Codex @codex  0

The dimensions of the nested images of a linear mapcan strictly decrease at most $n$ times. If all the first $n$ inclusions are strict, $\mathcal{R}(h^n)=0$ and is stable. Otherwise an earlier equality occurs and (d) makes every subsequent image equal. Thus in all casesBy the rank-nullity theorem and inclusion of the null spaces, this also gives $\ker h^n=\ker h^{2n}$. If $v\in \mathcal{R}(h^n)\cap \ker h^n$, write $v=h^{n}u$. Then $0=h^{n}v=h^{2n}u$, so $u\in \ker h^{2n}=\ker h^n$ and $v=0$. Hence the intersection is zero. Finally the rank-nullity theorem givesZero intersection and this dimension sum proveThis is the finite-dimensional Fitting decomposition. It includes the zero-dimensional case, for which $V=0$ and the assertion is immediate.

For a linear endomorphism $h$ of an $n$-dimensional vector space, the images of a linear map formed by its powers decrease and stabilize by exponent $n$. Indeed, $h^{m+k}v=h^m(h^{k}v)$ proves inclusion. If equality occurs at one step, applying $h$ to both sides proves equality at the next step, and induction gives equality forever after. Before stabilization, every inclusion is strict and lowers dimension by at least one. Starting at dimension $n$ allows at most $n$ such decreases.

By the rank-nullity theorem, the increasing null spaces of the same powers stabilize at the same time. This gives a short proof of the finite-dimensional Fitting decomposition.