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Let $\varepsilon>0$ and $T:\ell^2\to \ell^2 $ the operator defined by

$$T(x_1, x_2, \ldots)=(\varepsilon x_1, x_1,x_2,\ldots)$$

Can you help me to calculate the spectrum of $T$, please?.

I think that $\sigma_p(T)={\varepsilon}$ if $\varepsilon\geq1$ and that $\sigma_p(T)=\emptyset$ otherwise, because $(T-\lambda I)(x_1,\ldots) =0$ for $(x_1,\ldots) \neq 0$ iff $\lambda \neq 0, x_1 \neq 0, \lambda=\varepsilon$ and $$x_{n+1}=\frac{x_1}{\varepsilon^n}, \ n>1.$$

Am I right?

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    $\begingroup$ What have you done so far ? $\endgroup$ – Victoria M Apr 6 at 20:43
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    $\begingroup$ What is special about $\epsilon > 1$ for your expectation regarding the point spectrum? $\endgroup$ – jawheele Apr 6 at 20:54
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Your conclusion regarding the point spectrum is true, with the slight alteration that $\sigma_p (T)=\{\epsilon\}$ when $\epsilon > 1$, rather than $\geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-\lambda I$ for $\lambda \neq \epsilon$.

Clearly $T$ is not surjective, so suppose $\lambda \neq 0$. We have $(T-\lambda I)(x) = ((\epsilon-\lambda)x_1,x_1-\lambda x_2,...) =y \implies x_1 = \frac{y_1}{\epsilon - \lambda}$, $x_{n+1} = \frac{x_{n}-y_{n+1}}{\lambda}$, so $$x_{n} =\frac{x_1-\sum_{k=2}^{n} \lambda^{k-2} y_k }{\lambda^{n-1}}$$ In particular, if $y_k$ is nonzero for only finitely many $k \in \mathbb{N}$, so $\exists$ $N \in \mathbb{N}$ such that $y_n =0$ for $n>N$, then $\forall n>N$ $x_n=\frac{x_N}{\lambda^{n-N}}$. Choosing $y \in \ell^2$ such that $x_N \neq 0$ ( $y_k = \delta_{N,k}$ works), we have $x \in \ell^2 \iff |\lambda| > 1$. That is to say, $T-\lambda I$ is not surjective for $|\lambda| \leq 1$.

Further, this shows that for $|\lambda|>1$, im$(T-\lambda I)$ includes all sequences $y \in \ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-\lambda I$ for $|\lambda|>1$, then, it suffices to show $x \in \ell^2$ when $y \in \ell^2$ and $y_1=0$. In this case, $x_n = -\sum_{k=2}^n \lambda^{k-n-1}y_k$, so $|x_n|^2 \leq \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2$, and thus

$$\sum_{n=1}^N |x_n|^2 \leq \sum_{n=1}^N \sum_{k=1}^n |\lambda|^{2(k-n-1)}|y_k|^2 = \sum_{k=1}^N |y_k|^2 \sum_{n=k}^{N} |\lambda|^{2(k-n-1)} \\ = \sum_{k=1}^N |y_k|^2 \sum_{n=1}^{N+1-k} |\lambda|^{-2n} \leq \frac{1}{|\lambda|^2-1}\sum_{k=1}^N |y_k|^2 \leq \frac{\|y\|^2}{|\lambda|^2-1}$$

Showing $x \in \ell^2$, and hence that $\sigma(T)=\{|\lambda| \leq 1\} \cup \{\epsilon\}$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.

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  • $\begingroup$ Thank you so much!. I think that if $\varepsilon=1$, then $(T-I)(1/2,1/2,1/2,\ldots)=(0,0,0, \ldots)$ and because of that $\sigma_p(T)=\{1\}$. $\endgroup$ – Kanmat Apr 7 at 14:16
  • $\begingroup$ @krenick $(1/2,1/2,...) \notin \ell^2$. $\endgroup$ – jawheele Apr 7 at 14:35
  • $\begingroup$ You are right. Sorry, I don't know why I was thinking about the series $\sum_{n=0}^\infty \frac{1}{2^n}$ instead of $\sum_{n=0}^\infty \frac{1}{2^2}$. $\endgroup$ – Kanmat Apr 7 at 15:03

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