= Solution
The <dimensions> of the nested <images of a linear map>
$$
V=\mathcal R(h^0)\supseteq\mathcal R(h)\supseteq\cdots
$$
can 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 cases
$$
\mathcal R(h^n)=\mathcal R(h^{2n}).
$$
By 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 $v=h^nu$. Then $0=h^nv=h^{2n}u$, so $u\in\ker h^{2n}=\ker h^n$ and $v=0$. Hence the <intersection> is zero. Finally the <rank-nullity theorem> gives
$$
\dim\mathcal R(h^n)+\dim\ker h^n=n.
$$
Zero <intersection> and this <dimension> sum prove
$$
\boxed{V=\mathcal R(h^n)\oplus\mathcal N(h^n).}
$$
This is the <finite-dimensional Fitting decomposition>. It includes the zero-dimensional case, for which $V=0$ and the assertion is immediate.
Back to article page
