I'm note sure this needs proof.

Let g(t)=sum_{i=0,..n}} f_i t^i, then

g(T)x = sum_{i=0,..n}} f_i T^i x

but T^i x= lambda^i x as x is an eigen vector corresponding to eigen value lambda, so:

g(T)x = sum_{i=0,..n}} f_i lambda^i x =g(lambda) x

(that T^i x= lambda^i x can be proven by induction if it is not blindingly

obvious to you )

RonL