Let $T$ be the bilateral shift operator, that is: $T: \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ such that $(T(x))_k=x_{k-1}$ (where $x_{k-1}$ means the $k-1$ coordinate of the sequence.
I have been able to prove that its spectrum is the unit circle (set of all $\lambda$s with unit euclidean norm) and I am then asked to show that I can find a compact operator $K$ such that the spectrum of $T+K$ is the unit ball, i.e. the set of all complex numbers with norm not greater than 1.
Previously I have been asked to prove that in general, if $T$ and $K$ are bounded and compact operators (respectively) from a Banach space $X$ to itself, then if $\lambda$ is in the spectrum of $T$ but is not an eigenvalue of finite multiplicity it follows that $\lambda$ is also in the spectrum of $T+K$.
Could someone please help me connect the dots? Thanks.
Best Answer
To my surprise, I realized how to do it when I though "well, certainly $K$ cannot be finite rank". It can be done with a rank-one $K$.
To simplify notation, let $\{e_n\}$ be the canonical basis. Then $$ Te_n=e_{n+1},\qquad n\in\mathbb N. $$ Let $K=-E_{1,0}$. That is, $Ke_0=-e_1$, $Ke_n=0$ for all $n\ne0$. Then $$ (T+K)e_n=\begin{cases}e_{n+1},&\ n\ne0\\[0.3cm] 0,&\ n=0\end{cases} $$ Consider the decomposition $H=H_0+H_1$, where $H_0,H_1$ are the pairwise orthogonal subspaces $$ H_0=\overline{\operatorname{span}}\{e_n:\ n\leq0\},\qquad H_1=\overline{\operatorname{span}}\{e_n:\ n\geq1\}. $$ Both are invariant for $T+K$. So $T+K$ can be considered as the direct sum of the two operators $T_0=(T+K)|_{H_0}$ and $T_1=(T+K)|_{H_1}$. The spectrum of a direct sum is the union of the spectra. And $T_1$ is the unilateral shift on $H_1$, while $T_0$ is the adjoint of the unilateral shift on $H_0$. The spectra of these operators are $\sigma(T_0)=\sigma(T_1)=\overline{\mathbb D}$, so $$\sigma(T+K)=\overline{\mathbb D}.$$