Suppose that $\lambda $ and $\mu $ are distinct eigenvalues of T.
Prove that $ ( T- \mu I)|_{G_{\lambda}(T) }$ is injective.
assume that $( T- \mu I)|_{G_{\lambda}(T) } v = ( T- \mu I)|_{G_{\lambda}(T) } u $
$u,v \in G_{\lambda} (T) \implies u-v \in G_{\lambda} (T) $
$$\implies( T- \mu I)|_{G_{\lambda}(T) } u = ( T- \mu I) u= ( T- \mu I) v $$ So we have that $( T- \mu I) (u-v)=0 \implies u-v \in E_{\mu } \subset G_{\mu}$
Since $T(u-v) = \mu ( u-v) \implies u-v=0 \implies u=v \implies$ one to one.
I am alittle leery about a couple of the steps is this actually correct?