[proofplan] We reduce the assertion to a finite-dimensional solvable Lie subalgebra of $\mathfrak{gl}(V)$, namely the image of the representation. The proof then proceeds by induction on the dimension of this image. The key structural input is that every nonzero finite-dimensional solvable Lie algebra has an ideal of codimension one; after applying the induction hypothesis to that ideal, we prove that its common eigenspace is stable under the remaining generator. Algebraic closedness gives an eigenvector for that last generator, and characteristic $0$ is used in the trace argument forcing the relevant commutator weights to vanish. [/proofplan] [step:Replace the representation by its finite-dimensional image] Define \begin{align*} \mathfrak h := \rho(\mathfrak g) \subseteq \mathfrak{gl}(V). \end{align*} Since $V$ is finite-dimensional over $F$, the associative algebra $\operatorname{End}_F(V)$ is finite-dimensional, and hence the Lie subalgebra $\mathfrak h$ is finite-dimensional. Since $\rho$ is a Lie algebra homomorphism, the derived series of $\mathfrak h$ is the image under $\rho$ of a quotient of the derived series of $\mathfrak g$; therefore $\mathfrak h$ is solvable. It is enough to prove that there exist a nonzero vector $v \in V$ and a linear functional $\mu: \mathfrak h \to F$ such that \begin{align*} T v = \mu(T)v \end{align*} for every $T \in \mathfrak h$. Indeed, then defining \begin{align*} \lambda: \mathfrak g &\to F \\ x &\mapsto \mu(\rho(x)) \end{align*} gives a linear functional on $\mathfrak g$, and \begin{align*} \rho(x)v = \lambda(x)v \end{align*} for every $x \in \mathfrak g$. Thus we prove the following finite-dimensional statement: if $\mathfrak h \subseteq \mathfrak{gl}(V)$ is a finite-dimensional solvable Lie algebra over $F$ and $V \neq \{0\}$, then $V$ has a common eigenvector for all elements of $\mathfrak h$. [guided] The representation may come from a Lie algebra $\mathfrak g$ that is not explicitly assumed finite-dimensional, but its action on $V$ always factors through the finite-dimensional Lie algebra \begin{align*} \mathfrak h := \rho(\mathfrak g) \subseteq \mathfrak{gl}(V). \end{align*} This is finite-dimensional because $\mathfrak{gl}(V)=\operatorname{End}_F(V)$ is finite-dimensional when $V$ is finite-dimensional. We also need solvability to pass from $\mathfrak g$ to $\mathfrak h$. Since $\rho$ is a Lie algebra homomorphism, commutators are preserved: \begin{align*} \rho([x,y]) = [\rho(x),\rho(y)] \end{align*} for all $x,y \in \mathfrak g$. Therefore each derived algebra of $\mathfrak h$ is contained in the image of the corresponding derived algebra of $\mathfrak g$. Since the derived series of $\mathfrak g$ eventually reaches $\{0\}$, the derived series of $\mathfrak h$ also eventually reaches $\{0\}$. Now suppose we prove the theorem for finite-dimensional solvable Lie subalgebras $\mathfrak h \subseteq \mathfrak{gl}(V)$. Then we obtain a nonzero vector $v \in V$ and a linear functional $\mu: \mathfrak h \to F$ such that $T v=\mu(T)v$ for all $T \in \mathfrak h$. Pulling $\mu$ back along $\rho$ gives \begin{align*} \lambda: \mathfrak g &\to F \\ x &\mapsto \mu(\rho(x)). \end{align*} This map is linear because both $\rho$ and $\mu$ are linear, and it satisfies \begin{align*} \rho(x)v = \mu(\rho(x))v = \lambda(x)v \end{align*} for every $x \in \mathfrak g$. Hence the original statement follows from the finite-dimensional