# Eigenspace of \lambda = 1 for a bounded linear operator and its adjoint

• December 3rd 2009, 04:24 PM
jprobst
Eigenspace of \lambda = 1 for a bounded linear operator and its adjoint
$H$ is a Hilbert space and $T \in B(H)$ is such that $||T||_{H} \leq 1$. Show that $Tx = x \iff T^{*}x = x$.

I have tried looking at this problem in several ways and I just cannot seem to see the significance of $||A|| \leq 1$ nor how to go about proving it. I understand how the adjoint is defined for a Hilbert space as a consequence of Reisz representation. I have tried to attack it using fixed point theorems. Any nudge in the right direction would be appreciated.
• December 3rd 2009, 08:03 PM
Jose27
If $\Vert T \Vert <1$ then $I-T$ is invertible and so the only possible eigenvector of $T$ would be $0$. For the case $\Vert T\Vert =1$ I don't know how to prove this yet.
• December 5th 2009, 01:30 AM
Opalg
Quote:

Originally Posted by jprobst
$H$ is a Hilbert space and $T \in B(H)$ is such that $||T||_{H} \leq 1$. Show that $Tx = x \iff T^{*}x = x$.

The important thing to know is that $\|T^*\|=\|T\|$. Therefore $\|T^*x\|\leqslant\|x\|$.

If $Tx=x$ then $\|T^*x-x\|^2 = \langle T^*x-x,T^*x-x\rangle = ...$ (multiply out the inner product and use the fact that $\langle T^*x,x\rangle = \langle x,Tx\rangle$) $... = \|T^*x\|^2 - \|x\|^2\leqslant0$. Hence $T^*x-x=0$.