How to show that if $\lambda$ is in compression (or residual) spectrum of an operator $A$, then $\bar{\lambda}$ is an eigenvalue of $A^*$? Thanks.
2026-04-08 14:06:12.1775657172
A question on Compression spectrum
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If $A-\lambda I$ is injective, $\lambda \in \rho(A)$ iff $A$ is surjective. There are two ways for $A$ to not be surjective in this case: either the range is dense but not closed (continuous spectrum) or the range is not dense. The range is not dense iff $$ \{ 0 \} \ne \mathcal{R}(A-\lambda I)^{\perp}=\mathcal{N}(A^{\star}-\overline{\lambda}I). $$