[proofplan] We prove the theorem by induction on $\dim_F L$. For the induction step, solvability gives a codimension-one ideal $I \trianglelefteq L$ containing $[L,L]$, so the induction hypothesis gives a common eigenvector for the action of $I$. We then enlarge this vector to the finite cyclic space generated by a complementary element $x \in L$, prove that this space is stable under $I$ and that every element of $I$ acts upper triangularly with one repeated diagonal scalar. Algebraic closedness supplies an eigenvector for $x$ inside this finite cyclic space, and the commutator condition forces that eigenvector to remain a common eigenvector for $I$. [/proofplan]

[step:Reduce the assertion to the existence of a common eigenvector]Let $\rho: L \to \operatorname{End}_F(V)$ denote the representation map defining the left $L$-module structure, so $xv$ means $\rho(x)(v)$ for $x \in L$ and $v \in V$. If there exist $0 \ne v \in V$ and $\lambda \in L^*$ such that $xv = \lambda(x)v$ for every $x \in L$, then the line $Fv := \{av : a \in F\}$ is stable under every $x \in L$, hence is a one-dimensional $L$-submodule. Conversely, if $Fv$ is a one-dimensional $L$-submodule with $0 \ne v \in V$, then for every $x \in L$ there is a unique scalar $\lambda(x) \in F$ such that $xv = \lambda(x)v$. Linearity of the action gives, for $a,b \in F$ and $x,y \in L$, \begin{align*} (ax + by)v &= a(xv) + b(yv) \\ &= a\lambda(x)v + b\lambda(y)v \\ &= \bigl(a\lambda(x) + b\lambda(y)\bigr)v. \end{align*} Since $v \ne 0$, uniqueness of scalar multiples of $v$ gives \begin{align*} \lambda(ax+by)=a\lambda(x)+b\lambda(y). \end{align*} Thus $\lambda \in L^*$.[/step]

[guided]The theorem states two equivalent conclusions, so we first check that they really match. Let $\rho: L \to \operatorname{End}_F(V)$ be the representation map for the given left $L$-module structure. Thus the notation $xv$ means $\rho(x)(v)$ for $x \in L$ and $v \in V$. Suppose first that there are $0 \ne v \in V$ and $\lambda \in L^*$ such that \begin{align*} xv=\lambda(x)v \end{align*} for every $x \in L$. Then every $x \in L$ sends every element $av \in Fv$ to \begin{align*} x(av)=a(xv)=a\lambda(x)v \in Fv, \end{align*} so the line $Fv:=\{av:a\in F\}$ is stable under the action of $L$. Since $v\ne 0$, this line is one-dimensional over $F$, hence it is a one-dimensional $L$-submodule. Conversely, suppose $Fv$ is a one-dimensional $L$-submodule with $0\ne v\in V$. Stability of $Fv$ means that for each $x\in L$ the vector $xv$ lies in $Fv$, so there is a scalar $\lambda(x)\in F$ satisfying $xv=\lambda(x)v$. This scalar is unique because $v\ne 0$. To prove that $\lambda$ is a member of $L^*$, we verify linearity. For $a,b\in F$ and $x,y\in L$, linearity of the representation gives \begin{align*} (ax+by)v &=a(xv)+b(yv) \\ &=a\lambda(x)v+b\lambda(y)v \\ &=\bigl(a\lambda(x)+b\lambda(y)\bigr)v. \end{align*} By uniqueness of the scalar multiplying the nonzero vector $v$, this implies \begin{align*} \lambda(ax+by)=a\lambda(x)+b\lambda(y). \end{align*} Therefore $\lambda:L\to F$ is linear, so $\lambda\in L^*$.[/guided]

[step:Choose a codimension-one ideal containing the derived algebra]We prove the existence of the eigenvector by induction on $n := \dim_F L$. If $n = 0$, then $L = 0$. Choose any $0 \ne v \in V$ and define the unique functional $\lambda: L \to F$ on the zero [vector space](/page/Vector%20Space). Then $xv = \lambda(x)v$ for every $x \in L$. Assume $n \ge 1$, and assume the theorem is known for all solvable Lie algebras over $F$ of dimension strictly smaller than $n$. Since $L$ is solvable, its derived algebra $[L,L]$ is a proper subspace of $L$. Indeed, if $[L,L]=L$, then the derived series $L^{(0)}:=L$ and $L^{(m+1)}:=[L^{(m)},L^{(m)}]$ would satisfy $L^{(m)}=L$ for every $m \ge 0$, contradicting solvability. Choose a codimension-one subspace $I \subset L$ such that \begin{align*} [L,L] \subset I \subset L. \end{align*} Then $I$ is an ideal of $L$: for every $x \in L$ and $y \in I$, the bracket $[x,y]$ belongs to $[L,L] \subset I$. Choose $x \in L \setminus I$. Then \begin{align*} L = I \oplus Fx \end{align*} as vector spaces.[/step]

