# Spectrum of a right shift operator.

I have some doubts on the following problem :

Let us consider $T : \ell^1(\mathbb N) \to \ell^1(\mathbb N)$by $(x_1,x_2..... ) \to (x_2, x_3 ........)$.

I want to find the eigen values and spectrum of T and also of $T' : \ell^\infty (\mathbb N)\to \ell^\infty(\mathbb N)$

let us consider $\lambda$ to be the eigen value , then $Tx=\lambda x$ for a $x \in \ell^1$ then we get $(x_2,x_3,......)=(\lambda x_1, \lambda x_2 ........)$ which holds equality if $x_1=x_2=.....=0$ , which means there is no eigen value for $T$ .

How do i find the spectrum of $T$ and $T'$ ? Thank you for your help.

-
@Martin oops oops ! :P – Theorem Jan 17 '13 at 19:31
The "iff" part is wrong. Consider: $(x,\lambda x,\lambda^2 x,...)$ – Thomas Andrews Jan 17 '13 at 19:32
Take Thomas's comment to find the eigenvalues and to finish off determine the norm of $T$. – Martin Jan 17 '13 at 19:33
@Martin : which means $\|T\| \le \frac{1}{1-\lambda}$ right ? – Theorem Jan 17 '13 at 19:36
There's an easier way to determine $\lVert T\rVert$. Compare $\lVert Tx\rVert$ and $\lVert x \rVert$. – Martin Jan 17 '13 at 19:40

## 1 Answer

The "iff" part is wrong. Consider: $(x_i)$ with $x_i=\lambda^i$. For what $\lambda$ is this in $\ell^1$? In $\ell^\infty$?

You can write out explicitly for $|\lambda|>1$ the inverse of $\lambda I -T$ by writing:

$$S_\lambda = (\lambda I - T)^{-1} = \lambda^{-1}(I-\lambda^{-1}T)^{-1} = \lambda^{-1}\sum_{k=0}^\infty \lambda^{-k} T^k$$

Writing $x=(x_i)$ and $(y_i)=S_\lambda x$, we get:

$$y_i = \sum_{k=0}^\infty \lambda^{-(k+1)} x_{i+k}$$

You need to show that if $x\in\ell^1$ (resp. $\ell^\infty$), then this series for $y_i$ coverges for all $i$, and $(y_i)\in\ell^1$ (resp. $\ell^\infty$.)

-
This solves everything since you know that the spectrum is compact and contained in the ball of radius $\lVert T \rVert$ around $0$. – Martin Jan 17 '13 at 19:36
@Thomas Andrews : the spectrum lies strictly inside the unit circle . But how can i say that all of it is the spectrum ? For $\ell^\infty$ there $\lambda \le 1$ . – Theorem Jan 17 '13 at 19:51
@Theorem I've added an explicit inverse for $\lambda I-T$ when $|\lambda|>1$. Alternatively, you can use what Martin said, that the spectrum is compact and bounded by the norm of $T$, which you've already shown is $1$. – Thomas Andrews Jan 17 '13 at 20:01
The interesting case is $\ell^1$, where the spectrum elements on the boundary do not correspond to eigenvalues. (In $\ell^\infty$ the boundary elements are also eigenvalues.) – Thomas Andrews Jan 17 '13 at 20:03
@ThomasAndrews : Thanks . But now we know that for $\lambda >1$ cannot be in the spectrum . but what if there is $\lambda \le 1$ such that $\lambda I -T$ is invertible. – Theorem Jan 17 '13 at 20:10