= Finite-dimensional Fitting decomposition
{title2=$V=\operatorname{im}h^n\oplus\ker h^n,\quad n=\dim V$}
Every <linear endomorphism> of a finite-dimensional <vector space> admits the displayed <direct sum of vector spaces>. It separates the part on which repeated application is invertible from the part eventually sent to zero.
\b[Proof.] <Stabilization of images of powers of a linear map> gives $\operatorname{im}h^n=\operatorname{im}h^{2n}$. The <rank-nullity theorem> and nested <null spaces> give $\ker h^n=\ker h^{2n}$. If $v=h^nu$ also satisfies $h^nv=0$, then $u\in\ker h^{2n}=\ker h^n$, so $v=0$. Thus the displayed summands have zero <intersection>. Their <dimensions> sum to $n$ by the <rank-nullity theorem>, proving the decomposition.
Both summands are preserved by $h$. On $\operatorname{im}h^n$, stabilization makes $h$ a <surjective linear map>, hence invertible in finite <dimension>. On $\ker h^n$, its $n$th power is zero, so the restriction is a <nilpotent linear map>.
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