[guided]The induction needs a smaller Lie algebra on which we can already find a common eigenvector. The natural candidate is a codimension-one ideal. We construct it from solvability. Define the derived series by $L^{(0)}:=L$ and $L^{(m+1)}:=[L^{(m)},L^{(m)}]$. Since $L$ is solvable, there is some $m \ge 0$ with $L^{(m)}=0$. If $[L,L]=L$, then $L^{(1)}=L$, and induction on $m$ would give $L^{(m)}=L$ for every $m$, contradicting $L^{(m)}=0$. Hence $[L,L]$ is a proper subspace of $L$. Because $L$ is finite-dimensional, we may choose a codimension-one subspace $I \subset L$ containing $[L,L]$. This subspace is automatically an ideal: if $a \in L$ and $b \in I$, then $[a,b] \in [L,L] \subset I$. Thus $I \trianglelefteq L$. Since $I$ has codimension one, we may choose $x \in L \setminus I$ and write every element of $L$ uniquely as $y + cx$ with $y \in I$ and $c \in F$; equivalently, $L = I \oplus Fx$ as vector spaces.[/guided]

[step:Find a common eigenvector for the ideal]The ideal $I$ is solvable because its derived series satisfies $I^{(m)} \subset L^{(m)}$ for every $m \ge 0$. Also $\dim_F I = n-1$. By the induction hypothesis applied to the finite-dimensional solvable Lie algebra $I$ and the nonzero finite-dimensional $I$-module $V$, there exist $0 \ne v_0 \in V$ and $\mu \in I^*$ such that \begin{align*} yv_0 = \mu(y)v_0 \end{align*} for every $y \in I$. Define the finite-dimensional subspace \begin{align*} W := \operatorname{span}_F\{x^k v_0 : k \in \mathbb{N}\cup\{0\}\} \subset V, \end{align*} where $x^0v_0:=v_0$ and $x^{k+1}v_0:=x(x^k v_0)$. Since $V$ is finite-dimensional, $W$ is finite-dimensional. By construction, $xW \subset W$.[/step]

[guided]We now use the induction hypothesis on the smaller Lie algebra $I$. First we verify its hypotheses. The ideal $I$ is finite-dimensional because it is a subspace of the finite-dimensional vector space $L$, and $\dim_F I=n-1<n$. It is solvable because its derived series is termwise contained in the derived series of $L$: the inclusion $I^{(m)}\subset L^{(m)}$ follows by induction from $I\subset L$ and the definition $A^{(m+1)}=[A^{(m)},A^{(m)}]$. Since $L$ is solvable, some $L^{(m)}$ is zero, and then $I^{(m)}=0$ as well. The vector space $V$ is nonzero and finite-dimensional by the theorem hypothesis, and restriction of the $L$-action to $I$ makes $V$ an $I$-module. Therefore the induction hypothesis applied to $I$ gives a vector $0\ne v_0\in V$ and a linear functional $\mu\in I^*$ such that \begin{align*} yv_0=\mu(y)v_0 \end{align*} for every $y\in I$. The next move is to close this vector under the complementary element $x$. Define \begin{align*} W:=\operatorname{span}_F\{x^k v_0:k\in\mathbb N\cup\{0\}\}\subset V, \end{align*} where $x^0v_0:=v_0$ and $x^{k+1}v_0:=x(x^k v_0)$. This is a subspace of the finite-dimensional vector space $V$, so $W$ is finite-dimensional. Its definition also gives $xW\subset W$, because applying $x$ to a finite linear combination of the vectors $x^k v_0$ gives a finite linear combination of the vectors $x^{k+1}v_0$.[/guided]

