Let $L$ be a Lie algebra with $L \subseteq \mathfrak{gl}(V) $ ($V$ finite dimensional over $\mathbb{C}$) and let $I$ be an abelian ideal of $L$. Given $x \in I, \lambda \in \mathbb{C} $, I am trying to show that the generalised eigenspace $\ker((x-\lambda \text{Id}_V)^{\dim V })$ is stable meaning for all $y \in L$ and all $v \in V$ we have $y(v)$ $\in \ker((x-\lambda \text{Id}_V)^{\dim V }) $.
I need to show that it $v \in \ker ((x-\lambda \text{Id}_V)^{\dim v} )$, $y \in L$ then $(x-\lambda \text{Id}_V)^{\dim V } (yv)=0 $ but I just can’t make any headway after this.
Any pointers?
This is just a proof of the formula in the comments. I'll use $\mathbf{1}$ for the identity in $\mathfrak{gl}(V)$ so as not to use $I$ which is the symbol for the ideal.
$$[x-\lambda \mathbf{1}, y] = [x, y] - \lambda [\mathbf{1}, y] = [x,y]$$ because $[\mathbf{1}, y] = 0$.
Assume by induction that $$[(x-\lambda \mathbf{1})^n, y] = n[x,y](x-\lambda \mathbf{1})^{n-1}$$ then $$\begin{align} [(x-\lambda \mathbf{1})^{n+1}, y] &= [(x-\lambda \mathbf{1})(x-\lambda\mathbf{1})^{n}, y]\\ &=[x-\lambda \mathbf{1}, y](x-\lambda \mathbf{1})^{n} + (x-\lambda \mathbf{1})[(x-\lambda\mathbf{1})^{n}, y]\\ &=[x,y](x-\lambda\mathbf{1})^{n}+(x-\lambda\mathbf{1})n[x,y](x-\lambda \mathbf{1})^{n-1}\\ &=[x,y]((x-\lambda\mathbf{1})^{n} + n (x-\lambda\mathbf{1})^{n})\\ &=(n+1)[x,y](x-\lambda\mathbf{1})^{n} \end{align}$$ where use was made that $[x,y], x\in I$ so they commute because $I$ is abelian (and so $[x,y], (x-\lambda\mathbf{1})$ also commute), and also the identity $[uv,w] = [u,w]v + u[v,w]$ in an associative algebra.