Lemma 3

Let g and h be two commuting endomorphisms of a vector space E. Then each eigenspace of g is closed under h.

Indeed, if $W_\lambda$ is the eigenspace of g corresponding to the eigenvalue $\lambda$, then for all $x \in W$, we have

which says that $h \cdot x \in W_\lambda$.

Let us now return to the proof of Prop. 13. Among all decompositions of E as a direct sum of nonzero subspaces each closed under all the elements of $\mathscr{S}$, choose one with the greatest number of components (the dimension of E is an upper bound for this number), say $E = \sum_{i \in I} E_i$. Let $u \in \mathscr{S}$ and let $E = \sum V_\alpha$ be the decomposition of E as the direct sum of the eigenspaces of $u$. By Lemma 3, each $V_\alpha$ is closed under $\mathscr{S}$, and hence so is each $V_i \cap E_i$; by Prop. 12 each $E_i$ is the direct sum of the $V_i \cap E_i$. The choice of the $E_i$ thus forces each $E_i$ to be contained in one of the $V_\alpha$; thus the restriction of $u$ to each $E_i$ is a homothety. Since this is true for all the elements of $\mathscr{S}$, it follows that $\mathscr{S}$ is diagonalisable.