[step:Show the cyclic space is stable under the ideal]We claim that $IW \subset W$. More precisely, for every $y \in I$ and every $k \ge 0$, \begin{align*} yx^k v_0 \in \operatorname{span}_F\{v_0,xv_0,\dots,x^k v_0\}. \end{align*} We prove this by induction on $k$. For $k=0$, \begin{align*} yv_0=\mu(y)v_0 \in Fv_0. \end{align*} Assume the assertion holds for $k-1$, with $k \ge 1$. Since $V$ is an $L$-module, the module identity gives \begin{align*} yx^k v_0 &= yx(x^{k-1}v_0) \\ &= xyx^{k-1}v_0 + [y,x]x^{k-1}v_0. \end{align*} Because $I$ is an ideal, $[y,x] \in I$. By the induction hypothesis applied to $y$ and to $[y,x]$, both $yx^{k-1}v_0$ and $[y,x]x^{k-1}v_0$ belong to $\operatorname{span}_F\{v_0,xv_0,\dots,x^{k-1}v_0\}$. Applying $x$ to the first of these vectors places it in $\operatorname{span}_F\{xv_0,x^2v_0,\dots,x^k v_0\}$. Therefore $yx^k v_0$ belongs to $\operatorname{span}_F\{v_0,xv_0,\dots,x^k v_0\}$. Thus $W$ is stable under every $y \in I$, and hence $W$ is an $L$-submodule because $L=I\oplus Fx$ and $xW\subset W$.[/step]

[guided]The point of introducing $W$ is that it is automatically stable under $x$, but we must still prove it is stable under $I$. Fix $y \in I$. We prove by induction on $k$ that applying $y$ to $x^k v_0$ never produces anything outside the span of the first $k+1$ vectors \begin{align*} v_0,\ xv_0,\ \dots,\ x^k v_0. \end{align*} For $k=0$, the assertion is exactly the eigenvector relation for the ideal: \begin{align*} yv_0=\mu(y)v_0. \end{align*} Now assume $k \ge 1$ and that the assertion has already been proved for $k-1$. The module identity says that the action of the Lie bracket is the commutator of the two actions: \begin{align*} [y,x]w = y(xw)-x(yw) \end{align*} for every $w \in V$. Applying this to $w=x^{k-1}v_0$ gives \begin{align*} yx^k v_0 &= yx(x^{k-1}v_0) \\ &= xyx^{k-1}v_0 + [y,x]x^{k-1}v_0. \end{align*} The first term is controlled by the induction hypothesis for $y$: since $yx^{k-1}v_0$ lies in the span of $v_0,\dots,x^{k-1}v_0$, applying $x$ sends it into the span of $xv_0,\dots,x^k v_0$. The second term is controlled because $I$ is an ideal, so $[y,x]\in I$, and the induction hypothesis applies again with $[y,x]$ in place of $y$. Hence both terms lie in $\operatorname{span}_F\{v_0,xv_0,\dots,x^k v_0\}$. This proves $IW \subset W$. Since $xW \subset W$ by the definition of $W$ and every element of $L$ has the form $y+cx$ with $y \in I$ and $c \in F$, the whole space $W$ is an $L$-submodule.[/guided]

[step:Triangularize the action of the ideal on the cyclic space]Choose a basis of $W$ of the form \begin{align*} e_0,\dots,e_m \end{align*} obtained by deleting dependent vectors from the ordered list \begin{align*} v_0,\ xv_0,\ x^2v_0,\dots \end{align*} without changing the order. The previous step implies that for every $y \in I$ and every basis vector $e_j$, the vector $ye_j$ lies in $\operatorname{span}_F\{e_0,\dots,e_j\}$. Thus the matrix of $y|_W$ in this basis is upper triangular. Moreover, the diagonal entry of $y|_W$ in position $j$ is $\mu(y)$. We prove the stronger congruence: for every $y\in I$ and every $k\ge 0$, \begin{align*} yx^k v_0 \equiv \mu(y)x^k v_0 \pmod{\operatorname{span}_F\{v_0,xv_0,\dots,x^{k-1}v_0\}}, \end{align*} with the convention that the span is $\{0\}$ when $k=0$. For $k=0$, this is exactly $yv_0=\mu(y)v_0$. Assume it holds at level $k-1$, where $k\ge 1$. The module identity gives \begin{align*} yx^k v_0 &=xyx^{k-1}v_0+[y,x]x^{k-1}v_0. \end{align*} By the induction hypothesis applied to $y$, \begin{align*} yx^{k-1}v_0=\mu(y)x^{k-1}v_0+r \end{align*} for some $r\in\operatorname{span}_F\{v_0,xv_0,\dots,x^{k-2}v_0\}$. Hence \begin{align*} xyx^{k-1}v_0=\mu(y)x^k v_0+xr, \end{align*} where $xr\in\operatorname{span}_F\{xv_0,x^2v_0,\dots,x^{k-1}v_0\}$. Since $[y,x]\in I$, the containment proved in the previous step gives \begin{align*} [y,x]x^{k-1}v_0\in\operatorname{span}_F\{v_0,xv_0,\dots,x^{k-1}v_0\}. \end{align*} Combining the two terms proves the congruence at level $k$. Taking $k=k_j$ for $e_j=x^{k_j}v_0$ shows that each $y|_W-\mu(y)\operatorname{id}_W$ is strictly upper triangular in this ordered basis. In particular, for every $y \in I$, the operator \begin{align*} N_y: W \to W, \qquad w \mapsto yw-\mu(y)w \end{align*} is nilpotent.[/step]

