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

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

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

I have tried looking at this problem in several ways and I just cannot seem to see the significance of $\displaystyle ||A|| \leq 1$ 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.

2. If $\displaystyle \Vert T \Vert <1$ then $\displaystyle I-T$ is invertible and so the only possible eigenvector of $\displaystyle T$ would be $\displaystyle 0$. For the case $\displaystyle \Vert T\Vert =1$ I don't know how to prove this yet.

3. Originally Posted by jprobst
$\displaystyle H$ is a Hilbert space and $\displaystyle T \in B(H)$ is such that $\displaystyle ||T||_{H} \leq 1$. Show that $\displaystyle Tx = x \iff T^{*}x = x$.
The important thing to know is that $\displaystyle \|T^*\|=\|T\|$. Therefore $\displaystyle \|T^*x\|\leqslant\|x\|$.

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