Can you help me with a Spectral and Point Spectrum Problem

Question

The Question: Let $T: l^2 \rightarrow l^2 $ be defined by $Tx = T(\zeta_j) = (\alpha_j \zeta_j)$, where $(\alpha_j)$ is dense in $[0,1]$. Find $\sigma_{p}(T)$ and $\sigma(T)$.

Here is what I have for the point spectrum:

By definition, I need to find the eigenvalues. $$ Tx = \lambda x$$ if and only if $$ T(\zeta_j) = \lambda (\zeta_j)$$ if and only if $$(\alpha_j\zeta_j) = (\lambda \zeta_j)$$ if and only if $$(\alpha_j - \lambda)(\zeta_j) = 0.$$

This is what I have so far. I have no idea how to compute the eigenvalues. I am guessing that $\lambda = \alpha_j$ for some $j \in \mathbb{N}. $

Can you please help me on this problem? Thank you very much!

Answer 1

Since $\zeta \neq 0$ there is at least one $j$ with $\zeta_j \neq 0$. This gives $\lambda =\alpha_j$. Thus every eigen value is one of the numbers $\alpha_j$. Every $\alpha_j$ is an eigen value since $Te_j=\alpha_j e_j$ (where $(e_j)$ is the standard orthonormal basis). Hence the point spectrum is exactly $\{\alpha_j: j \geq 1\}$.

The spectrum is closed and it contains a dense subset of $[0,1]$. So $[0,1] \subset \sigma(T)$. Let us show that equality holds by showing that if $\lambda \notin [0,1]$ then $T-\lambda I $ is invertible. Let $S(\zeta_j)=(\frac 1 {\alpha_j -\lambda} \zeta_j)$. I will let you verify that $S$ is a bouded operator and $S=(T-\lambda I)^{-1}$. Hint: there exists $\epsilon >0$ such that $|\alpha_j -\lambda | \geq \epsilon$ for all $j$.

Answer 2

1. Let $ \lambda \in \sigma_p(T)$ and $x=(\zeta_1, \zeta_2,...) \ne (0,0,...)$ be a corresponding eigenvektor. Then we have $(\alpha_j - \lambda)(\zeta_j) = 0$ for all $j$. Since $x \ne 0$, there is $j_0$ such that $ \zeta_{j_0} \ne 0$, hence $ \lambda = \alpha_{j_0}.$

2. Let $(e_j)$ be the usual ortonormal basis of $l^2$. Then we have $Te_j= \alpha_je_j.$

Consequence: $ \sigma_p(T)= \{\alpha_1,\alpha_2,...\}.$

Now it is your turn to show that $\sigma(T)=\overline{\{\alpha_1,\alpha_2,...\}}=[0,1].$