[proofplan] The proof is a direct computation from the $\mathfrak{sl}_2(\mathbb C)$ commutation relations. We take a vector of weight $\lambda$ and use the module property that Lie brackets act as commutators of the corresponding endomorphisms of $V$ to move $h$ past $e$ and $f$. The relation $[h,e]=2e$ shifts the $h$-eigenvalue upward by $2$, while $[h,f]=-2f$ shifts it downward by $2$. [/proofplan] [step:Use the commutator relation to show that $e$ raises the weight by $2$] Let $v\in V_\lambda$. By the definition of $V_\lambda$, we have \begin{align*} h v=\lambda v. \end{align*} Since $V$ is an $\mathfrak{sl}_2(\mathbb C)$-module, the Lie bracket acts by commutators of the corresponding endomorphisms of $V$. Applying the relation $[h,e]=2e$ to $v$ gives \begin{align*} h(e v)-e(h v)=2e v. \end{align*} Using $h v=\lambda v$, this becomes \begin{align*} h(e v)-e(\lambda v)=2e v. \end{align*} By complex linearity of the action, \begin{align*} e(\lambda v)=\lambda e v. \end{align*} Therefore \begin{align*} h(e v)=(\lambda+2)e v. \end{align*} Thus $e v\in V_{\lambda+2}$. Since $v\in V_\lambda$ was arbitrary, $e(V_\lambda)\subseteq V_{\lambda+2}$. [guided] Let $v\in V_\lambda$. The meaning of this assumption is that $v$ is an eigenvector of the action of $h$ with eigenvalue $\lambda$, or possibly $v=0$; in either case the defining equation is \begin{align*} h v=\lambda v. \end{align*} We want to prove that $e v$ has weight $\lambda+2$. That means we must prove the defining equation for membership in $V_{\lambda+2}$: \begin{align*} h(e v)=(\lambda+2)e v. \end{align*} The available information about how $h$ interacts with $e$ is the standard $\mathfrak{sl}_2(\mathbb C)$ relation \begin{align*} [h,e]=2e. \end{align*} Because $V$ is a module over the [Lie algebra](/page/Lie%20Algebra), this bracket relation becomes an operator identity on $V$. Applying it to the vector $v$ gives \begin{align*} h(e v)-e(h v)=2e v. \end{align*} Now substitute the weight equation $h v=\lambda v$ into the middle term: \begin{align*} h(e v)-e(\lambda v)=2e v. \end{align*} The action of $e$ is complex-linear, so $e(\lambda v)=\lambda e v$. Hence \begin{align*} h(e v)-\lambda e v=2e v. \end{align*} Adding $\lambda e v$ to both sides gives \begin{align*} h(e v)=(\lambda+2)e v. \end{align*} This is exactly the defining condition that $e v\in V_{\lambda+2}$. The zero-vector case is included automatically, since if $e v=0$, then $h(e v)=0=(\lambda+2)e v$. Since the argument works for every $v\in V_\lambda$, we conclude that \begin{align*} e(V_\lambda)\subseteq V_{\lambda+2}. \end{align*} [/guided] [/step] [step:Use the commutator relation to show that $f$ lowers the weight by $2$] Let $v\in V_\lambda$. Again $h v=\lambda v$. Applying the module form of the relation $[h,f]=-2f$ to $v$ gives \begin{align*} h(f v)-f(h v)=-2f v. \end{align*} Substituting $h v=\lambda v$ and using complex linearity of the action of $f$, we obtain \begin{align*} h(f v)-\lambda f v=-2f v. \end{align*} Hence \begin{align*} h(f v)=(\lambda-2)f v. \end{align*} Thus $f v\in V_{\lambda-2}$. Since $v\in V_\lambda$ was arbitrary, $f(V_\lambda)\subseteq V_{\lambda-2}$. Combining the two inclusions proves the theorem. [/step]