image case. [/guided] [/step] [step:Choose a codimension-one ideal in a nonzero solvable Lie algebra] We claim that if $\mathfrak h$ is a nonzero finite-dimensional solvable Lie algebra over $F$, then $\mathfrak h$ contains an ideal $\mathfrak k \trianglelefteq \mathfrak h$ with \begin{align*} \dim_F(\mathfrak h/\mathfrak k)=1. \end{align*} Since $\mathfrak h$ is solvable and nonzero, its derived algebra $[\mathfrak h,\mathfrak h]$ is a proper subspace of $\mathfrak h$; otherwise the derived series would be constant and nonzero. Thus the abelianization \begin{align*} \mathfrak a := \mathfrak h/[\mathfrak h,\mathfrak h] \end{align*} is a nonzero finite-dimensional [vector space](/page/Vector%20Space) over $F$. Choose a hyperplane $\mathfrak b \subseteq \mathfrak a$, meaning a subspace with $\dim_F(\mathfrak a/\mathfrak b)=1$. Let \begin{align*} \pi: \mathfrak h &\to \mathfrak a \end{align*} be the quotient map, and define \begin{align*} \mathfrak k := \pi^{-1}(\mathfrak b). \end{align*} Because $\mathfrak a$ is abelian, every subspace of $\mathfrak a$ is an ideal of $\mathfrak a$. Hence $\mathfrak k$ is an ideal of $\mathfrak h$, and \begin{align*} \mathfrak h/\mathfrak k \cong \mathfrak a/\mathfrak b \end{align*} has dimension $1$ over $F$. [guided] We need a way to reduce the dimension of $\mathfrak h$ while preserving enough Lie algebra structure to apply induction. The right object is a codimension-one ideal. Since $\mathfrak h$ is solvable, its derived series eventually reaches $\{0\}$. If $[\mathfrak h,\mathfrak h]=\mathfrak h$, then every subsequent derived algebra would again equal $\mathfrak h$, contradicting solvability because $\mathfrak h \neq \{0\}$. Therefore \begin{align*} [\mathfrak h,\mathfrak h] \subsetneq \mathfrak h. \end{align*} The quotient \begin{align*} \mathfrak a := \mathfrak h/[\mathfrak h,\mathfrak h] \end{align*} is therefore a nonzero finite-dimensional vector space over $F$. It is also an abelian Lie algebra, because all commutators have been quotiented out. Choose a hyperplane $\mathfrak b \subseteq \mathfrak a$, so \begin{align*} \dim_F(\mathfrak a/\mathfrak b)=1. \end{align*} Let \begin{align*} \pi: \mathfrak h &\to \mathfrak a \end{align*} be the quotient map, and define \begin{align*} \mathfrak k := \pi^{-1}(\mathfrak b). \end{align*} Because $\mathfrak a$ is abelian, every subspace of $\mathfrak a$ is an ideal: the bracket of any element of $\mathfrak a$ with any element of $\mathfrak b$ is $0$. The inverse image of an ideal under a Lie algebra homomorphism is an ideal, so $\mathfrak k \trianglelefteq \mathfrak h$. Finally, the quotient map induces an isomorphism \begin{align*} \mathfrak h/\mathfrak k \cong \mathfrak a/\mathfrak b, \end{align*} and the latter has dimension $1$. Thus $\mathfrak k$ is an ideal of codimension one in $\mathfrak h$. [/guided] [/step] [step:Apply induction to the codimension-one ideal] We prove the finite-dimensional image statement by induction on $d:=\dim_F \mathfrak h$. If $d=0$, then $\mathfrak h=\{0\}$, and every nonzero vector $v \in V$ is a common eigenvector, with $\mu(0)=0$. Assume $d\geq 1$ and that the result is known for all solvable Lie subalgebras of $\mathfrak{gl}(V)$ of dimension strictly smaller than $d$. Choose an ideal $\mathfrak k \trianglelefteq \mathfrak h$ with $\dim_F(\mathfrak h/\mathfrak k)=1$. Since $\mathfrak k$ is a subalgebra of the solvable Lie algebra $\mathfrak h$, it is solvable. By the induction hypothesis, there exist a nonzero vector $v_0 \in V$ and a linear functional $\mu: \mathfrak k \to F$ such that \begin{align*} A v_0 = \mu(A)v_0 \end{align*} for every $A \in \mathfrak k$. Define the common $\mu$-eigenspace of $\mathfrak k$ by \begin{align*} W := \{w \in V : A w = \mu(A)w \text{ for every } A \in \mathfrak k\}. \end{align*} Then $W$ is a nonzero linear subspace of $V$ because $v_0 \in W$. [guided] We now begin the induction. The base case is the zero Lie algebra: if $\mathfrak h=\{0\}$, then there is no nonzero operator to constrain the vector, so any nonzero $v \in V$ works. For the induction step, suppose $\dim_F \mathfrak h=d\geq 1$. From the previous step, choose an ideal $\mathfrak k \trianglelefteq \mathfrak h$ such that \begin{align*} \dim_F(\mathfrak h/\mathfrak k)=1. \end{align*} This implies $\dim_F \mathfrak k=d-1$, so the induction hypothesis applies to $\mathfrak k$. The subalgebra $\mathfrak k$ is solvable because its derived series is contained in the derived series of $\mathfrak h$. The induction hypothesis gives a nonzero vector $v_0 \in V$ and a linear functional $\mu: \mathfrak k \to F$ satisfying \begin{align*} A v_0 = \mu(A)v_0 \end{align*} for every $A \in \mathfrak k$. We collect all vectors with this same system of eigenvalues by defining \begin{align*} W := \{w \in V : A w = \mu(A)w \text{ for every } A \in \mathfrak k\}. \end{align*} This is a linear subspace: if $w_1,w_2 \in W$ and $c_1,c_2 \in F$, then for every $A \in \mathfrak k$, \begin{align*} A(c_1w_1+c_2w_2) &= c_1Aw_1+c_2Aw_2 \\ &= c_1\mu(A)w_1+c_2\mu(A)w_2 \\ &= \mu(A)(c_1w_1+c_2w_2). \end{align*} It is nonzero because $v_0 \in W$. [/guided] [/step] [step:Show the common eigenspace of the ideal is stable under the remaining generator] Choose $X \in \mathfrak h$ whose image spans the one-dimensional quotient $\mathfrak h/\mathfrak k$. Then every element of $\mathfrak h$ has the form $A+cX$ with $A \in \mathfrak k$ and $c \in F$. We first prove that \begin{align*} \mu([X,A])=0 \end{align*} for every $A \in \mathfrak k$. Fix $A \in \mathfrak k$. Since $\mathfrak k$ is an ideal, $[X,A]\in \mathfrak k$. Let $m \in \mathbb N$ be the smallest positive integer for which the vectors \begin{align*} v_0, Xv_0, \dots, X^m v_0 \end{align*} are linearly dependent. Define \begin{align*} U := \operatorname{span}_F\{v_0, Xv_0,\dots, X^{m-1}v_0\}. \end{align*} Then $U$ is nonzero and finite-dimensional, and by minimality of $m$ the listed spanning vectors form a basis of $U$. The minimal dependence relation expresses $X^m v_0$ as an element of $U$, so $X(U)\subseteq U$. For each $B \in \mathfrak k$, we prove by induction on $j$ that \begin{align*} B X^j v_0-\mu(B)X^jv_0 \in \operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\} \end{align*} for $0\leq j\leq m-1$, where the span is interpreted as $\{0\}$ for $j=0$. The case $j=0$ is exactly $Bv_0=\mu(B)v_0$. If the assertion holds for $j-1$, then using the commutator identity $BX=X B+[B,X]$ gives \begin{align*} B X^jv_0 &= X B X^{j-1}v_0 + [B,X]X^{j-1}v_0. \end{align*} The first term equals $\mu(B)X^jv_0$ modulo $\operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\}$ by the induction hypothesis, and the second term lies in $\operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\}$ because $[B,X]\in \mathfrak k$ and the induction assertion applies to $[B,X]$. Thus the induction is complete. Consequently $U$ is invariant under every $B \in \mathfrak k$, and the matrix of $B|_U$ in the ordered basis \begin{align*} (v_0,Xv_0,\dots,X^{m-1}v_0) \end{align*} is upper triangular with diagonal entries all equal to $\mu(B)$. Applying this to $B=[X,A]$, we obtain \begin{align*} \operatorname{tr}(([X,A])|_U)=m\,\mu([X,A]). \end{align*} On the other hand, since $U$ is invariant under both $X$ and $A$, the restriction of the commutator is the commutator of the restrictions: \begin{align*} ([X,A])|_U = X|_U A|_U - A|_U X|_U. \end{align*} The trace of a commutator of finite-dimensional endomorphisms is $0$, so \begin{align*} m\,\mu([X,A])=0. \end{align*} Since $m\geq 1$ and $F$ has characteristic $0$, this implies $\mu([X,A])=0$. Now let $w\in W$ and $A\in\mathfrak k$. Since $[A,X]\in \mathfrak k$ and $\mu([A,X])=-\mu([X,A])=0$, we have \begin{align*} A(Xw) &= X(Aw)+[A,X]w \\ &= X(\mu(A)w)+\mu([A,X])w \\ &= \mu(A)Xw. \end{align*} Thus $Xw\in W$, so $X(W)\subseteq W$. [guided] The quotient $\mathfrak h/\mathfrak k$ is one-dimensional, so choose an element $X \in \mathfrak h$ whose coset spans it. Then every element of $\mathfrak h$ can be written as $A+cX$ with $A \in \mathfrak k$ and $c \in F$. Since $\mathfrak k$ is an ideal, commutators with $X$ stay inside $\mathfrak k$: \begin{align*} [X,A]\in \mathfrak k \end{align*} for every $A \in \mathfrak k$. The main point is to prove that the common eigenspace $W$ is preserved by $X$. If $w\in W$, then for $A\in\mathfrak k$, \begin{align*} A(Xw)=X(Aw)+[A,X]w. \end{align*} The first term is $\mu(A)Xw$. The second term would obstruct $Xw$ from lying in $W$, so we must prove \begin{align*} \mu([A,X])=0. \end{align*} Fix $A\in\mathfrak k$. We prove this vanishing by a trace argument on a finite-dimensional $X$-cyclic subspace. Let $m\in\mathbb N$ be the smallest positive integer for which \begin{align*} v_0, Xv_0, \dots, X^m v_0 \end{align*} are linearly dependent. Define \begin{align*} U:=\operatorname{span}_F\{v_0,Xv_0,\dots,X^{m-1}v_0\}. \end{align*} By minimality, the displayed spanning vectors form a basis of $U$, and the first dependence relation expresses $X^m v_0$ in $U$. Hence $X(U)\subseteq U$. Next we show that every $B\in\mathfrak k$ also preserves $U$, and more precisely has upper triangular matrix with diagonal entries $\mu(B)$ in the basis \begin{align*} (v_0,Xv_0,\dots,X^{m-1}v_0). \end{align*} For $j=0$, the assertion is \begin{align*} Bv_0=\mu(B)v_0, \end{align*} which holds because $v_0\in W$. Suppose the assertion is known for $j-1$. Using $BX=XB+[B,X]$, we compute \begin{align*} B X^jv_0 &= X B X^{j-1}v_0 + [B,X]X^{j-1}v_0. \end{align*} By the induction hypothesis applied to $B$, the vector $BX^{j-1}v_0-\mu(B)X^{j-1}v_0$ lies in $\operatorname{span}_F\{v_0,\dots,X^{j-2}v_0\}$; applying $X$ sends it into $\operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\}$. Since $[B,X]\in\mathfrak k$, the induction hypothesis applied to $[B,X]$ shows that $[B,X]X^{j-1}v_0$ also lies in $\operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\}$. Therefore \begin{align*} B X^jv_0-\mu(B)X^jv_0 \in \operatorname{span}_F\{v_0,\dots,X^{j-1}v_0\}. \end{align*} This proves the triangular form. Apply this conclusion to $B=[X,A]$. Since the diagonal entries of $([X,A])|_U$ are all $\mu([X,A])$, we have \begin{align*} \operatorname{tr}(([X,A])|_U)=m\,\mu([X,A]). \end{align*} But $U$ is invariant under both $X$ and $A$, so \begin{align*} ([X,A])|_U=X|_U A|_U-A|_U X|_U. \end{align*} The trace of a commutator of two endomorphisms of a finite-dimensional vector space is $0$, because $\operatorname{tr}(PQ)=\operatorname{tr}(QP)$. Hence \begin{align*} m\,\mu([X,A])=0. \end{align*} Since $m\geq 1$ and the field has characteristic $0$, multiplication by $m$ is injective in $F$, so \begin{align*} \mu([X,A])=0. \end{align*} Finally, let $w\in W$ and $A\in\mathfrak k$. Then $[A,X]=-[X,A]$, so $\mu([A,X])=0$. Therefore \begin{align*} A(Xw) &= X(Aw)+[A,X]w \\ &= X(\mu(A)w)+\mu([A,X])w \\ &= \mu(A)Xw. \end{align*} This holds for every $A\in\mathfrak k$, so $Xw\in W$. Thus $X(W)\subseteq W$. [/guided] [/step] [step:Choose an eigenvector for the remaining generator and finish] The subspace $W$ is nonzero and finite-dimensional. Since $X(W)\subseteq W$, the restriction \begin{align*} X|_W: W &\to W \end{align*} is a linear endomorphism. Because $F$ is algebraically closed, $X|_W$ has an eigenvalue $\alpha\in F$ and a nonzero eigenvector $v\in W$ such that \begin{align*} Xv=\alpha v. \end{align*} For every $A\in\mathfrak k$, since $v\in W$, \begin{align*} Av=\mu(A)v. \end{align*} Define \begin{align*} \nu: \mathfrak h &\to F \end{align*} by writing each $T\in\mathfrak h$ uniquely as $T=A+cX$ with $A\in\mathfrak k$ and $c\in F$, and setting \begin{align*} \nu(T):=\mu(A)+c\alpha. \end{align*} This is a well-defined linear functional because $\mathfrak h=\mathfrak k\oplus FX$ as vector spaces. For $T=A+cX$, \begin{align*} Tv &= Av+cXv \\ &= \mu(A)v+c\alpha v \\ &= \nu(T)v. \end{align*} Thus $v$ is a common eigenvector for $\mathfrak h$. Pulling $\nu$ back along $\rho$ as in the first step gives the desired linear functional on $\mathfrak g$, completing the proof of Lie's theorem. [guided] We have reduced the problem to the nonzero subspace $W$, which is stable under the remaining operator $X$. Thus the restriction \begin{align*} X|_W: W &\to W \end{align*} is a well-defined linear endomorphism of a nonzero finite-dimensional vector space. Here algebraic closedness is used. The characteristic polynomial of $X|_W$ has a root $\alpha\in F$, so $X|_W$ has a nonzero eigenvector $v\in W$ satisfying \begin{align*} Xv=\alpha v. \end{align*} Because $v\in W$, it is already a common eigenvector for every element of $\mathfrak k$: \begin{align*} Av=\mu(A)v \end{align*} for all $A\in\mathfrak k$. Now every element $T\in\mathfrak h$ has a unique decomposition \begin{align*} T=A+cX \end{align*} with $A\in\mathfrak k$ and $c\in F$, because $\mathfrak h/\mathfrak k$ is one-dimensional and $X\notin\mathfrak k$. Define \begin{align*} \nu: \mathfrak h &\to F \end{align*} by \begin{align*} \nu(A+cX):=\mu(A)+c\alpha. \end{align*} This is linear by construction from the direct-sum decomposition $\mathfrak h=\mathfrak k\oplus FX$. For $T=A+cX$, we compute \begin{align*} Tv &=(A+cX)v \\ &=Av+cXv \\ &=\mu(A)v+c\alpha v \\ &=\nu(T)v. \end{align*} Therefore $v$ is a common eigenvector for all elements of $\mathfrak h$. Returning to the original representation, define \begin{align*} \lambda: \mathfrak g &\to F \\ x &\mapsto \nu(\rho(x)). \end{align*} Then $\lambda$ is linear and \begin{align*} \rho(x)v=\lambda(x)v \end{align*} for every $x\in\mathfrak g$. This is exactly the claimed common eigenvector for the representation of $\mathfrak g$ on $V$. [/guided] [/step]