# Spectrum of $T(x_1, x_2, \ldots)=(\varepsilon x_1, x_1,x_2,\ldots)$ in $\ell^2$ for $\varepsilon>0$

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?

• What have you done so far ? – Victoria M Apr 6 at 20:43
• What is special about $\epsilon > 1$ for your expectation regarding the point spectrum? – jawheele Apr 6 at 20:54

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.

• 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 $\sigma_p(T)=\{1\}$. – Kanmat Apr 7 at 14:16
• @krenick $(1/2,1/2,...) \notin \ell^2$. – jawheele Apr 7 at 14:35
• 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 $\sum_{n=0}^\infty \frac{1}{2^2}$. – Kanmat Apr 7 at 15:03