 Solution

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.