Theorem : Let T be a linear operator on a vector space V and let $\lambda$ be an eigenvalue of T. Then : For any scalar $\mu \not = \lambda$, the restriction of T - $\mu$I to $K_\lambda$ is one - to -one. Where $K_\lambda$ is the generalized eigenspace of T.
Proof: Let x $\in K_\lambda$ and (T - $\mu$I)(x) = 0. By way of contradiction suppose that $x \not = 0$. Let p be the smallest integer for which (T-$\lambda I)^p$(x)= 0 and let y = (T- $\lambda I)^{p-1}$(x). Then (T - $\lambda I$)(y) = (T - $\lambda I)^p$ (x) = 0 and hence y $\in E_\lambda$. Furthermore, (T - $\mu I$)(y) = (T - $\mu I)$(T - $\lambda I)^{p-1} (x)$ = 0 and hence, y $\in E_\mu$. But $E_\mu \cap E_\lambda$ = {0} and thus y = 0, contrary to the hypothesis. So x = 0 and the restriction of T - $\mu I$ to $K_\lambda$ is one to one.
Note $E_\lambda$ is the eigenspace of T. I'm not sure how this proof implies injectiveness, can someone please explain.