For $T:V\to V$ on an $n$-dimensional space, $V=\ker(T^n)\oplus\operatorname{range}(T^n)$ (Fitting decomposition). If $x=T^ny$ and $T^nx=0$, then $T^{2n}y=0$; stabilized kernels imply $T^ny=0$, so $x=0$. Rank–nullity for $T^n$ makes dimensions add to $n$. The finite-dimensional hypothesis is essential.
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