Proposition 2

a) Suppose that the sequence $(N_p)$ is stationary. Then we have $N_{\infty}\cap I_{\infty}= 0$, the restriction of $u$ to $I_{\infty}$ is injective, and $u$ induces a nilpotent endomorphism of $N_{\infty}$.

b) Suppose that the sequence $(I_p)$ is stationary. Then we have M = $N_{\infty}+ I_{\infty}$ and $u(I_{\infty}) = I_{\infty}$.

c) (“Weyr–Fitting decomposition”—sometimes called the “Fitting decomposition”) Suppose that the sequences $(N_p)$ and $(I_p)$ are stationary. Then M is a direct sum of submodules $N_{\infty}$ and $I_{\infty}$ that are stable under $u$, and $u$ induces a nilpotent endomorphism of $N_{\infty}$ and an automorphism of $I_{\infty}$.

Let $p$ be a natural number such that $N_p= N_{\infty}$, and let $v=u^p$. By construction, $v$ and $v^2$ have the same kernel $N_{\infty}$, and $I_p$ is the image of $v$. For $x$ in $N_{\infty}\cap I_p$, there exists a $y\in M$ such that $x=v(y)$ and therefore $v^2(y) =v(x) = 0$; hence, we have $y\in N_{\infty}$ and consequently $x= 0$. In particular, we have $N_{\infty}\cap I_{\infty}= 0$. Since the kernel $N_1$ of $u$ is contained in $N_{\infty}$, the restriction of $u$ to $I_{\infty}$ is injective; on the other hand, we have $u^p(N_{\infty}) = 0$. This proves a).

Let $q$ be a natural number such that $I_q= I_{\infty}$, and let $w=u^q$. Then $w$ and $w^2$ have the same image $I_{\infty}$, and $N_q$ is the kernel of $w$. Let $x\in M$. We have $w(x)\in I_{\infty}$, so there exists a $y\in M$ such that $w(x) =w^2(y)$; we then have $x-w(y)\in N_q$, whence $M = N_q+ I_{\infty}$ and a fortiori $M = N_{\infty}+ I_{\infty}$. We have $u(I_{\infty}) =u(I_q) = I_{q+1}= I_{\infty}$. This proves b).

Assertion c) follows immediately from a) and b).