 Finite-dimensional Fitting decomposition

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.

Proof. Stabilization of images of powers of a linear map gives $\mathrm{im}{h}^n=\mathrm{im}{h}^{2n}$. The rank-nullity theorem and nested null spaces give $\ker h^n=\ker h^{2n}$. If $v=h^{n}u$ also satisfies $h^{n}v=0$, 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 $\mathrm{im}{h}^n$, 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.