I'm working on the following problem.
Let $(e_n)$ be a complete orthonormal sequence in a Hilbert space $H$, and let $(\lambda_n)$ be a bounded complex sequence. Let $T$ be a bounded linear operator on $H$ such that $Te_n=\lambda_ne_n$ for each $n$. Find $\sigma(T)$, the spectrum of $T$.
What I've done so far: I know that if $\lambda\in\sigma(T)$ then $T-\lambda I$ is not invertible.
I think that $\{\lambda_n\}_n\subset\sigma(T)$, because $(T-\lambda_nI)e_n=Te_n-\lambda e_n=0$, which means $e_n\in\ker(T-\lambda_n I)$, so $T-\lambda_n I$ is not injective (which means it is not invertible.)
But I don't know if there are other members of $\sigma(T)$. Any help would be appreciated!
Best Answer
Since $\sigma (T)$ is closed we see that the closure of $\{\lambda_n : n \geq 1\}$ is contained in it. If $\lambda $ does not belong to this closure there is a positive distance $r$ from $\lambda$ to this set. Thus $|\lambda-\lambda_n| \geq r$ for all $n$. I will let you check tat $T-\lambda I$ has a bounded inverse in this case so $\lambda$ cannot be in the spectrum. [Note that there is an obviosu formula for the inverse of $T-\lambda I$]