Spectrum of the operator $Tx=(x_2,0,x_2-x_3,x_3-x_4,...)$ in $\ell^2$

Question

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 $n\ge3$.

1. Find the adjoint operator $\textrm{T}^{\dagger}$ and $||\textrm{T}||$.
2. Find the point spectrum $\sigma_P(T)$ and $\sigma_P(\textrm{T}^{\dagger})$
3. 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 $\sigma_\rho(\textrm{T})$.

Answer 1

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 $$ \sigma_P(T^*)=\{0,-1\}\cup (-1+\mathbb D). $$