5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ @Martin oops oops ! :P $\endgroup$
    – Theorem
    Jan 17, 2013 at 19:31
  • $\begingroup$ The "iff" part is wrong. Consider: $(x,\lambda x,\lambda^2 x,...)$ $\endgroup$
    – Thomas Andrews
    Jan 17, 2013 at 19:32
  • $\begingroup$ Take Thomas's comment to find the eigenvalues and to finish off determine the norm of $T$. $\endgroup$
    – Martin
    Jan 17, 2013 at 19:33
  • $\begingroup$ @Martin : which means $\|T\| \le \frac{1}{1-\lambda}$ right ? $\endgroup$
    – Theorem
    Jan 17, 2013 at 19:36
  • 1
    $\begingroup$ There's an easier way to determine $\lVert T\rVert$. Compare $\lVert Tx\rVert$ and $\lVert x \rVert$. $\endgroup$
    – Martin
    Jan 17, 2013 at 19:40

1 Answer 1

Reset to default
2
$\begingroup$

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. $\ell^\infty$.)

$\endgroup$
10
  • $\begingroup$ This solves everything since you know that the spectrum is compact and contained in the ball of radius $\lVert T \rVert$ around $0$. $\endgroup$
    – Martin
    Jan 17, 2013 at 19:36
  • $\begingroup$ @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 $\lambda \le 1$ . $\endgroup$
    – Theorem
    Jan 17, 2013 at 19:51
  • $\begingroup$ @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 $1$. $\endgroup$
    – Thomas Andrews
    Jan 17, 2013 at 20:01
  • 1
    $\begingroup$ I didn't exactly mean to write out that the series should be taken as converging as an operator, but rather to then work out what the "obvious" operator would be when applied to $x=(x_i).$ So if $S$ is the naive sum of $\lambda^{-1}\sum_{k=1}^{\infty}\lambda^{-k}T^k,$ then $(Sx)_i=\sum_{k=1}^{\infty}\lambda^{-k-1}x_{i+k},$ which, for each $i,$ converges and is bounded if $(x_i)\in\ell^{\infty}.$ It's a little more work to show $\sum_i |(Sx)_i|<+\infty$ if $x\in \ell^1.$ I'd guess the operators converge in some sense, but I don't have the answer to that off the top of my head. @MSIS $\endgroup$
    – Thomas Andrews
    Sep 20, 2022 at 0:33
  • 1
    $\begingroup$ Basically, you use the naive series to define the inverse, but then have to prove the operator $S$ has the properties that you want - sending $\ell^1$ to $\ell^1$ and $\ell^\infty$ to $\ell^{\infty},$ and being an inverse of $\lambda I-T.$ $\endgroup$
    – Thomas Andrews
    Sep 20, 2022 at 0:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .