# Operator: $Tx =\sum_{n=2}^\infty x_{n-2} \frac{1}{2^{n-2}} e_{n}$. Which is $\rho(T)?$

Let $X$ be a separable Hilbert space and let $\{e_{k}: k\geq1\}$ be a orthonormat base in $X$. Let $T \in \mathcal{B}{(X)}$ defined by: $$Te_{k}=2^{-k}e_{k+2}, k\in\mathbb{N}^{*}.$$

Find out the expression of the operator $T$.

I know only what $e_{k}$ means. $T(0\ldots1,0,0\ldots)=(0,0,\ldots,0,0,2^{-k},0,0\ldots)$.

Cooper.hat found out the expression of $Tx$ and this expression is:

$$Tx =\sum_{n=2}^\infty x_{n-2} \frac{1}{2^{n-2}} e_{n}.$$ Now how can I find $\rho(T)$? $\rho(T)$ is resolvent set of $T$. It's about matrices, or not. It is an infinite dimensional space, I think is not possible to attach a matrix.

Thanks for help :)

Thanks :)

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Why did you write $T(0,\ldots,1,0,0,\ldots)=(0,0,\ldots,0,0,1,0,0\ldots)$? What happened with the powers of two? –  Sigur Feb 4 '13 at 22:54
@Iuli: A polite suggestion: it looks like you're trying to learn functional analysis given the similarity of your recent string of questions. Rather then outsourcing your education to stackexchange, would you like some accessible references? I say this because the questions you've asked thus far are short steps from the core curriculum. For example, this set of notes is very accessible: www-m12.ma.tum.de/web/poetzsch/… –  Alex R. Feb 4 '13 at 22:58
The basic idea is that you know the image of every element in a basis. So, for an arbitrary element $x$, you can write $x$ in terms of the basis and then use the linearity and continuity of $T$ to find $Tx$. This is what's done in copper.hat's answer. –  David Mitra Feb 4 '13 at 23:18

If $x = \sum_{n=0}^\infty x_n e_n$, then $Tx = T(\sum_{n=0}^\infty x_n e_n) = \sum_{n=0}^\infty x_n T e_n = \sum_{n=0}^\infty x_n \frac{1}{2^n} e_{n+2} = \sum_{n=2}^\infty x_{n-2} \frac{1}{2^{n-2}} e_{n}$, or, in other words, the coordinate representation of $Tx$ is the basis $e_n$ is $(0,0,x_0, \frac{x_1}{2}, \frac{x_2}{2^2},...)$.

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For me is not very clear how did you choose $x=\sum_{n=0}^{\infty}{x_{n}e_{n}}$. How did you think? –  Iuli Feb 4 '13 at 23:10
If $x \in \mathbb{H}$, and $e_n$ is a complete orthonormal Hilbert basis, then if we let $x_n =\langle x,e_n \rangle$, then we have $x = \sum_n x_n e_n$. It is analogous to the finite dimensional case. –  copper.hat Feb 4 '13 at 23:12
Deloused, hopefully... I think precision is particularly important from a pedagogical perspective. –  copper.hat Feb 4 '13 at 23:18

The spectrum $\sigma(T)$ is the closure of all points $z$ such that $T-zI$ is not invertiable. The resolvent $\rho(T)$ on the other hand consists of all points not in the spectrum, i.e. $$\rho(T)=\mathbb C / \sigma(T)$$

The spectrum is the closure of $\{1,1/2,...,1/2^n,...\}$ which gives us the resolvent

Edit, i dont know how to do "not"s in tex, sry

Edit: sry, didnt notice that the operator was a rotation, can't help you with this one

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