Corollary 1

Let E be a separated locally convex space and let $u\in \mathscr{L}(E)$. In order that $u$ be a Riesz endomorphism, it is necessary and sufficient that there exist an element $v$ of $\mathscr{L}(E)$ which commutes with $u$ and such that $1_E-u\circ v$ is compact.

Since every continuous linear mapping of finite rank is compact, the condition is necessary (III, p. 47, prop. 6 d)). Let us prove that it is sufficient. Let $v$ be an element of $\mathscr{L}(E)$ which commutes with $u$ and such that the endomorphism $h= 1_E-u\circ v$ of E is compact. There exist a Banach space G, a continuous linear mapping $p: E\rightarrow G$ and a compact linear mapping $q: G\rightarrow E$ such that $h=q\circ p($III, p. 5, prop. 5). The endomorphism $p\circ q$ of G is compact. According to prop. 2, the linear mapping $1_G-p\circ q$ is a Riesz endomorphism of G. But then $u\circ v= 1_E-q\circ p$ is a Riesz endomorphism of E (III, p. 49, prop. 10, g)). Since $u$ and $v$ commute, $u$ is a Riesz endomorphism of E (III, p. 49, prop. 9).