Lemma 6

If $v \in \mathscr{L}(E)$, then $\mathrm{Sp}\,\sigma(v) \subset \mathrm{Sp}\,v + \mathrm{Sp}\,v + \cdots + \mathrm{Sp}\,v$, where the sum comprises $n+1$ terms.

We define elements $v_0, v_1, \ldots, v_n$ of $\mathscr{L}(F)$ by writing, for all $f \in F$,

Then $\sigma(v) = \sum_{i=0}^n v_i$ and the $v_i$ are pairwise permutable. Let A be the total closed subalgebra of $\mathscr{L}(F)$ generated by the $v_i$; it is commutative (Spectral Theories, Chapter I, § 1, no. 4) and $\mathrm{Sp}_{\mathscr{L}(F)} v' = \mathrm{Sp}_A v' \subset \sum_{i=0}^n \mathrm{Sp}\,v_i$ (Spectral Theories, Chapter I, § 3, Proposition 3 (ii)). Now, if $\lambda \in \mathbf{C}$ is such that $v - \lambda$ is invertible, clearly the $v_i - \lambda_i$ are invertible and hence $\mathrm{Sp}\,v_i \subset \mathrm{Sp}\,v$ for all $i$.