Lemma 3

Let E be a vector space and let $u$ be an endomorphism of E which possesses an index. Suppose that the index of $u$ is zero. Let N be the nilespace of $u$ and I its conilespace.

a) The space E is then the direct sum of the subspaces N and I (Weyr-Fitting decomposition, cf. A, VIII, §2, n$^o2$, p. 25).

b) The endomorphism $u$ induces, by passing to the subspaces, a nilpotent endomorphism of N and an automorphism $u_I$ of I.

c) Let $v$ denote the endomorphism of E which is zero on N and which coincides with $u^{-1}_I$ on I. One has $u\circ v=v\circ u$ and the endomorphism $1_E-u\circ v$ is the projector with image N and kernel I.

d) Every endomorphism of E which commutes with $u$ stabilises N and I, and commutes with $v$.

The endomorphism $u$ satisfies the properties of the preceding lemma.

Let $n\in \mathbf{N}$ be such that Ker($u^n$) $= N$. Then Ker($u^m$) $=$ Ker($u^n$) for every integer $m\geqslant n$; since ind($u^m$) $=m$ ind($u$) $= 0$, this implies Im($u^m$) $=$ Im($u^n$) for $m\geqslant n$, hence I = Im($u^n$). It follows from A, VIII, p. 25, prop. 2 that assertions a) and b) are satisfied.

For every $x\in N$, one has $v(x) = 0$ and $u(x)\in N$, hence $u(v(x)) = 0$ and $v(u(x)) = 0$. For every $x\in I$, one has $u(v(x)) =v(u(x)) =x$. This implies that $u\circ v=v\circ u$ and that $1_E-u\circ v$ is the projector with image N and kernel I.

Let $w$ be an endomorphism of E which commutes with $u$. Let $n\in \mathbf{N}$ be such that N = Ker($u^n$) and I = Im($u^n$). Let $x\in N$; one has $u^n(w(x)) =w(u^n(x)) = 0$, hence $w(x)\in N$. Let $x\in I$; there exists $y\in E$ such that $x=u^n(y)$; then, $w(x) =w(u^n(y)) =u^n(w(y))\in I$. This proves that $w$ stabilises N and I.

Let us prove finally that $w$ and $v$ are permutable. For every $x$ in N, one has $v(w(x)) = 0 =w(v(x))$. Let $x\in I$; the elements $v(w(x))$ and $w(v(x))$ of E belong to I and their images under $u$ are both equal to $w(x)$, hence they are equal. By linearity, this implies that $v$ and $w$ commute.