5

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.

Theorem
  • 7,979

1 Answers1

2

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. $\ell^\infty$.)

Thomas Andrews
  • 177,126
  • 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
  • I've show that $\lambda$ is an eigenvalue if $0<|\lambda|<1$. That means all such $\lambda$ must be in the spectrum. Again, since the spectrum is compact, it must be closed, so in particular, it must contain the closure the non-zero elements of the open unit ball, which is the full closed unit ball. – Thomas Andrews Jan 17 '13 at 20:10
  • (Note, you are failing to include the absolute values, which are necessary. $\lambda = -2$ is not in the spectrum.) – Thomas Andrews Jan 17 '13 at 20:14
  • @ThomasAndrews: Sorry for necropost: In what topology/Space does $\lambda^{-1}\sum_{k=0}^\infty \lambda^{-k} T^k$ converge? some Operator space topology? – MSIS Sep 19 '22 at 23:54
  • 1
    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 – Thomas Andrews Sep 20 '22 at 00:33
  • 1
    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 $\lambda I-T.$ – Thomas Andrews Sep 20 '22 at 00:37