Spectrum of operator in $l_2$ Sorry guys, I have a problem with this exercise.

Let T an operator in $l_2$ Hilbert space:
  $$
(\textrm{T}x)_1 = x_2 ,
$$
  $$
(\textrm{T}x)_2 = 0 ,
$$
  $$
(\textrm{T}x)_n = x_{n-1} - x_n     
$$
  With $n\ge3$.



*

*Find the adjoint operator $\textrm{T}^{\dagger}$ and $||\textrm{T}||$.

*Find the point spectrum $\sigma_P(T)$ and $\sigma_P(\textrm{T}^{\dagger})$

*Find the residual spectrum $\sigma_{\rho}(\textrm{T})$ and $\sigma_{\rho}(\textrm{T}^{\dagger})$


The first step is simple, I have found the $\textrm{T}^{\dagger}$ :
$$
(y,\textrm{T}x) = (\textrm{T}^{\dagger}y,x)  
$$
But
$$
(y,\textrm{T}x) = y_1^*x_2 + y_3^*(x_2-x_3) + y_4^*(x_3-x_4) + ....  
$$
$$
= (y_1^*+y_3^*)x_2 + (y_4^*-y_3^*)x_3 + ....
$$
So the $\textrm{T}^{\dagger}$ is defined by:
$$
(\textrm{T}^{\dagger}x)_1 = 0 ,
$$
$$
(\textrm{T}^{\dagger}x)_2 = x_1 + x_3 ,
$$
$$
(\textrm{T}^{\dagger}x)_n = x_{n+1} - x_n     
$$
The $||\textrm{T}||$ is:
$$
||\textrm{T}x||^2 = (\textrm{T}x,\textrm{T}x) = |x_2|^2 + |x_2|^2 + |x_3|^2 + |x_3|^2 + |x_4|^2 + |x_4|^2 + ... 
$$
$$
||\textrm{T}x||^2 \le 2 \sum\limits_{n=1}\limits^{\infty} |x_n|^2 
$$
Finally the $||\textrm{T}|| =  \sqrt{2}$.
But in the second and third step I have found problem, because I have tried to find the $\sigma_P(\textrm{T})$ :
$$
\lambda x_1 = x_2
$$
$$
\lambda x_2 = 0
$$
$$
\lambda x_3 = x_2-x_3
$$
$$
\lambda x_n = x_{n-1}-x_n
$$
But I don't understand the condition that the $\lambda$, so the $\sigma_P(\textrm{T})$, have  to satisfies. And then I have no idea for the $\sigma_\rho(\textrm{T})$.
 A: In the calculation for the norm of $T$, your evaluation of $(Tx,Tx)$ is wrong. And, in any case, you don't show how the norm would be achieved. 
Actually, $\|T\|=2$. If you try with $x=(0,1,-1,1,-1,\ldots,1,-1,0,\ldots)$ (with $n$ nonzero entries) you'll get that 
$$
\|x\|=\sqrt n,\ \ \ \|Tx\|=(1+4(n-1))^{1/2},
$$
so
$$
\|T\|\geq\frac{\|Tx\|}{\|x\|}=\left(4-\frac3n\right)^{1/2}.
$$
It follows that $\|T\|\geq2$. Via the triangle inequality one can easily see that $\|T\|\leq2$. So $\|T\|=2$. 
For the eigenvalues, assume first that $\lambda=0$. If $Tx=0$, you get that $x_2=0$ (first equation), $x_3=0$ (third equation), etc. So the eigenvalues for $\lambda=0$ are $(t,0,0,\ldots)$, $t\in\mathbb C$. 
For $\lambda\ne0$, your second equation gives $x_2=0$. Then the first gives $x_1=0$. The third is $\lambda x_3=-x_3$; if $\lambda\ne-1$, then $x_3=0$ and we can continue to $x_n=0$ for all $n$. This means that $\lambda\ne-1$ cannot be an eigenvalue. When $\lambda=-1$, the equations become $-x_n=x_{n-1}-x_n$, so $x_{n-1}=0$; so $x=0$ and $-1$ is not an eigenvalue either. In conclusion, 
$$
\sigma_P(x)=\{0\}.
$$
I'll leave $\sigma_P(T^*)$ to you. Finally, for the residual spectrum you have to show/use that
$$
\sigma_\rho(T)=\overline{\sigma_P(T^*)}\setminus\sigma_P(T).
$$
Edit: eigenvalues of $T^*$.
For the eigenvalues of $T^*$, you have the equations
$$
\lambda x_1=0,\ \ \lambda x_2=x_1+x_3,\ \ \lambda x_n=x_{n+1}-x_n,\ \ n\geq3.
$$
If $\lambda=0$, you get $x_3=x_4=x_5=\cdots$ which forces them all to be zero. Then $x_1=0$ by the second equation and we obtain that $(0,1,0,0,\ldots)$ is an eigenvector for $\lambda=0$. 
When $\lambda\ne0$, we immediately get $x_1=0$. The general equation becomes 
$$
x_{n+1}=(1+\lambda)x_n.
$$
If $x_3=0$, this forces $x=0$. When $\lambda=-1$ we get an eigenvector. If $x_3\ne0$, we get 
$$
x_{n+1}=(1+\lambda)^nx_3,\ \ n\geq3.
$$
For this to yield a sequence in $\ell^2$ we need $|1+\lambda|<1$. This is the open ball of radius one, centered at $-1$. Thus
$$
\sigma_P(T^*)=\{0,-1\}\cup (-1+\mathbb D).
$$