[guided]We now record exactly how elements of $I$ act on $W$. The previous step did more than prove stability: it proved that $y$ sends the $k$-th cyclic vector $x^k v_0$ into the span of the cyclic vectors up to level $k$. This is the triangularity mechanism. Choose a basis $e_0,\dots,e_m$ of $W$ by passing through the ordered list \begin{align*} v_0,\ xv_0,\ x^2v_0,\dots \end{align*} and keeping a vector precisely when it is linearly independent of the ones already kept. This preserves the filtration by initial spans. Hence, for every $y \in I$, the matrix of $y|_W$ in this basis is upper triangular. We also need the diagonal entries, not only triangularity. We prove the needed congruence by induction on $k$: \begin{align*} yx^k v_0 \equiv \mu(y)x^k v_0 \pmod{\operatorname{span}_F\{v_0,xv_0,\dots,x^{k-1}v_0\}}, \end{align*} where the lower span is interpreted as $\{0\}$ when $k=0$. The base case is the eigenvector relation $yv_0=\mu(y)v_0$. For the inductive step, assume the congruence is known at level $k-1$. The Lie module identity gives \begin{align*} yx^k v_0 &=xyx^{k-1}v_0+[y,x]x^{k-1}v_0. \end{align*} The induction hypothesis writes $yx^{k-1}v_0=\mu(y)x^{k-1}v_0+r$ with $r$ in the span of the cyclic vectors below level $k-1$. Applying $x$ sends $r$ into the span below level $k$. The commutator term is also below level $k$ because $[y,x]\in I$ and the previous stability induction applies to every element of $I$. Therefore the only contribution at level $k$ is $\mu(y)x^k v_0$, which proves the congruence. Thus each diagonal entry is $\mu(y)$. Consequently the operator $N_y: W \to W$ defined by \begin{align*} N_y(w)=yw-\mu(y)w \end{align*} has a strictly upper triangular matrix in the chosen basis. Multiplying strictly upper triangular matrices shifts nonzero entries at least one diagonal farther above the main diagonal, so the $(m+1)$-st power of such a matrix is zero. Hence $N_y$ is nilpotent.[/guided]

[step:Choose an eigenvector for the complementary element]Define the common $I$-weight space inside $W$ by \begin{align*} K_I:=\{w\in W: yw=\mu(y)w \text{ for every } y\in I\}. \end{align*} This space is nonzero because $v_0\in K_I$. We prove that $K_I$ is stable under $x|_W$. First fix $z\in [L,L]$. Since $[L,L]\subset I$, the triangular description from the previous step applies to $z|_W$: it is upper triangular with every diagonal entry equal to $\mu(z)$. Hence \begin{align*} \operatorname{tr}(z|_W)=(\dim_F W)\mu(z). \end{align*} In particular, for $y\in I$ the commutator $[y,x]$ belongs to $[L,L]$, and \begin{align*} [y,x]|_W=y|_W x|_W-x|_W y|_W. \end{align*} Using cyclic invariance of trace for finite matrices, which follows from $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for endomorphisms $A,B$ of a finite-dimensional vector space, we get \begin{align*} \operatorname{tr}([y,x]|_W) &=\operatorname{tr}(y|_W x|_W)-\operatorname{tr}(x|_W y|_W) \\ &=0. \end{align*} Therefore \begin{align*} 0=(\dim_F W)\mu([y,x]). \end{align*} Since $\operatorname{char}F=0$ and $\dim_F W$ is a positive integer, multiplication by $\dim_F W$ is injective in $F$, so $\mu([y,x])=0$. Now let $w\in K_I$ and $y\in I$. The Lie module identity gives \begin{align*} y(xw) &=x(yw)+[y,x]w \\ &=x(\mu(y)w)+\mu([y,x])w \\ &=\mu(y)xw. \end{align*} Since this holds for every $y\in I$, we have $xw\in K_I$. Thus $K_I$ is stable under $x|_W$. The space $K_I$ is finite-dimensional and nonzero. The endomorphism $x|_{K_I}:K_I\to K_I$ has a characteristic polynomial of positive degree; since $F$ is algebraically closed, that polynomial has a root $\alpha\in F$. Therefore $x|_{K_I}-\alpha\operatorname{id}_{K_I}$ is not invertible, so there exists $0\ne v\in K_I$ such that \begin{align*} xv=\alpha v. \end{align*} By the definition of $K_I$, this same vector also satisfies \begin{align*} yv=\mu(y)v \end{align*} for every $y\in I$.[/step]

