Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@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
up vote 2 down vote accepted

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. $\ell^\infty$.)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.