Fixed point of an adjoint operator Let $H$ Hilbert space and $T: H → H$ linear operator continuous. We define $T^*: H → H$ such as 
$\langle T^*x\,,y\rangle = \langle x\,,Ty\rangle $
We suppose $||T|| \lt 1 $, I want to proof that $T$ and $T^*$ has the same fixed points, (for $x_0 \in H$, $Tx_0 = x_0 \iff T^*x_0 = x_0$). 
It seems simple but I do not see it clearly, I would also like to see an example of when this is not true (for example, when $||T|| \ge 1 $)
Ideas and suggestions are appreciated, many thanks!
 A: If $\|T\|<1$ and $Tx=x$ with $x\ne0$, then 
$$
\|x\|=\|Tx\|\leq\|T\|\,\|x\|<\|x\|.
$$
This is a contradiction, so the only fixed point is $x=0$. As $\|T^*\|=\|T\|$, the same reasoning applies to $T^*$. 
When $\|T\|\geq1$, this is not true anymore. For instance consider 
$$
T=\begin{bmatrix} 1&0\\1&0\end{bmatrix}.
$$
Then the fixed points of $T$ are 
$$
\left\{\begin{bmatrix} t\\ t\end{bmatrix}:\ t\in\mathbb C\right\},
$$
while the fixed points of $T^*$ are 
$$
\left\{\begin{bmatrix} t\\ 0\end{bmatrix}:\ t\in\mathbb C\right\}.
$$
A: this semester i've been doing the functional analysis subject, and i had to do this exercise.
The statement is wrong!
The real statement is to prove that there are the same fixed points
with ||T||<=1.
And the solution is wrong too!, because if ||T|| = 1 they have the same fixed points. I let here a link where i solved the problem, and as you said:
''Thank you! I was thinking to decompose T∗x0=λx0+z with λ∈R and ⟨x0,z⟩=0), but you way is to much better and simple!''
We needed to use this!
Please look at the answer, is in catalan but with google translate it's easy to understand.
I hope it helps anybody that wanted to find the solution of this problem!!!
Here is the link:
Exercise
Is located in "apartat b"
