A linear transformation is injective on its generalized range space.

Created almost 9 years ago, updated about 8 hours ago

For finite-dimensional $V$, $T$ restricts to an automorphism of its generalized range $R=\operatorname{range}(T^n)$. Stabilization gives $T(R)=R$, so the restriction is surjective and hence injective in finite dimension. On the complementary generalized kernel, the restriction is nilpotent.

Like how this is organised?

Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.