My understanding is
$[x,y] \in L_{\lambda + \mu} \Rightarrow (\delta - (\lambda +\mu )I_L)^m [x,y]=0$
Somehow, the $m$ has changed into a $1$ and I believe that is because of the 'acts diagonalisably'. Is this somehow saying that the normal $\lambda$-eigenspace of $\sigma$, i.e. $m=1$, is the generalised eigenspace $L_\lambda$ of $\delta$, i.e. $m\geq 1$?
If so, I want to understand why acting diagonalisably means this is true? Or if it isn't why this is justified?
When you restrict $\delta$ to $L_{\lambda}$, you have $\delta_{\lambda}=\sigma_{\lambda}+\mu_{\lambda}$, you have a basis of $L_{\lambda}$ where the matrix of $\delta_{\lambda}$ is an upper triangular matrix $A_{\lambda}$, you have $A_{\lambda}=B_{\lambda}+C_{\lambda}$ where $B_{\lambda}$ is $\lambda I_{L_{\lambda}}$ where $I_{L_{\lambda}}$ is the matrix of the identity of $L_{\lambda}$. The matrix $B_{\lambda}$ is the matrix of $\sigma_{\lambda}$ in this basis.