ has trace $0$, while $b$ acts as the scalar $f(\lambda_i)$. Therefore \begin{align*} 0=\operatorname{tr}(xb) = \sum_{i=1}^m d_i \lambda_i f(\lambda_i). \end{align*} [/step] [step:Apply rational linear functionals to show every eigenvalue is zero] The element \begin{align*} \sum_{i=1}^m d_i \lambda_i f(\lambda_i) \end{align*} belongs to $E$, so applying the $\mathbb Q$-linear map $f:E\to\mathbb Q$ gives \begin{align*} 0 = f\left(\sum_{i=1}^m d_i \lambda_i f(\lambda_i)\right) = \sum_{i=1}^m d_i f(\lambda_i)^2. \end{align*} Here $d_i$ is a positive integer and $f(\lambda_i)\in\mathbb Q$. Since the embedding $\mathbb Q\hookrightarrow F$ is injective in characteristic $0$, this equality is an equality in $\mathbb Q$. Hence each $f(\lambda_i)=0$. Because $f:E\to\mathbb Q$ was arbitrary, no nonzero $\lambda_i$ can occur: if some $\lambda_i\neq 0$, extend $\lambda_i$ to a $\mathbb Q$-basis of $E$ and choose a $\mathbb Q$-linear functional sending $\lambda_i$ to $1$. Thus \begin{align*} \lambda_1=\cdots=\lambda_m=0. \end{align*} So every eigenvalue of $x$ is zero, and therefore $x$ is nilpotent as an endomorphism of $V$. [/step] [step:Conclude solvability from nilpotence of the derived algebra] We have shown that every element of $[\mathfrak h,\mathfrak h]$ is a nilpotent endomorphism of $V$. By [Engel's theorem](/theorems/3753) (citing a result not yet in the wiki: Engel's Theorem), the finite-dimensional linear Lie algebra $[\mathfrak h,\mathfrak h] \leq \mathfrak{gl}(V)$ is nilpotent. A nilpotent Lie algebra is solvable. Let $\mathfrak h^{(0)}:=\mathfrak h$ and define the derived series by \begin{align*} \mathfrak h^{(r+1)} := [\mathfrak h^{(r)},\mathfrak h^{(r)}]. \end{align*} Since $\mathfrak h^{(1)}=[\mathfrak h,\mathfrak h]$ is solvable, there exists $r\in\mathbb N$ such that \begin{align*} (\mathfrak h^{(1)})^{(r)}=\{0\}. \end{align*} But for every $r\geq 0$, \begin{align*} \mathfrak h^{(r+1)} = (\mathfrak h^{(1)})^{(r)}. \end{align*} Thus $\mathfrak h^{(r+1)}=\{0\}$ for some $r$, so $\mathfrak h$ is solvable. [/step]