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?

  • 3
    $\begingroup$ What have you done so far ? $\endgroup$ – Victoria M Apr 6 at 20:43
  • 1
    $\begingroup$ What is special about $\epsilon > 1$ for your expectation regarding the point spectrum? $\endgroup$ – 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.

  • $\begingroup$ 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 $\sigma_p(T)=\{1\}$. $\endgroup$ – Kanido mat 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 $\sum_{n=0}^\infty \frac{1}{2^2}$. $\endgroup$ – Kanido mat Apr 7 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.