[guided]We need a vector that is an eigenvector for both pieces of the decomposition $L=I\oplus Fx$. The space $W$ is stable under $x$, so algebraic closedness guarantees eigenvectors for $x$ on any nonzero $x$-stable subspace. The correct subspace is the common eigenspace for $I$: \begin{align*} K_I:=\{w\in W: yw=\mu(y)w \text{ for every } y\in I\}. \end{align*} This subspace is nonzero because $v_0\in K_I$. We must verify that $K_I$ is stable under $x$. Let $w\in K_I$ and let $y\in I$. Using the Lie module identity, \begin{align*} y(xw) &= x(yw)+[y,x]w \\ &= x(\mu(y)w)+[y,x]w \\ &= \mu(y)xw+[y,x]w. \end{align*} Thus $xw$ will again lie in $K_I$ once we know $[y,x]w=0$ for every $y\in I$. Because $[y,x]\in [L,L]\subset I$, the vector $w\in K_I$ gives \begin{align*} [y,x]w=\mu([y,x])w. \end{align*} So it remains to prove $\mu([y,x])=0$. The finite-dimensional space $W$ is stable under both $x$ and $y$, so the trace of the commutator of their restrictions is defined and equals zero: \begin{align*} \operatorname{tr}\bigl([y|_W,x|_W]\bigr) = \operatorname{tr}(y|_W x|_W-x|_W y|_W) =0. \end{align*} On the other hand, $[y,x]\in I$, and the triangular form proved above says that $[y,x]|_W$ is upper triangular with every diagonal entry equal to $\mu([y,x])$. Therefore \begin{align*} 0 = \operatorname{tr}([y,x]|_W) = (\dim_F W)\mu([y,x]). \end{align*} Since $F$ has characteristic zero and $\dim_F W$ is a positive integer, multiplication by $\dim_F W$ is injective in $F$. Hence $\mu([y,x])=0$. Therefore $K_I$ is stable under $x$. Since $K_I$ is a nonzero finite-dimensional vector space over the algebraically closed field $F$, the endomorphism $x|_{K_I}:K_I\to K_I$ has an eigenvector. Choose $0\ne v\in K_I$ and $\alpha\in F$ with \begin{align*} xv=\alpha v. \end{align*} Because $v\in K_I$, we also have \begin{align*} yv=\mu(y)v \end{align*} for every $y\in I$.[/guided]

[step:Assemble the common weight on all of $L$]Define a linear functional $\lambda:L\to F$ by \begin{align*} \lambda(y+cx):=\mu(y)+c\alpha \end{align*} for $y\in I$ and $c\in F$. This is well-defined and linear because $L=I\oplus Fx$. For any element $z\in L$, write uniquely $z=y+cx$ with $y\in I$ and $c\in F$. Then \begin{align*} zv &=(y+cx)v \\ &= yv+c\,xv \\ &= \mu(y)v+c\alpha v \\ &= \lambda(y+cx)v \\ &= \lambda(z)v. \end{align*} Thus $0\ne v\in V$ is a common eigenvector for the action of $L$, and the line $Fv$ is a one-dimensional $L$-submodule. This completes the proof.[/step]

[guided]The decomposition $L=I\oplus Fx$ lets us define the desired weight separately on the ideal part and on the complementary direction. Define $\lambda:L\to F$ by \begin{align*} \lambda(y+cx):=\mu(y)+c\alpha \end{align*} for $y\in I$ and $c\in F$. This formula is well-defined because each element of $L$ has a unique expression $y+cx$ with $y\in I$ and $c\in F$. It is linear because $\mu:I\to F$ is linear and the coordinate $c$ in the direct-sum decomposition is linear. Now take any $z\in L$ and write $z=y+cx$ with $y\in I$ and $c\in F$. The vector $v$ was chosen so that $yv=\mu(y)v$ for every $y\in I$ and $xv=\alpha v$. Therefore \begin{align*} zv &=(y+cx)v \\ &=yv+c\,xv \\ &=\mu(y)v+c\alpha v \\ &=\lambda(y+cx)v \\ &=\lambda(z)v. \end{align*} This proves that $v$ is a common eigenvector for the whole Lie algebra $L$. Since $v\ne 0$, the line $Fv$ is a one-dimensional $L$-submodule by the equivalence established at the beginning of the proof.[/guided]