[proofplan] The Casimir element lies in the center of $U(\mathfrak g)$, so its action on any $\mathfrak g$-module commutes with the action of every element of $\mathfrak g$. Since $F$ is algebraically closed and $V$ is finite-dimensional, the linear operator $C_V$ has an eigenvalue. The corresponding eigenspace is a nonzero $\mathfrak g$-submodule of $V$, and simplicity forces that eigenspace to be all of $V$. [/proofplan] [step:Show that the Casimir operator is a $\mathfrak g$-module endomorphism] Let $\rho: \mathfrak g \to \operatorname{End}_F(V)$ denote the given representation, and let $\widetilde{\rho}: U(\mathfrak g) \to \operatorname{End}_F(V)$ be its extension to the universal enveloping algebra. Define the [linear map](/page/Linear%20Map) \begin{align*} C_V: V &\to V \\ v &\mapsto \widetilde{\rho}(C)(v). \end{align*} We prove that $C_V$ commutes with the $\mathfrak g$-action. Let $x \in \mathfrak g$ and $v \in V$. Since $C \in Z(U(\mathfrak g))$, the elements $x$ and $C$ commute inside $U(\mathfrak g)$, so $xC = Cx$. Applying the algebra homomorphism $\widetilde{\rho}$ gives \begin{align*} \rho(x) C_V(v) &= \widetilde{\rho}(x)\widetilde{\rho}(C)(v) \\ &= \widetilde{\rho}(xC)(v) \\ &= \widetilde{\rho}(Cx)(v) \\ &= \widetilde{\rho}(C)\widetilde{\rho}(x)(v) \\ &= C_V(\rho(x)v). \end{align*} Thus $C_V$ is a $\mathfrak g$-module endomorphism of $V$. [/step] [step:Find a nonzero eigenspace of the Casimir operator] Because $V$ is finite-dimensional over the algebraically closed field $F$, the characteristic polynomial of the linear operator $C_V \in \operatorname{End}_F(V)$ splits over $F$ and has a root. Hence there exists $\lambda \in F$ such that the eigenspace \begin{align*} E_\lambda := \ker(C_V - \lambda \operatorname{id}_V) \end{align*} is nonzero. [guided] The purpose of this step is only to produce a nonzero invariant candidate subspace. Since $V$ is finite-dimensional, the characteristic polynomial of $C_V$ is a nonconstant polynomial over $F$ unless $V = 0$. A simple module is nonzero by convention, so $\dim_F V \geq 1$. Because $F$ is algebraically closed, this characteristic polynomial has a root $\lambda \in F$. Equivalently, the operator $C_V - \lambda \operatorname{id}_V$ is not invertible, so its kernel is nonzero. We therefore define \begin{align*} E_\lambda := \ker(C_V - \lambda \operatorname{id}_V). \end{align*} This is the subspace of all vectors $v \in V$ satisfying $C_V(v) = \lambda v$. [/guided] [/step] [step:Prove that the eigenspace is a $\mathfrak g$-submodule] We show that $E_\lambda$ is stable under the action of $\mathfrak g$. Let $v \in E_\lambda$ and $x \in \mathfrak g$. Since $C_V$ is a $\mathfrak g$-module endomorphism, \begin{align*} C_V(\rho(x)v) &= \rho(x)C_V(v) \\ &= \rho(x)(\lambda v) \\ &= \lambda \rho(x)v. \end{align*} Therefore $\rho(x)v \in E_\lambda$. Since this holds for every $x \in \mathfrak g$ and every $v \in E_\lambda$, the subspace $E_\lambda$ is a $\mathfrak g$-submodule of $V$. [/step] [step:Use simplicity to force the eigenspace to be the whole module] The subspace $E_\lambda$ is a nonzero $\mathfrak g$-submodule of the simple $\mathfrak g$-module $V$. By simplicity, the only $\mathfrak g$-submodules of $V$ are $0$ and $V$, so $E_\lambda = V$. Hence every $v \in V$ satisfies \begin{align*} C_V(v) = \lambda v = \lambda \operatorname{id}_V(v). \end{align*} Therefore $C_V = \lambda \operatorname{id}_V$, as required. [/step]