[proofplan] We compute the action of a Lie bracket $[x,y]$ on the generator $v$ in two ways. The module identity expresses this action as the commutator of the two endomorphisms by which $x$ and $y$ act on $V$. Since both endomorphisms are scalar multiplication operators on the one-dimensional module $Fv$, their commutator is zero. The nonzero vector $v$ then forces $\lambda([x,y])=0$, and linearity extends this from brackets to their $F$-linear span. [/proofplan] [step:Compute the action of a bracket on the one-dimensional generator] Fix $x,y \in L$. Since $V$ is an $L$-module, the action satisfies \begin{align*} [x,y]\cdot v = x\cdot (y\cdot v) - y\cdot (x\cdot v). \end{align*} Using $y\cdot v = \lambda(y)v$, $x\cdot v = \lambda(x)v$, and $F$-linearity of the action in the vector argument, we obtain \begin{align*} [x,y]\cdot v &= x\cdot(\lambda(y)v) - y\cdot(\lambda(x)v) \\ &= \lambda(y)(x\cdot v) - \lambda(x)(y\cdot v) \\ &= \lambda(y)\lambda(x)v - \lambda(x)\lambda(y)v \\ &= 0, \end{align*} because multiplication in the field $F$ is commutative. [guided] Fix arbitrary elements $x,y \in L$. The point is to use the compatibility condition defining an $L$-module: the Lie bracket in $L$ acts as the commutator of the corresponding linear operators on $V$. Applied to the vector $v$, this gives \begin{align*} [x,y]\cdot v = x\cdot (y\cdot v) - y\cdot (x\cdot v). \end{align*} Now use the hypothesis that every element of $L$ acts on the generator $v$ by a scalar. Specifically, \begin{align*} y\cdot v = \lambda(y)v, \qquad x\cdot v = \lambda(x)v. \end{align*} Because the action is $F$-linear in the vector argument, scalars may be pulled out: \begin{align*} [x,y]\cdot v &= x\cdot(\lambda(y)v) - y\cdot(\lambda(x)v) \\ &= \lambda(y)(x\cdot v) - \lambda(x)(y\cdot v) \\ &= \lambda(y)\lambda(x)v - \lambda(x)\lambda(y)v. \end{align*} Since $F$ is a field, scalar multiplication in $F$ is commutative, so $\lambda(y)\lambda(x)=\lambda(x)\lambda(y)$. Therefore \begin{align*} [x,y]\cdot v = 0. \end{align*} [/guided] [/step] [step:Convert the zero action into vanishing of the character on brackets] By the defining property of $\lambda$, applied to the element $[x,y] \in L$, we also have \begin{align*} [x,y]\cdot v = \lambda([x,y])v. \end{align*} Combining this with the previous step gives \begin{align*} \lambda([x,y])v = 0. \end{align*} Since $v \ne 0$ and $V$ is a [vector space](/page/Vector%20Space) over the field $F$, scalar multiplication by $\lambda([x,y])$ kills $v$ only if \begin{align*} \lambda([x,y]) = 0. \end{align*} Thus $\lambda$ vanishes on every bracket $[x,y]$ with $x,y \in L$. [/step] [step:Extend the vanishing from brackets to the derived Lie algebra] Let $z \in [L,L]$. By definition of the $F$-linear span, there exist $m \in \mathbb{N}$, scalars $a_1,\dots,a_m \in F$, and elements $x_1,\dots,x_m,y_1,\dots,y_m \in L$ such that \begin{align*} z = \sum_{i=1}^{m} a_i [x_i,y_i]. \end{align*} Using linearity of $\lambda: L \to F$ and the bracket vanishing proved above, \begin{align*} \lambda(z) &= \lambda\left(\sum_{i=1}^{m} a_i [x_i,y_i]\right) \\ &= \sum_{i=1}^{m} a_i \lambda([x_i,y_i]) \\ &= \sum_{i=1}^{m} a_i \cdot 0 \\ &= 0. \end{align*} Therefore $\lambda(z)=0$ for every $z \in [L,L]$, so $\lambda([L,L])=0$. [/step]