Answer to Question #252470 in Functional Analysis for Bhakta

Question

let X be a normed space and T belongs to B ( X).show that if a is eigen value of T then |a|<||T||

Answer 1

From definition we have ||T(x)||$\le||T||\cdot ||x||$ for any x$\in X$

Let a be eigenvalue of T, then $\exist x\in X,||x||\ne0$ such that $T(x)=a\cdot x$

Therefore ||T(x)||=||a$\cdot$ x||=|a|$\cdot$ ||x||$\le ||T||\cdot| |x||$

Dividing both parts on value ||x||>0 we will have:

|a|$\le$ ||T||, proof is done.

Statement |a|<||T|| is not true in the general case because if $X_0$ ={$\lambda\cdot x,\lambda \in R$ }, $T_{X0}(y)=a\cdot y,y\in X_0$ we have ||T||=|a|.