[proofplan] The proof is a direct application of [Lie's theorem](/theorems/3754). [Lie's theorem](/theorems/3802) supplies a nonzero vector whose span is stable under the action of the solvable Lie algebra. That span is a nonzero $L$-submodule of the irreducible module $V$, so irreducibility forces it to be all of $V$. [/proofplan] [step:Apply Lie's theorem to obtain a one-dimensional invariant subspace] Let \begin{align*} \rho: L &\to \mathfrak{gl}(V) \end{align*} denote the representation associated to the given $L$-module structure, so that $x \cdot v = \rho(x)(v)$ for $x \in L$ and $v \in V$. The hypotheses needed for [Lie's theorem](/theorems/3803) are satisfied: $F$ is algebraically closed of characteristic $0$, $L$ is finite-dimensional and solvable over $F$, and $V$ is a nonzero finite-dimensional $F$-[vector space](/page/Vector%20Space) carrying a representation of $L$. Therefore, by Lie's theorem (citing a result not yet in the wiki: Lie's Theorem), there exists a nonzero vector $v \in V$ and a linear functional \begin{align*} \lambda: L &\to F \end{align*} such that \begin{align*} \rho(x)(v) = \lambda(x)v \end{align*} for every $x \in L$. Define the one-dimensional subspace \begin{align*} W := Fv \subset V. \end{align*} For every $x \in L$ and every $a \in F$, bilinearity of the module action gives \begin{align*} x \cdot (av) = a(x \cdot v) = a\rho(x)(v) = a\lambda(x)v \in W. \end{align*} Hence $W$ is an $L$-submodule of $V$. Since $v \neq 0$, we have $W \neq 0$ and $\dim_F W = 1$. [/step] [step:Use irreducibility to identify the invariant line with the whole module] Because $V$ is irreducible as an $L$-module, its only $L$-submodules are $0$ and $V$. The subspace $W$ constructed above is a nonzero $L$-submodule of $V$, so irreducibility implies \begin{align*} W = V. \end{align*} Therefore \begin{align*} \dim_F V = \dim_F W = 1. \end{align*} This proves the theorem. [/step]